This forms a number of triangles. What proportion of area is in the triangles below the zigzag pattern?

This is a brief excerpt from my book Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing for aspiring mathematicians who want to learn how to write proofs. Compelling mathematical statements with interesting, elementary proofs.

Proof and the Art of Mathematics has been awarded the 2024 Daniel Solow Author’s Award by the Mathematical Association of America, awarded for outstanding and impactful contributions to mathematics education.

Theorem. The triangles formed by any zigzag pattern in a rectangle carve out exactly half the area of the rectangle.

Proof: We assume at first in this proof that the zigzag pattern never doubles back, so that the zigzag path proceeds steadily to the right. In this case, all the triangles have acute base angles, and the vertex of each triangle occurs above its base. Let us construct a vertical line on each vertex used in the zigzag pattern. This carves the rectangle into smaller rectangles, as illustrated below, and the main point is that each such rectangle is cut exactly in half diagonally by the corresponding zig or zag across the rectangle.

Thus, exactly half the area of the rectangle is above the zigzag and half below, as desired. This proves the theorem for the acute-triangle case.

Let us now drop the no-doubling-back assumption and consider the general case, where we allow zigzag patterns that sometimes move backward (but not crossing themselves), like this:

This kind of case breaks the previous proof, because the constructed rectangles would overlap. Nevertheless, we can mount an alternative argument. The point is that if the original rectangle has width w and height h, then each triangle has height h and thus area ½bh, where b is the base of that triangle. Since the triangle bases add up to the total width w, it follows that the sum of the areas of the triangles is ½wh, which is exactly half the area of the rectangle, as desired. □

Alternative proof: Another way for us to argue in the general case is to observe that moving the top vertices of the triangles along the top edge of the rectangle is an area-preserving transformation because it preserves the base and height of each triangle. We can transform the skewed-triangle zigzags to the acute-triangle case, for example, simply by moving the top vertex of every triangle to be above its base. More directly, however, let us simply combine all the triangles into one by moving their top vertices all the way to the right like this:

After this transformation, they form the region below the diagonal of the rectangle, which is exactly half the area. Since the transformation is area preserving, the original triangles must also have exactly half the area. □

I find it interesting to notice that the theorem remains true even for zigzag patterns involving infinitely many zigs and zags. There is no need in the argument to assume that there were only finitely many triangles.

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Read more in the book itself:

See also the supplement:

More generally, mathematicians define that a permutation of a set is a one-to-one correspondence of the set with itself. In the case of a finite set, the permutations in effect describe all the different ways that we might enumerate the elements of the set.

This is an excerpt from chapter 5 of my book Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing for aspiring mathematicians who want to learn how to write proofs, in the context of interesting mathematical statements having interesting elementary proofs.

Proof and the Art of Mathematics has recently been awarded the 2024 Daniel Solow Author’s Award by the Mathematical Association of America, awarded for outstanding and impactful contributions to mathematics education.

For any natural number n, we define the factorial number n! by recursion:

In this way, one can see that n! is simply the product of all the numbers from n down to 1, as in 5! = 5⋅4⋅3⋅2⋅1 = 120.

Theorem. For any integer n ≥ 0, the number of permutations of n objects is the factorial n!.

Proof: We prove the theorem by common induction. The statement is true when n = 0, since there is exactly one arrangement of zero objects, the empty arrangement. If the reader finds it confusing to consider permutations of the empty set, it may help to observe that the theorem is also true when n = 1, since there is exactly one way to permute one object also, and 1! = 1.

Now, suppose by induction that the theorem is true for a number n, and consider n + 1. If we have n + 1 objects

to be rearranged, let us first rearrange the first n objects

By the induction hypothesis, there are n! many ways to do that. Next, we place the final object a_n+1​ into the list. But where? There are precisely n + 1 many slots into which it might be placed: before all them, between two of them, or after all of them. Every rearrangement of the n + 1 objects can be realized by first rearranging the first n objects and then placing the final object into one of the slots. So we will have (n + 1)⋅n! many rearrangements of n + 1 objects, and this is precisely (n + 1)!, as desired. So by induction, the theorem is true for all numbers n. □

Why did we define 0! = 1? Does that seem unnatural? Students sometimes suggest or expect that 0! should be 0. Would that be a good idea? As mathematicians, we are free to define our functions however we want. Which definition is best?

Notice that if we set 0! to be 0, then it would break the identity (n + 1)! = (n + 1)⋅n! in the case n = 0; we might have to start adding exceptions to our theorems on account of this. Indeed, the previous theorem itself provides a very good reason to define 0! = 1, since as we observed, there is exactly one arrangement of the empty set, the empty arrangement. For this reason, among others, mathematicians have agreed that it is best to define 0! = 1.

For integers 0 ≤ k ≤ n, we define

pronounced, “n choose k,” as the number of ways to choose k items from a set of n items, disregarding the order.

Theorem. The number of ways to choose k items from a set of n items, for 0 ≤ k ≤ n, is given by the formula

We shall give several proofs.

First proof: Consider the following process: enumerate all n items, and then take just the first k in that enumeration. We will often get the same set of k elements, so there is considerable multiple counting in this process. But how much? The number of enumerations of n objects is precisely n!, by the previous theorem. For a given enumeration of the objects, there are k! many permutations of the first k elements of it, and (n - k)! many permutations of the other elements. So there are k!(n - k)! many ways to permute the elements in the enumeration, while having the same first k elements. So this is the factor that we have double counted by, and so the number of ways to choose k elements from n is precisely

as desired. □

The second proof will make use of the following lemma:

Lemma:

Proof: Consider a set A with n + 1 objects, and we wish to count the number of ways of choosing k + 1 objects from A. Fix a particular element a ∈ A. Now, some of the size-(k + 1) subsets of A have the element a, and some may not. There are exactly

many size-(k + 1) subsets of A that have the element a, since once you choose a, then you must choose k additional elements from A∖{a}, a set of size n. Similarly, there are exactly

many ways to choose k + 1 elements from A without using the element a, since in this case one is really choosing from amongst the other n elements. So the total number of ways of choosing k + 1 elements from A is the sum of these two, and we have proved the lemma. □

Second proof: We prove the theorem by induction on n. If n = 0, then also k = 0, and it is easy to verify that there is exactly one way to choose nothing from nothing, so

which fits the formula

and so the anchor case is proved. Suppose now that the formula is correct for a number n, using any k, and consider the next number, n + 1. Since again there is only one way to choose nothing from an n element set, we see that

which is equal to

verifying the theorem in this case. Next, we consider

By lemma 1, this is equal to

By the induction hypothesis, we know that

and

Putting all of this together, we may now compute

This verifies the required instance of the theorem, and so we have proved it by induction for all natural numbers. □

Continue reading more about this and many other topics in Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing for aspiring mathematicians who want to learn how to write proofs.

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This is a quick excerpt from chapter 5 of my book Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing for aspiring mathematicians who want to learn how to write proofs.

Proof and the Art of Mathematics has been awarded the 2024 Daniel Solow Author’s Award by the Mathematical Association of America, awarded for outstanding and impactful contributions to mathematics education.

Theorem. Every positive integer n can be expressed as a sum of one or more positive integers in precisely 2^n-1 many ways.

For example, the number 4 can be expressed in the following eight ways:

For the purpose of this theorem, we allow the sum to have only one term, if desired, or many terms, and we pay attention to the order of the summands, but we do not use parentheses or pay attention to associativity.

Proof: Place n ones down in a row, like a picket fence:

There are n − 1 spaces between the ones. For each such space, let us imagine placing either a + sign in it or leaving it empty, in all the various possible ways to do this. Since there are n − 1 spaces and two choices for each space, there are precisely 2^n-1 many different ways to place these signs. Each such placement gives rise to a distinct way to represent n as a sum, if we interpret each contiguous block of ones as a single summand, so that

for example, represents 1 + 3 + 2. And conversely, every representation of n as a sum corresponds to such a marking. So there are 2^n-1 ways to represent n as a sum. □

A fence-post error is a common kind of mathematical error conflating fence posts with the spaces between them, such as in a list of numbers, an array, or an interval of time, and arises because there is one fewer space between the fence posts than there are fence posts. For example, if you arrive at a hotel on Sunday and leave the following Saturday, then you were present at the hotel on all seven days of the week, but you stayed only six nights. There are five integers from 8 to 12, even though these numbers differ by only four. If you were president of the club from the beginning of 2014 to the end of 2016, then you were president for three years: 2014, 2015, and 2016.

In exercise 11, the reader will give an alternative proof of theorem 1 by induction.

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Selection from chapter’s end:

Exercises

11. Give an alternative proof of the theorem by induction. [Hint: Consider how to generate representations of n + 1 from representations of n.]

I encourage readers of Infinitely More to post their solution to the exercise in the comments.

Can you vacate the shaded corner area? Please give it a try. One can make a good start, but then the outer stones begin to block one's path. You have to move these other stones out of the way to make room. Is it possible to get all three stones out of the corner? I have provided an online interactive version of the game, for trying out strategies and ideas, at this link, and see also my blog post at this link.

Interlude

This is a brief excerpt from chapter 5 of my book Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing for aspiring mathematicians who want to learn how to write proofs. The book is filled with compelling mathematical statements having interesting elementary proofs.

Proof and the Art of Mathematics has been awarded the 2024 Daniel Solow Author’s Award by the Mathematical Association of America, awarded for outstanding and impactful contributions to mathematics education.

Did you succeed? Perhaps one begins to think that it is impossible. But how could one ever prove such a thing?

Theorem. It is impossible to make moves in the Escape! game so as to vacate the shaded corner region.

Proof: Let us assign weights to the squares in the lattice according to the following pattern: We give the corner square weight 1/2​, the next diagonal of squares 1/4​ each, and then 1/8​, and so on throughout the whole playing board. Every square gets a corresponding weight according to the indicated pattern.

The weights are specifically arranged so that making a move in the game preserves the total weight of the occupied squares. That is, the total weight of the occupied squares is invariant as one makes moves, because moving a stone with weight 1/2^k will create two stones of weight 1/2^k+1​, which adds up to the same. Since the original three stones have total weight

it follows that the total weight remains 1 after every move in the game. Meanwhile, let us consider the total weight of all the squares on the board. If you consider the bottom row only, the weights add to

which is the geometric series with sum 1. The next row has total weight

which adds to 1/2​. And the next adds to 1/4​ and so on. So the total weight of all the squares on the board is

which is 2. This leaves a total weight of 1 for the unshaded squares. The subtle conclusion is that after any finite number of moves, only finitely many of those other squares are occupied, and so some of them remain empty. So after only finitely many moves, the total weight of the occupied squares off the original L-shape is strictly less than 1. Since the total weight of all the occupied squares is exactly 1, this means that the L-shape has not been vacated. So it is impossible to vacate the original L-shape in finitely many moves. □

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From chapter’s end:

Mathematical Habits

Make conjectures.

Use your mathematical insight, based on examples or suggestive reasoning, to guess the answer to a mathematical question or the mathematical fact that would explain a given mathematical phenomenon. Test your conjecture by checking whether it is consistent with known facts or examples. Try to prove your conjecture.

Exercises

12. In the Escape! game, is it possible to fill as much as desired of the plane outside of the original L-shape? For example, can one make sure for any given n that all squares in the n × n square, except possibly the original L-shape, have a stone?

13. Consider a version of the Escape! game on a finite n × n board, modifying the rules so that stones that would have been placed outside this region are simply not placed. For which values of n can you vacate the yellow corner? For which values of n can you vacate the entire board?

14. Show that in the n × n version of Escape!, you cannot lose: every sequence of legal moves leads eventually to a completely vacated board. [Hint: Show first that a nonempty board always has a legal move; next, show by induction from the lower left that every square can be activated only finitely many times. So there is no infinite play, and therefore the board must become empty at a finite stage of play.]

15. (Challenge) Show that the number of steps to vacate the n × n board does not depend on the particular sequence of moves that are made. The game always ends with an empty board in exactly the same number of steps regardless of how the plays are made.

Credits. The Escape! game is due to Maxim Kontsevich and was discussed on the Numberphile video series at this link.

Theorem 37. Every 2ⁿ × 2ⁿ grid of unit squares, with one square removed, can be tiled by L-shaped tiles consisting of three unit squares.

Proof: We prove this by induction on n. The claim is clear when n = 0, since in this case we have a 1 × 1 grid, with only one square, and when we remove it, we can tile what remains with no tiles at all. (The case n = 1, or a 2 × 2 grid, is similarly trivial, since when you remove one of the squares, what is left is perfectly covered by a single tile.)

Suppose now by induction that the statement is true for a number n, and consider a 2ⁿ⁺¹ × 2ⁿ⁺¹ grid of unit squares, with one unit square removed. Since 2ⁿ⁺¹ is even, we may divide the large square into four 2ⁿ × 2ⁿ quadrants. The omitted square resides in one of them. Place a single tile near the center, oriented so that it covers one corner square from each of the other three quadrants, as indicated:

If we now think of the smaller quadrants separately, we have four smaller quadrants, each with one square removed. By the induction hypothesis, these can each be tiled with the L-shaped tiles. Together with our initial center tile, we thereby produce a tiling of the whole square. So all such grids with one square omitted can be tiled. □

Consider next the collection of shapes, similar to those in the popular game Tetris, that can be formed from four unit squares. There are five such shapes, pictured above.

Question 38: Can you arrange these shapes to form a rectangle?

For example, can you tile a 4 × 5 rectangle with these shapes, using one tile of each type? Give it a try!

Interlude

This is a brief excerpt from chapter 5 of my book Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing for aspiring mathematicians who want to learn how to write proofs. The book is filled with compelling mathematical statements having interesting elementary proofs.

Proof and the Art of Mathematics has been awarded the 2024 Daniel Solow Author’s Award by the Mathematical Association of America, awarded for outstanding and impactful contributions to mathematics education.

Did you try it? The answer is no, one cannot tile a rectangle with these shapes.

Theorem 39. The collection of tiles, with one of each tile shape type above, cannot tile a rectangle.

Proof: Suppose that we could tile a rectangle using exactly one of each type of those tile shapes. Since we have five shapes, each with area four, the total area is twenty, and so the possible size rectangles are 4 × 5, 2 × 10, or 1 × 20. One can see easily that the latter two rectangles are impossible, but our argument works generally, and so there is no need to argue separately for that case. Imagine that we have such a tiling of the rectangle. Let us place a chessboard pattern on the rectangle, and note that there are equal numbers of dark and light squares (but see also exercise 9).

This chessboard pattern will induce a chessboard pattern on each of the pieces. Namely, if we could arrange the pieces to form the rectangle, then they would each inherit a light-square/dark-square pattern from the global chessboard pattern.

The thing to notice, however, is that each of the individual tiles would have exactly two dark squares and two light squares, with the exception of the final inverted T-shaped piece, which has three dark squares and one light square, or possibly three light squares and one dark square, depending on how it should happen to land on the chessboard pattern. This means that there cannot be a tiling of the rectangle, because on the one hand, the number of light and dark squares on the rectangle is equal, but when you think about the individual tiles, the problematic green tile will prevent the light and dark squares from balancing. So there can be no tiling. □

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Selections from the chapter’s end:

Mathematical Habits

Recognize when you have a proof and when you do not. This is an important, difficult step in one's mathematical development. Unfortunately, beginners are sometimes satisfied by an argument that experienced mathematicians will say is nonsense. Perhaps the attempted argument makes unwarranted assumptions, or misuses terms, or does not logically establish the full conclusion, or perhaps it makes any number of other mathematical errors, without the beginner realizing this. Therefore, be honest with yourself in your work; be skeptical about your argument; avoid talking nonsense. If you are not sure whether you have a proof, then you probably do not. Verify that your arguments logically establish their conclusions. Do not offer hand-waving arguments that avoid difficult but essential details. Never offer arguments that you do not understand.

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Exercises

6. Is the conclusion of the theorem 37 true for n × n squares in general? Why or why not?

7. Explain how the proof of theorem 37 provides a construction method for producing the desired tiling. (Namely, once a given square is omitted, then perform the division into quadrants, place the one tile, and iterate with the smaller squares.) What tiling do you get this way for the 16 × 16 grid shown in the illustration?

8. Prove that the collection of shapes pictured before theorem 39 is indeed the collection of all shapes in the plane, freely allowing rotations and reflections, that can be formed from four unit squares by joining them at vertices along their edges.

9. Prove or refute the following statement: If you place a chessboard pattern on an n × m rectangular grid, there will be an equal number of dark and light squares. Generalize your answer as much as you can.

10. In the context of theorem 39, suppose that you have three pieces of each shape. Can you now tile a rectangle? Can you tile a 10 × 10 square with five tiles of each type?

There are a variety of ways that we might do this. For example, for the bar pictured above, we could first make the three long breaks, making four 8 × 1 sticks, and then would break off one square at a time from those sticks. This would make 3 + 4⋅7 breaks altogether. Alternatively, we could first make all the short breaks, and then break off individual squares from the resulting 1 × 4 sticks, resulting in 7 + 8⋅3 breaks.

Question: What is the most efficient method of breaking the chocolate bar into squares, using the fewest total number of breaks?

This is a quick fun excerpt from chapter 5 of my book Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing for aspiring mathematicians who want to learn how to write proofs. I have strived to fill the book with interesting mathematical statements having interesting elementary proofs.

Proof and the Art of Mathematics has been awarded the 2024 Daniel Solow Author’s Award by the Mathematical Association of America, awarded for outstanding and impactful contributions to mathematics education.

Let us be honest in our counting of what a break means; we are not allowed to break two pieces off at once or break off an empty piece. To break a piece of chocolate means to take a single connected piece of chocolate and separate it into two nonempty pieces by cutting along one of the lines between the squares, following the line all the way across. How would you break the chocolate bar? Does it matter how you do it? In fact, it does not.

Theorem. Breaking a chocolate bar into individual squares always takes exactly the same number of steps, regardless of the breaking protocol that is followed.

Proof: Notice that each time we break a rectangle along an edge, we make two rectangles, each a bit smaller than the original. Each break increases the number of rectangles by exactly one. If the original piece of chocolate has n small squares, therefore, then after n - 1 breaks, regardless of how the breaks are performed (as long as each break creates exactly one new piece), there will be n pieces. And in this case, each piece must be a single small square. So all methods of breaking the chocolate bar use the same number of breaks, which is one less than the number of squares in the bar. □

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Selection from the chapter’s end:

Mathematical Habits

Generalize. After proving a statement, seek to prove a more general statement. Weaken the hypothesis or strengthen the conclusion. Apply the idea of the argument in another similar-enough circumstance. Unify our understanding of diverse situations. Seek the essence of a phenomenon.

Exercises

4. Prove that if the initial chocolate bar is a rectangle, then one always has rectangular pieces after every stage of the breaking process.

5. Generalize the chocolate bar theorem to nonrectangular chocolate bars.

Now, I have a question about whether you might be able to achieve a certain feature in your pattern of pointing at each other.

Question: Can you arrange it so that every person is altogether more often pointed at than pointing?

In other words, could we all be pointed at strictly more times than we point at others? Ponder the problem on your own, before reading further.

Interlude

This is a brief excerpt from chapter 5 of Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing for aspiring mathematicians who want to learn how to write proofs. The book is filled with compelling mathematical statements having interesting elementary proofs.

Proof and the Art of Mathematics has been awarded the 2024 Daniel Solow Author’s Award by the Mathematical Association of America, awarded for outstanding and impactful contributions to mathematics education.

Theorem. The answer is no, it is not possible to have a nonempty finite set of people pointing at each other in such a way that every person is more often pointed at than pointing.

Let us give several different proofs.

First proof: Suppose that we have a finite arrangement of people pointing at each other or themselves. For each person, let us say their pointed-at score is the number of times someone is pointing at them, and their pointing score is the number of times they are pointing at someone, including all instances of multiple pointing and self-pointing in both of these scores. Let A be the sum total of all the pointed-at scores, and let P be the sum total of all the pointing scores. I claim that P = A. The reason is that every instance of someone pointing is also simultaneously an instance of someone being pointed at, simply viewed from the other person's perspective, at the other end of the finger. Every instance of pointing adds exactly one to P and also exactly one to A. If every person were more often pointed at than pointing, however, then it would follow that P < A, since P would be the sum of a finite sequence of numbers, each of which is smaller than the corresponding summands giving rise to A. Since P = A, this cannot happen. □

Second proof: We prove the theorem by induction on the number of people. That is, no set of n people can form a counterexample. This statement is true for n = 1 person, since the person can point only at herself, and if she does so k times, then she will be both pointing and pointed at k times equally. Suppose now that the statement is true for all groups of size n, and consider a group of n + 1 people. Suppose that we have an arrangement of the n + 1 people for which everyone is more often pointed at than pointing. Let us call one of those people “Horatio.” In particular, Horatio is more often pointed at than pointing. Thus, we may simply remove Horatio from the group of people and direct some of the people who were pointing at him to point instead at those to whom Horatio had pointed. Since Horatio was more often pointed at than pointing, there are enough people who had been pointing at Horatio to cover his pointing commitment. After this rearrangement of the pointing, anyone left still pointing at where Horatio had been may simply lower his or her finger. In this way, we arrive at a new configuration, with one fewer person and hence of size n, but that still satisfies that everyone left is more often pointed at than pointing. This contradicts the induction assumption that there is no such group of size n, and so we have completed the induction step. So there can be no such group of people of any finite size. □

Third proof: Suppose we are part of a finite group of people pointing at each other, and everyone is more often pointed at than pointing. Let us instruct everyone to pay one dollar each to the people to whom they point, for each instance of pointing; and let us assume that we all have enough cash on hand to do this. The curious thing to notice is that, after the payments, because everyone is more often pointed at than pointing, it follows that every person will take in more money than they paid out. We made money! And we could do it again and make more money, and again and again, as many times as we desire. We could make millions of dollars simply by exchanging it like this. Since this is clearly impossible, as the total dollar holdings of the group does not change as money is exchanged within it, there can be no such pointing arrangement. □

I find the third proof very clear, though I recognize that it is essentially similar to the first proof, if one simply thinks of the pointing-at and pointing scores as measured in dollars. Perhaps the reason it is so clear is that it replaces the abstract quantity-preserving argument of the first proof with something much easier to grasp, namely, the fundamental fact that we cannot get more money as a group simply by exchanging money within our group. Such anthropomorphizing arguments or metaphors can often be surprisingly effective in simplifying a mathematical idea. We leverage our innate human experience in order to understand more easily what would otherwise be a complex mathematical matter. Our human experience with the difficulty of getting money makes the final conclusion of the third argument obvious.

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Selections from the end of the chapter:

Mathematical Habits

Use metaphor. Express your mathematical issues metaphorically in terms of a familiar human experience, if doing so makes them easier to understand. Find evocative terminology that represents your mathematical quantities or relationships in familiar terms, if doing so supports the mathematical analysis.

Exercises

1. Suppose that a finite group of people has some pattern of pointing at each other, with each person pointing at some or all or none of the others or themselves. Prove that if there is a person who is more often pointed at than pointing, then there is another person who is less often pointed at than pointing.

2. Suppose that you could control who follows whom on Twitter. Could you arrange it so that every person has more followers than people they follow? For example, some extremely famous people currently have many millions of followers, and one might hope to reassign most of those followers in such a way that everyone will be more followed than following.

3. Show that if there are infinitely many people, then it could be possible for every person to be more pointed at than pointing. Indeed, can you arrange infinitely many people, such that each person points at only one person but is pointed at by infinitely many people? How does this situation interact with the money-making third proof of theorem?

Credits. The pointing-at painting is David and Charles Colyear by Sir Godfrey Kneller, in the public domain via Wikimedia Commons.

This is a free extended excerpt from chapter 4 of my book Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing, for aspiring mathematicians who want to learn how to write proofs.

Proof and the Art of Mathematics teaches the art of proof-writing in the diverse contexts of a variety of mathematical subjects, but always with mathematical theorems and statements that carry their own inherent interest and fascination—compelling mathematical results with interesting, elementary proofs.

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Buckets of Fish via nested induction

Let us consider now a somewhat more fanciful instance of induction, which also illustrates the idea of nested induction. Consider the game I call Buckets of Fish, played with two players on the beach.

The game begins with finitely many buckets arranged in a row, each bucket containing some finite number of fish, with a large supply of additional fish available nearby, as many as needed, fresh off the boats. Each player, in turn, takes a fish from one bucket and adds to each of the buckets to the left, if any, as many fish from the extra supply as he or she likes. If we label the buckets 1, 2, 3, and so on, from left to right, then it would be a legal move for a player to take one fish out of bucket 4 and add ten fish (from the extra supply) to bucket 1, no fish to bucket 2, and ninety-four fish (from the extra supply) to bucket 3. The winner is whoever takes the very last fish from the buckets, leaving them empty.

Since huge numbers of fish can often be added to the buckets during play, but only one fish removed, a skeptical reader could reasonably wonder whether we'll really always even get a winner. By adding fish suitably, can the players prolong the game indefinitely? Or does every play of the game Buckets of Fish necessarily come to an end?

The answer is that, indeed, every play of the game must eventually come to an end. Regardless of how the players might conspire to add fish to the buckets during play, even with an endless supply of fish from the boats, nevertheless they will eventually run out of fish in the buckets and one of the players will take the last fish. I shall give several proofs.

Theorem. Every play of the game Buckets of Fish ends in finitely many moves. All the fish in the buckets, including all the new fish that may have been added during play, will eventually run out by some finite stage during play.

Before proving this theorem, let me first prove a weaker claim, namely, that either player may force the game to end in finitely many moves. The way to do this is simply to take fish always only out of the rightmost bucket. Since that bucket is never replenished, it follows that eventually it will become empty, and then there will be in effect strictly fewer buckets. By now taking fish only from the new rightmost bucket, we can eventually ensure that that bucket also becomes empty, and so on. In this way, by induction, we can ensure that the number of buckets with fish eventually reduces to just one bucket, which is eventually emptied. Thus, either player can play so as to ensure that the game ends in finitely many moves. But this is a weaker claim than made in the theorem, which states that actually all ways of playing the game will end in finitely many moves.

First proof. We prove the claim by (nested) induction on the number of buckets. If there is only one bucket, then there are no buckets to the left of it, and consequently there is no possibility in this case to add fish to the game; so if the one bucket contains k fish, then the claim clearly ends in k moves.

Assume by induction that all plays using at most n buckets end in finitely many moves, and suppose that we have a game situation with n + 1 buckets, with k fish in bucket n + 1. We now prove by induction on k that all such games terminate. This argument is therefore an instance of nested induction, because we are currently inside our proof by induction on n, in the induction step of that proof, and in order to complete it, we are undertaking a separate full induction on k. If k = 0, then there are no fish in bucket n + 1, and so the game amounts really to a game with only n buckets, which terminates in finitely many steps by our induction hypothesis on n.

So, let us assume that all plays with k fish in bucket n + 1 terminate in finitely many moves. Consider a situation where there are k + 1 many fish in that bucket. I claim that eventually one of those fish must be taken, since otherwise all the moves will be only on the first n buckets, and all plays on only n buckets terminate in finitely many moves. So at some point, one of the players will take a fish from bucket n + 1, possibly adding additional fish to the earlier buckets. But this produces a situation with only k fish in bucket n + 1, which by our induction assumption on k we know will terminate in finitely many steps.

So we have proved that no matter how many fish are in bucket n + 1, the game will end in finitely many moves, and so the original claim is true for n + 1 buckets. Thus, the theorem is true for any finite number of buckets. ☐

Second proof. Let me now give another proof. We want to prove that there is no infinitely long play of the game Buckets of Fish. Suppose toward contradiction that there is a way for the players to conspire to produce an infinite play, starting from some configuration of some finite number n of buckets, each with finitely many fish in them.

Fix the particular infinitely long play. Let m be the rightmost bucket from which infinitely often a fish was taken during that infinite course of play. It follows, for example, that m < n, since bucket n can be used only finitely often, as it never gets replenished. Since bucket m starts with only finitely many fish in it, and each time it is replenished, it is replenished with only finitely many fish, it follows that in order to have been used infinitely many times, it must also have been replenished infinitely often. But each time it was replenished, it was because there was some bucket farther to the right that had been used. Since there are only finitely many buckets to the right of bucket m, it follows that one of them must have been used infinitely often. This contradicts the choice of m as the rightmost bucket that was used infinitely often. ☐

Let me mention briefly a third proof of the theorem, making use of the concept of transfinite ordinals, with which some readers might be familiar. (If you are not familiar with ordinals yet, please do not worry about it; you might learn about them in an introductory set theory class at your university.) Specifically, we associate with each Buckets-of-Fish position a certain ordinal. With the position

for example, we associate the ordinal

In general, the number of fish in each bucket of a position becomes the coefficient of the corresponding power of ω, using higher powers for the buckets farther to the right. The key observation to make is that these associated ordinals strictly descend for every move of the game, since one reduces a higher-power coefficient and increases only lower-power coefficients. Since there is no infinite descending sequence of ordinals, it follows that there is no infinite play in the game Buckets of Fish. This idea also shows that the ordinal game values of positions in this game are bounded above by ω^ω, and every ordinal less than ω^ω is realized by some position.

So now we know that the game will always end. But how shall we play? What is the winning strategy? Say you are faced with buckets having fish in the following amounts:

What is your winning move? Please give it some thought; we shall return to the topic in a later chapter, The Theory of Games, where we will provide the winning strategy for Buckets of Fish.

Every number is interesting

I should like to conclude this chapter with an informal “proof” that every natural number is interesting. For example, 0 is interesting, since it is the additive identity, which is an interesting property for a number to have. Similarly, the number 1 is the multiplicative identity, which also is quite interesting. The number 2 is the first prime number, 3 is the first odd prime number, 4 is the first composite number, and so on.

I claim that every number is interesting. To prove this, observe that if there were any uninteresting numbers, then by the least-number principle, there would have to be a number n that was the smallest uninteresting number. But that is an extremely interesting property for a number to have! This therefore contradicts the assumption that n was uninteresting, and so there must not be any uninteresting numbers. So we may rejoice: every number is interesting! The reader will criticize this argument in the exercises.

Habits

• Understand induction deeply. Grasp the idea that mathematical induction is about the impossibility of minimal counterexamples. Learn how this manifests in the various forms of induction — common induction, strong induction, the least-number principle — and understand deeply why these are valid.

• Use induction flexibly. Do not insist upon a rigid format for proofs by induction but, rather, adopt a flexible approach, choosing whichever valid form of induction fits your argument best.

• Distinguish sharply between example and proof. Key examples can be extremely valuable and insightful, but examples do not prove a universal statement. Examples can offer evidence in favor of a mathematical conjecture, but proving a universal claim will require a general argument. Examples can prove an existential statement, however, and they can refute a universal statement, in which case they are called counterexamples.

Exercises

1. Show by induction that 2ⁿ < n! for all n ≥ 4.

2. Show by induction that

3. Show by induction that f₀ + ··· + f_n = f_n+2 - 1 in the Fibonacci sequence.

4. Show by induction that f_n < 2ⁿ in the Fibonacci sequence.

5. Consider an alternative Fibonacci sequence, starting with 0, 2. Can you prove the analogue of the theorem about summing the squares of the sequence?Generalize the result as far as you can.

6. In analogy with the triangular numbers, define the pentagonal numbers and the hexagonal numbers. Can you find and prove a formula for the nth pentagonal number and the nth hexagonal number?

7. Show by induction that a finite set with n elements has exactly 2ⁿ many subsets.

8. Prove by induction that the following equation holds for every positive integer n.

9. Prove that

10. Prove that

11. Show that every natural number has a unique base 3 representation.

12. Prove that if you divide the plane into regions using finitely many straight lines, then the regions can be colored with two colors in such a way that adjacent regions have opposite colors. [Hint: Use induction on the number of lines.]

13. True or false: every flying polka-dotted elephant in this room is smoking a cigar.

14. Criticize the following “proof.” Claim. All horses are the same color. Specifically, every finite set of horses is monochromatic. Proof. We argue by induction. The statement is clearly true for sets of size 1. Assume by induction that all sets of n horses are monochromatic, and consider a set of size n + 1. The first n horses are all the same color. The last n horses are all the same color. Because of the overlap, this means that all n + 1 horses are the same color. So by induction, all finite sets of horses are all the same color, and so all horses are the same color. ☐

15. Criticize the following “proof.” Claim.

    Proof. Let

    and consider the statement asserting S(n) < ∞. This statement is true for n = 1, since 1/1 = 1 < ∞. And if it is true for n, then it is true for n + 1, since the next sum

    is equal to the previous sum

    plus the next term 1/(n + 1), or in other words, S(n + 1) = S(n) + 1/(n + 1), and this is the sum of two finite numbers. So S(1) is true, and S(n) implies S(n + 1). So we have therefore proved

    by induction. ☐

16. Prove the following common induction principle variation: Assume that A is a set of integers with some k ∈ A, and furthermore, that whenever n ∈ A, then also n + 1 ∈ A. Then A contains all integers above k.

17. Prove that the least-number principle, the common induction principle, and the strong induction principle are all equivalent. Assuming any one of them, you can prove the others.

18. Criticize the argument that every number is interesting, given at the end of the chapter.

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This is a free extended excerpt from chapter 4 of my book Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing, for aspiring mathematicians who want to learn how to write proofs.

Proof and the Art of Mathematics teaches the art of proof-writing in the diverse contexts of a variety of mathematical subjects, but always with mathematical theorems and statements that carry their own inherent interest and fascination—compelling mathematical results with interesting, elementary proofs.

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The principle of mathematical induction lies at the core of number theory and mathematics generally, serving as a powerful unifying central principle in axiomatic developments of the subject, the main engine used to prove essentially all of the fundamental facts of arithmetic and more. So let us explore this core principle a little more fully. We have already begun to use it informally in the previous chapters.

The least-number principle

The core idea of induction is expressed by the following axiomatic principle:

Least-number principle. If there is a natural number with a certain property, then there is a smallest number with that property.

Equivalently, every nonempty set A of natural numbers has a least element. This principle asserts that the natural numbers are well-ordered, for a well-order is one for which every nonempty set has a least member.

The reader may have noticed that we made use of this principle in several previous arguments. We used it, for example, in chapter 1 in the revised proof that √2 is irrational, when we chose the numerator to be as small as possible, and we also used it in the proof that every rational number can be placed into lowest terms. We used the least-number principle again in the geometric proof that √2 is irrational, when we took the integer-sided squares to be as small as possible for which one had double the area of another, and we used it when proving the Euclidean algorithm and when proving the fundamental theorem of arithmetic. So although this chapter is titled, “Mathematical Induction,” in fact we have already been making inductive arguments in the earlier chapters without saying so.

One should not think of mathematical induction as an obscure proof method used only to prove certain curious recursive sums. Rather, mathematical induction is a fundamental principle, a bedrock idea upon which all the basic facts of number theory rest. In the axiomatic developments of number theory from first principles, such as in the Peano or Dedekind axiomatizations, for example, the principle of mathematical induction is used in almost every fundamental proof.

Common induction

In mathematical practice, elementary applications of induction often make use of certain idiosyncratic formulations of the least-number principle. Perhaps the most common formulation of the induction principle in elementary arguments is the following:

Theorem (Common induction principle). Suppose that A is a set of natural numbers for which 0 ∈ A, and furthermore, whenever n ∈ A, then also n + 1 ∈ A. Then every natural number is in A.

Another way to express the principle is that if 0 has a certain property, and the property is necessarily propagated from each natural number n to the next number n + 1, then indeed every natural number has the property.

The notation x ∈ A appearing in the principle here means that x is an element of the set A. The claim that 0 ∈ A is commonly referred to as the anchor of the induction, and the implication n ∈ A ⇒ n + 1 ∈ A is called the induction step. Because of the anchor, we know 0 is in the set. By applying the induction step to that case, we see that 1 also is in the set. And by applying the induction step to that case, we see that 2 is in the set, and so on. Each number in the set causes the next number to be in the set. Like a train of dominoes, every number falls in turn into the set.

The first domino is pushed, and each falling domino causes the next to fall; in the end, of course, every domino falls. This image might help us see the fundamental correctness of the common induction principle; we shall nevertheless later prove the principle, using the least-number principle, and this is why I have stated the common induction principle as a theorem rather than as an axiom.

Alternative formulations of the common induction principle allow you to anchor the induction at a number other than 0. For example, if A is a set of integers for which 1 ∈ A, and furthermore, whenever n ∈ A then also n + 1 ∈ A, then every positive integer n ≥ 1 is in A. We could also anchor the induction at 5, or at -3, or whatever, and conclude that all numbers from this anchor and above n ≥ 5 are in the set. You will be asked to prove this in the exercises. The principle of mathematical induction is robust and takes many useful forms.

Several proofs using induction

Before proving the common induction principle, let us first get a little practice using it. Consider the triangular numbers, which are the numbers of the form 1 + 2 + 3 + ··· + n. These numbers are called triangular because this many objects can be arranged in the form of a triangle, as pictured here, where one adds one more layer each time as n increases.

Theorem. The sum of the first n positive integers is n(n + 1)/2, for any positive integer n. In other words,

Thus, the nth triangular number is n(n + 1)/2.

Proof. We prove the theorem by common induction. The statement is true at the anchor n = 1, since 1 = 1(1 + 1)/2, so the number 1 is in the set of numbers for which the statement is true. Consider now the induction step. We assume that the statement is true for a number n and consider n + 1. Since the statement was true for the number n, we know that 1 + 2 + ··· + n = n(n + 1)/2. If we now add n + 1 to both sides of this equation and apply some elementary algebra, we see that

This is exactly what it means for the statement to be true in the case of n + 1, so we have proved that if the statement is true for n, then it is also true for n + 1. Therefore, we conclude by the principle of induction that the statement is true for all positive integers. ☐

We could have anchored the induction at n = 0 rather than n = 1, because the statement is also true for n = 0, as the sum of the first zero many positive integers is zero, which is equal to 0 · 1/2. In light of this, instead of stating the theorem as “for every positive integer n,” we could have stated the theorem as “for any natural number n.”

Theorem. The sum of the first n odd numbers is n², for any natural number n. In other words,

Proof. Let us prove the theorem by common induction. The statement is easy to see when n = 1, since 1 = 1². But again, we may actually take n = 0 as the base case in the original statement, since the sum of the first zero many odd numbers is 0, since there are not any, and 0² = 0 as well, so it is true for n = 0. (Perhaps the case n = 0 is a little confusing in the case of the displayed equation, however, since perhaps it is not clear what the left side is supposed to indicate when n = 0. To my way of thinking, though, the only sensible interpretation of the left-side expression when n = 0 is that there are no terms in it at all, giving a sum of 0, which is indeed precisely what we want.) For the induction step, assume that the statement is true for a number n, and consider n + 1. Since the statement is true for n, we know that 1 + 3 + 5 + ··· + (2n - 1) = n². If we add 2(n + 1) - 1 to both sides, which is the same as 2n + 1, we get 1 + 3 + 5 + ··· + (2n + 1) = n² + 2n + 1, and this further simplifies to (n + 1)². Thus, the statement is true for n + 1. So we have proved that the statement is true at the anchor, and the truth propagates from each number n to n + 1. So by the principle of induction, we conclude that the statement is true for every positive integer. ☐

We shall have an alternative proof of the theorem in a later chapter, entitled Proofs Without Words.

Theorem. The sum of the powers of 2 up to 2ⁿ is 1 less than the next power of 2. In other words,

Proof. We prove the theorem by common induction. The statement is true for n = 0, since 1 = 2¹ - 1. Assume that the statement is true for a number n, and consider n + 1. Since the statement was true for n, we have

By adding 2ⁿ⁺¹ to both sides, we get

which is equal to

which is what is desired for the case of n + 1. So the statement is true for all n by induction. ☐

The Fibonacci sequence is the sequence

Can you see the pattern? The sequence starts out with zero and one and then each next number is the sum of the previous two. We can therefore describe the sequence by the following recursive rules:

Theorem. The Fibonacci numbers obey the following identity:

Proof. We prove the theorem by common induction. The statement is true for n = 0, since (f₀)² = 0 = 0 · 1 = f₀ f₁. Assume that the statement is true for a number n, and consider n + 1. Since the statement is true for n, we have

Let us add (f_n+1)² to both sides, concluding that

which is equal to

By applying the recursive definition, this is equal to f_n+2 f_n+1. So we have established that

and so the statement is true for n + 1. Thus, by induction, it is true for all natural numbers n. ☐

An alternative visual proof of this identity will be given in the chapter on proofs without words.

Theorem. The number 6ⁿ - 1 is a multiple of 5 for every natural number n.

Proof. We prove the theorem by common induction. The statement is true when n = 0. Suppose by induction that the statement is true for a number n, and consider n + 1. Since 6ⁿ - 1 is a multiple of 5, we have 6ⁿ - 1 = 5k for some integer k. In other words, 6ⁿ = 5k + 1. Multiplying both sides by 6, we see that 6ⁿ⁺¹ = 6(5k + 1) = 30k + 6, and consequently, 6ⁿ⁺¹ - 1 = 30k + 5 = 5(6k + 1), which is a multiple of 5. So the statement is true for n + 1, and we have thereby proved by induction that the statement holds for all natural numbers. ☐

The previous theorem can also be easily proved with modular arithmetic, for those who are familiar with it, because 6 modulo 5 is 1, and so the expression 6ⁿ - 1 is equivalent to 1ⁿ - 1 mod 5, which is 0 mod 5. Thus, 6ⁿ - 1 is a multiple of 5.

Theorem. The number n³ - n is a multiple of 6, for every natural number n.

Proof. We prove the theorem by common induction. The statement is true when n = 0, since 0³ - 0 = 0, and this is a multiple of 6. Suppose by induction that the statement is true for a number n. So n³ - n is a multiple of 6. Observe that

The number n(n + 1) is always even, since either n or n + 1 is even (as with the arguments of chapter 2), and so 3n(n + 1) is a multiple of 6. So the next number (n + 1)³ - (n + 1) arises from the previous number n³ - n by adding a multiple of 6. So if the previous number was a multiple of 6, then so will be the next number. So the statement is true for n + 1, and we have thereby proved by induction that the statement is true for all natural numbers. ☐

This statement was also proved without induction in the exercises of chapter 2.

Theorem. The inequality n² < 3ⁿ holds for every natural number.

Proof. We prove the theorem by common induction. First, the statement is easy to verify in the cases n = 0, n = 1, and n = 2, as the reader may easily check. Assume by induction that the statement is true for a number n, and consider n + 1. Since we have verified the claim up to 2 already, we may also assume that 2 ≤ n. Since the statement is true for this number n, we know that n² < 3ⁿ. Using the fact that 2 ≤ n, we may argue with the following chain of inequalities:

And so the statement is true for n + 1. Thus, by mathematical induction, it holds for all natural numbers. ☐

I should like to call attention to the fact that in the proof of the theorem we made at first what might appear to be three different anchor cases, considering not only n = 0 but also n = 1 and n = 2. Why did we do this? Could we have skipped that part? Although with common induction one generally needs only one anchor, in this particular argument we actually needed those extra cases, because the argument we gave in the induction-step part of the proof began to work only once 2 ≤ n. One way to think about it is that we just established the cases n = 0 and n = 1 separately, but the real anchor was n = 2, since the induction step of the argument works only for 2 ≤ n.

Proving the induction principle

So we have stated the common induction principle, and we have used it in a few arguments. But can we prove the principle itself? Yes, in fact it is easy to prove the common induction principle from the least-number principle.

Proof of common induction principle. We prove the common induction principle using the least-number principle. Assume that A is an inductive set of natural numbers, which means that 0 ∈ A and whenever n ∈ A, then also n + 1 ∈ A. We want to prove that every number is in A. If not, then there must be some numbers not in A, and by the least-number principle there must therefore be a smallest number that is not in A. Let us call this number m. The assumption on A ensures that m ≠ 0. Consequently, m ≥ 1, and so m - 1 is a natural number. Since m - 1 is smaller than m and m is the smallest number not in A, it must be that m - 1 is in A. Therefore, by the induction assumption on A, it follows that (m - 1) + 1 is also in A. But this is m itself, and so m is in A after all, contradicting our earlier assumption that it was not. So A must contain every natural number. ☐

Strong induction

Although the common induction principle is indeed quite common, there are sometimes situations where one wants to prove something by induction, but the induction implication goes naturally not from n to n + 1 but, rather, from some other smaller number to n + 1, or from several smaller numbers. In these cases, the following stronger formulation of the induction principle is useful.

Theorem (Strong induction principle). Assume that A is a set of natural numbers with the property that, for every natural number n, if every number smaller than n is in A, then n itself is in A. Then every natural number is in A.

One thing to notice about the strong induction formulation is that it has no anchor case! How can this induction principle be correct? But it is correct, I assure you, and the reason has to do with what is called vacuous truth. Specifically, in the case n = 0, there are no natural numbers smaller than n, and so it is vacuously true that all such numbers are in A, since this assertion can have no counterexample. So the strong induction assumption vacuously implies that 0 ∈ A, and the induction engine is thereby started.

Proof. This argument is similar to the proof of the common induction principle from the least-number principle. Namely, assume that A satisfies the strong induction property mentioned in the strong induction principle. If not every natural number is in A, then there would be some least natural number n that is not in A. But in this case, every number smaller than n would be in A and so n itself would have to be in A after all, by the strong induction assumption on A. This contradicts our assumption on n, and so there cannot be any natural numbers not in A. In other words, every natural number must be in A, as desired. ☐

Several of our informal uses of mathematical induction that we gave in chapter 3 had essentially used strong induction. For example, to prove that every positive integer has a prime factorization, you suppose that every number smaller than n has such a factorization and then observe that either n is prime, in which case it is its own factorization, or else n = ab for smaller numbers a,b < n, which by the induction assumption have prime factorizations, which can be put together to make a prime factorization of n. In this argument, you see, we did not move from n to n + 1 but, rather, reduced the problem on n to instances of the problem for smaller numbers a and b. Several of the other results in chapter 3 can similarly be set out as proofs by strong induction.

Let us now give a few more applications of the principle of induction. Consider the case of binary representation of numbers. Perhaps you know that the number 6 is represented in binary as 110, since 6 = 4 + 2, and the number 19 is represented as 10011, since 19 = 16 + 2 + 1. Perhaps you know that every natural number has a binary representation. But how could we prove this? Furthermore, how do we know that different binary representations always represent different numbers?

Theorem. Every natural number can be expressed as the sum of a unique set of distinct powers of 2.

Proof. Assume by strong induction that the statement is true for all numbers smaller than n. Since 0 is the empty sum, we may assume that n ≥ 1. Let 2^k be the largest power of 2 that fits inside n, so that 2^k ≤ n. Since n - 2^k is smaller than n, it has a unique representation as a sum of distinct powers of 2. Note that 2^k cannot appear in the sum for n - 2^k, since otherwise n would be at least 2^k + 2^k = 2^k+1, but we had chosen 2^k to be the largest power of 2 that fits inside n. Thus, by adding 2^k to the sum for n - 2^k, we obtain n as a sum of distinct powers of 2, establishing the existence claim of the theorem. For uniqueness, note that 2^k must occur in any representation of n as a sum of distinct powers of 2, since no larger powers ever occur by the choice of k, and if we use only smaller powers, then even if we use all of them the sum will be at most 1 + 2 + 4 + ··· + 2^k-1, which by an earlier theorem is less than 2^k and so therefore cannot sum to n. So 2^k is used in every representation of n as a sum of powers of 2, and the uniqueness claim therefore follows from the uniqueness of the representation of n - 2^k. ☐

Corollary. Every natural number has a unique binary representation.

Proof. Every natural number has a unique binary representation, since the binary representation is simply a convenient compact notation for indicating which powers of 2 one is including in the sum, and by the previous theorem this collection is unique. ☐

For example, the binary number 101101 is a compact notation for the sum

which is 45 in decimal notation.

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Infinitely many primes

Let us turn now to another classical result, the fact that there are infinitely many primes. This is a classic argument, often attributed to Euclid, known for thousands of years.

Theorem. There are infinitely many prime numbers.

Proof. Suppose that you have a list of finitely many prime numbers: p₁, p₂, …, p_n.

Let N = (p₁p₂ ··· p_n) + 1, the result of multiplying them together and adding 1. Observe that if you should divide N by any particular prime number p_i on your list, then there will be a remainder of 1. In particular, this number N is not divisible by any prime number on your list. But N is a product of primes, as every natural number is, and so there must be a prime that is not on the list. Thus, no finite list of numbers includes all the primes, and so there must be infinitely many of them. ☐

Sometimes one sees this argument given as a proof by contradiction, like this:

Proof. Assume toward contradiction that there are only finitely many primes, p₁, p₂, ···, p_n. Multiply them all together and add one N = (p₁p₂ ··· p_n) + 1. This new number N is not a multiple of any p_i, and so its prime factorization must involve new primes, not on the list, a contradiction. ☐

Is it better to give a proof by contradiction or directly? This second proof seems perfectly fine. Euclid had not used proof by contradiction but, rather, proved that every finite list of primes can be extended. Mathematicians often prefer direct proofs over proofs by contradiction. One reason, to my way of thinking, is that direct proofs usually carry more information about the mathematical background — they paint a fuller picture of mathematical reality. When one proves an implication p ⇒ q directly, one assumes p and derives further consequences p₁, p₂, and so on, before ultimately concluding q. Thus, one has learned a whole context about what it is like in the p worlds. Similarly, with a proof by contraposition, one assumes the negation of the conclusion ¬q and derives consequences about what it is like in the worlds without q, before finally concluding ¬p, negating the hypothesis. But in a proof by contradiction, in contrast, one assumes something that ultimately does not hold in any world; the argument often seems to tell us little beyond the brute fact of the implication p ⇒ q itself.

Next, let us prove that although there are infinitely many prime numbers, nevertheless there are also long stretches of numbers with no prime numbers at all.

Theorem. There are arbitrarily large intervals in the positive integers containing no prime numbers.

Proof. Consider any positive integer n ≥ 2. Notice that n! + k is a multiple of k, whenever 1 ≤ k ≤ n, since in this case it is the sum of two multiples of k. By considering 2 ≤ k ≤ n, we have therefore found an interval of positive integers, from n! + 2 up to n! + n, containing no prime numbers. This interval (inclusive) has n-1 numbers in it, and so there are arbitrarily large intervals of nonprimes in the positive integers. ☐

It is natural to inquire about the density of the prime numbers. Let π(n) be the prime-counting function, the number of prime numbers p ≤ n.

The figure above illustrates the density of primes, with a narrow blue line for each prime number up to five thousand, spaced out on the number line (some selected primes are indicated in red). The pattern is irregular, with some intervals having primes clumped near each other and other intervals more sparsely populated, with perhaps a barely perceptible decay in the density as one moves to higher numbers. Since the previous theorem shows that there are large primeless intervals, we should expect to find increasing white patches, without any blue lines, if the figure were continued to larger numbers.

In contrast to the previous result, let us now provide a lower bound on the number of prime numbers. Define that a positive integer r is square-free if it is not a multiple of any square number bigger than 1; equivalently, as the reader will show in the exercises, a positive integer is square-free if all the primes in its prime factorization have exponent 1.

Theorem (Erdős). The prime-counting function π has a lower bound provided by the inequality

Proof. Every natural number can be factored as rs², where r is square-free, as the reader will verify in the exercises. How many square-free numbers are there less than n? Well, every square-free number is a product of distinct primes, and so it is determined by the set of primes that divide it. So the number of square-free numbers up to n will be at most 2^π(n), which is the number of sets of primes up to n. Second, the number of squares up to n is at most √n, since if s² ≤ n then s ≤ √n, and s² is determined by s. So the number of numbers up to n having the form rs², where r is square-free, is at most 2^π(n)√n. Since there are n positive integers less than or equal to n, we conclude that

Dividing both sides by √n leads to

and then taking the logarithm (base 2) enables us to conclude that

which amounts to the inequality

as desired. ☐

An arithmetic progression is a sequence of numbers of the form p, p + e, p + 2e, p + 3e, and so on. Two numbers are relatively prime if they have no nontrivial common factor.

Theorem. There are arbitrarily long arithmetic progressions consisting of relatively prime numbers.

Proof. Fix any number d, and let e = d!, the factorial of it. Let p be a prime number larger than e, and consider the numbers

in other words, the numbers of the form p + ei, where i < d. These form an arithmetic progression of length d and period e. Let me show that these numbers are all relatively prime. Suppose that q is a prime factor of two of the numbers, p + ei and p + ej, where i < j < d. It follows that q divides the difference of the two numbers, which is equal to e( j - i ). Since i and j are both less than d and e = d!, it follows that all of the prime factors of e( j - i ) are at most d, so q ≤ d. In this case, q divides e and hence ei, and since it also divides p + ei, it must be that q divides p, which is prime. So q = p, contrary to our choice of p to be larger than e. ☐

It had been a long-standing open question, since at least 1770, whether one could find arbitrarily long arithmetic progressions of prime numbers (not merely relatively prime). This much harder question was finally settled in the affirmative in a celebrated 2004 result of Ben Green and Terence Tao. The Green-Tao theorem states that for every natural number d, there is an arithmetic progression of length d consisting entirely of prime numbers.

Habits

• Choose variable names well. Choose sensible variable names that help remind you of their meaning. But also try to follow established variable-naming conventions, when possible, in order to hook into your readers' expectations about what kind of quantity your variable represents. For example, many mathematicians use variables n and m to represent natural numbers or integers, while p and q are probably prime numbers, the variables x and y frequently represent real numbers, z is often a complex number, and f and g are likely functions. Conventions can vary between particular mathematical specializations.

• Use technical words with accuracy and precision. Recognize that mathematical words often carry extremely precise meanings, far more explicit and detailed than the meanings conveyed in ordinary language by those words. Use technical vocabulary strictly to carry these more specific meanings.

• Define your technical terms. Provide explicit formal definitions for your mathematical terms, and use the words in accordance with their definitions. Clarify undefined terms; do not allow vagueness and ambiguity to slip into your analysis simply because you have not taken the trouble to define your terms.

• Try proving implications directly. When proving a statement of the form, “if p, then q,” try starting your argument by writing, “Assume p,” and then argue for q. This method is called a direct proof of the implication.

• Try proving implications via the contrapositive. When proving a statement of the form “if p, then q,” consider the contrapositive, “if not q, then not p.” This is logically equivalent to the original statement and sometimes admits a smoother argument, particularly when q has a negative form, in which case “not q” becomes a positive assumption. Write, “To prove the contrapositive, assume not q.” And then argue for not p.

Exercises

1. Prove that a positive integer is prime if and only if it has exactly two positive integer divisors.

2. Show that a positive integer p > 1 is prime if and only if, whenever p divides a product ab of integers, then either p divides a or p divides b.

3. Prove a stronger version of Bézout's lemma, namely, that for any two integers a and b, the smallest positive number d expressible as an integer linear combination of a and b is precisely the greatest common divisor of a and b.

4. Read Timothy Gower's blog post, “Why isn't the fundamental theorem of arithmetic obvious?” and his follow-up post, “Proving the fundamental theorem of arithmetic.” Write a summary.

5. Show that if we count the number 1 as prime, then the uniqueness claim of the fundamental theorem of arithmetic would be false. (And this is one good reason not to count 1 as amongst the prime numbers.)

6. Prove that if one prime divides another, then they are equal. (This was used in the proof of the fundamental theorem of arithmetic.)

7. Mathematician Evelyn Lamb fondly notes of any large prime number presented to her that it is “one away from a multiple of 3!” And part of her point is that this is true whether you interpret the exclamation point as an exclamation or instead as a mathematical sign for the factorial. Prove that every prime larger than 3 is one away from a multiple of 3!. [Hint: Consider the remainder of the number modulo 6.]

8. Prove that if a, b, and c are positive integers and the product abc is a multiple of 6, then one of the numbers ab, ac, or bc is a multiple of 6.

9. Define that an integer is even if it is a multiple of 2; and otherwise it is odd. Show that every odd number has the form 2k + 1 for some integer k. Conclude that any two consecutive integers consist of one even number and one odd number.

10. Prove or refute the following: For any list of primes p₁, p₂, ···, p_n, the number (p₁p₂ ··· p_n) + 1 is prime.

11. Criticize the following “proof.” Claim. There are infinitely many primes. Proof. Consider any number n. Let n! be the factorial of n, and consider the number p = n! + 1. So p is larger than n, and it has a remainder of 1 dividing by any number up to n. So p is prime, and we have therefore found a prime number above n. So there are infinitely many prime numbers. ☐

12. Prove that a positive integer is square-free if and only if all of the exponents in its prime factorization are 1.

13. Prove that every positive integer m can be factored as m = rs², where r is square-free. [Hint: Let r be the product of the primes having odd exponent in the prime factorization of m. Alternatively, let s be the largest number such that s² divides m, and argue that the quotient is square-free.]

14. Prove or refute the following: The lower bound on π(n) provided by Erdős’s theorem is best possible, after applying the ceiling function in the integers. Make a clear theorem statement about what this means and about what exactly you are proving.

Please enjoy my musical video collaboration with Hannah Hoffman proving the infinitude of primes:

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This is a free extended excerpt from chapter 3 of my book Proof and the Art of Mathematics, an introduction to the art and craft of proof-writing, for aspiring mathematicians who want to learn how to write proofs.

Proof and the Art of Mathematics teaches the art of proof-writing in the diverse contexts of a variety of mathematical subjects, but always with mathematical theorems and statements that carry their own inherent interest and fascination—compelling mathematical results with interesting, elementary proofs.

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Number theory

Number theory is celebrated by mathematicians as a pure form of abstract thought, a distillation of reason. Carl Friedrich Gauss called it the “queen of mathematics,” while G. H. Hardy, in A Mathematician's Apology, admired its pure, abstract isolation, praising the fact that it was unencumbered by the physical world, without use or application.

And yet, despite this, in a strange and surprising twist of fate, the theory has in contemporary times found key practical applications; deep number-theoretic ideas, for example, lie at the core of cryptography and internet security. Our dreamy iridescent theory of numbers, it turns out, supposedly without use or application, does in fact have important applications, vital for commerce and communication, so much so that number-theoretic ideas are now firmly established via cryptography in the foundations of our economy.

Prime numbers

Let us develop some elementary number theory, beginning with the prime numbers. What is a prime number? At this question, I imagine, perhaps a nervous laugh goes through the classroom — of course we are all familiar with prime numbers, right? A helpful student suggests, tentatively, that a number is prime if the only divisors of the number are 1 and itself. Is this a good definition? Let us quibble. One minor issue is that numbers can have negative divisors. We want 3 to be prime, to be sure, but does the factorization 3 = (-3) · (-1) mean that it has -3 and -1 also as divisors? Let us therefore interpret the student's proposal to refer only to positive divisors. Similarly, we do not want (7/2) · 2 to rule out 7 as prime, and so we should interpret the student's proposal as referring only to positive integer divisors.

A more serious issue, however, is the question of whether the number 1 should count as prime. Is 1 prime? Since 1 has no positive integer divisors other than 1 and itself, it would seem that, according to the student's proposal, we would say that, indeed, 1 is prime. Nevertheless, this would go against the advice of many mathematicians, who have come to the conclusion that we do not actually want to include 1 amongst the prime numbers. On this view, the student's suggestion is not quite exactly right.

I should like to emphasize that, as mathematicians, we are free to define our terms to mean whatever we want. We could define our notion of prime number so that 1 turns out to be prime, or we could have defined the notion differently, so that 1 would not count as prime. There is no issue here of discovering whether or not the number 1 is prime; we can define our technical terms as we wish. Rather, what is at stake here is what we want the word prime to mean and whether we want to use this word in a way that makes the number 1 prime or makes it not prime.

All things considered, mathematicians have concluded that it is more convenient in many contexts if we say that 1 is not prime. To give one example, if the number 1 was counted as prime, then it would no longer be true that every positive integer had a unique prime factorization, as in 12 = 3 · 2², since we could always add more 1s the factorization, like this: 12 = 3 · 2² · 1 · 1 · 1. If 1 counted as prime, we would find ourselves often having to add exceptions to many theorems about prime numbers, to say in effect, “except for 1.” Ultimately, it is more convenient simply to say that 1 does not count as prime. So let us make our official definition as follows:

Definition. An integer p is prime if p > 1 and the only positive integer divisors of p are 1 and p.

Students of abstract algebra might recognize that this definition is actually not concerned with primality but with irreducibility. Specifically, in the algebraic structures known as rings, an element is defined to be irreducible if it admits no nontrivial factorization, as in the definition above. An element p of a ring is prime, in contrast, if it is nonzero, it has no multiplicative inverse, and whenever p divides a product ab, then either p divides a or p divides b. Although in the class of rings known as integral domains every prime element is irreducible, the converse is not always true, and many rings and integral domains have irreducible elements that are not prime. Nevertheless, it will follow from Euclid’s lemma below that in the positive integers the prime numbers are the same as the irreducible numbers, and so it will be fine for us to stick with the traditional definition above as our official definition of what it means to be prime.

The fundamental theorem of arithmetic

The reader has likely had experience factoring numbers as a product of prime numbers. Twelve factors as 12 = 3 · 2², and the number 8470 can be factored first as the product 10 · 847, and then further broken into primes as 2 · 5 · 7 · 11 · 11. Let me ask an innocent question:

Question. For a given number, will we always get the same prime factorization?

Of course, I mean the same here in the sense that 3² · 5 · 7 counts as the same factorization as 5 · 3 · 7 · 3 — we have simply rearranged the same prime factors. Are prime factorizations always unique in this sense? Perhaps. Can we imagine that, for extremely large numbers, the outcome of the prime factorization might depend on how we had proceeded to compute it? Perhaps. If we had started with a slightly different product at the first step, might we have found ourselves ultimately with a different prime factorization in the end?

The number 11543, for example, can be factored as 7 · 17 · 97, and these factors are each prime. Do we know that there is not also some other way to factor it into primes? For a comparatively small number like this, we might hope to try out all the other candidates and be convinced this way. That method is actually impractical, even for numbers such as 11543, and for much larger numbers like 653465345453435463456534655534354652, it is downright infeasible. Do we actually know that this number has a unique prime factorization?

When factoring numbers, to be sure, we routinely refer to the prime factorization. But perhaps we should be saying merely that we have a prime factorization? Can there be more than one? On my view, this is a serious question, more difficult than it might initially appear. The fact of the matter is that the usual naive treatment of prime factorization simply does not touch on the question of uniqueness. We have become familiar with the uniqueness of prime factorizations largely by observing many instances of prime factorization, without ever encountering a situation where a number admits more than one factorization. Does this experience constitute proof? No, of course not.

Meanwhile, prime factorizations are indeed unique, and this fact is the first deep theorem of number theory, known as the fundamental theorem of arithmetic:

Fundamental theorem of arithmetic. Every positive integer can be expressed as a product of primes, and furthermore, this factorization is unique up to a rearrangement of these prime factors.

We shall prove this theorem presently, but before doing so, let me remark on a certain issue arising in the statement of this theorem and a habit that mathematicians have. Namely, the theorem says that every positive integer is a product of primes; but is this right? What about the number 5? Yes, it is prime, but is it a product of primes?

Mathematicians will say yes, 5 is a product of primes, a product consisting of just one factor, the number 5 itself. You might reply, that is not a product at all! Should not the theorem say, “Every positive integer is either prime or a product of primes”? Mathematicians will insist, nevertheless, that it is a good idea to consider 5 and the other prime numbers as products of primes, degenerate products if you will, products consisting of just one factor.

The reason is that our mathematical theories often become more robust when we incorporate the trivial or degenerate instances of a phenomenon into the fundamental definitions. Every square counts also as a rectangle; every equilateral triangle is also isosceles. This practice often leads to a smoother mathematical analysis in the end. A rectangle is precisely a quadrilateral with four right angles, for example, but this would not be true if we did not count squares also as rectangles. In the same way, we allow ourselves to refer to the product of just one number, without being multiplied by any other number.

One sometimes hears that 5ⁿ means that we multiply 5 by itself n times. This is not really accurate, however, if what is meant is that we have n multiplications, since 5² means 5 × 5, which is only one act of multiplication; we would be more correct, therefore, to say that n refers to the number of factors, rather than to the number of times we are multiplying. This is a fence-post error, discussed further in a later chapter. So we regard 5¹ as a product consisting of just one factor. Similarly, we take 5⁰ = 1 as a product with no factors at all, the empty product. This is the sense in which 1 is a “product” of prime numbers.

The thing to notice here is that, even if we had stated the theorem as, “Every positive integer is either prime or a product of primes,” it still would not be correct without this empty product convention, since the number 1 is a positive integer, but it cannot be expressed as a product of primes except as the empty product. Without the empty and singleton product considerations, therefore, we would have to say, “Every integer greater than 1 is either prime or a product of primes.” But is it not both simpler and more elegant to state the theorem as we have above in the fundamental theorem? We have included the primes themselves each as a product with only one factor and the number 1 as the empty product.

So let us now prove the fundamental theorem, which makes two essentially different claims: an existence claim and a uniqueness claim. Namely, the existence claim is that every positive integer admits at least one prime factorization, and the uniqueness claim is that every number admits at most one prime factorization, in the sense that any two factorizations of the same number are rearrangements of one another. Let us begin with the existence claim, since this is perhaps more familiar, as well as easier to prove.

Fundamental theorem of arithmetic (existence). Every positive integer can be expressed as a product of prime numbers.

Proof. Let us prove that every positive integer has the property. The number 1 is expressed as the empty product, and so we have made a start. Suppose that all the numbers up to some number n have the property, where n > 1, and now consider whether n itself has the property. If n happens to be prime, then indeed, it is expressible as a product of just one prime factor, itself, and so it would have the property. Otherwise, n is not prime, and so we may factor it as n = ab for some numbers a and b, both smaller than n. Because we assumed that the property holds up to n, we know that both a and b are expressible as products of primes. So a = p₁ ··· p_k and b = q₁ ··· q_r for primes p_i and q_j, where 1 ≤ i ≤ k and 1 ≤ j ≤ r. By simply combining these products, we can now realize n also as a product of primes:

So the property does indeed hold at n. We have therefore proved that whenever the property holds up to a number n, then it holds at the number n. In other words, there can be no minimal counterexample to the property; thus, there can be no counterexample at all. So every number has the property. ☐

We used the method of minimal counterexamples, by which one shows that a property holds of all positive integers by showing that there can be no smallest counterexample to the property, and hence no counterexample at all. This method is closely related to (and essentially identical to) the method of mathematical induction, which we shall explore more fully in a later chapter.

Some mathematicians may have preferred to cover the method of induction before proving the fundamental theorem of arithmetic, but my preference was to mount these simple minimal-counterexample arguments in this chapter, using them in part as an introduction to inductive reasoning. In chapter 4, we shall give a more thorough general account of the theory of mathematical induction.

Euclidean division algorithm

In order to prove the uniqueness part of the fundamental theorem of arithmetic, we shall rely on some classic elementary number theory, which we now develop. Let us begin with the familiar fundamental principle that we can always divide integers, possibly with remainder, in such a way that the remainder is less than the number by which we are dividing. This fact is known as the Euclidean division algorithm. This terminology, quite old and firmly established, perhaps conflates the mathematical fact that the quotient and remainder exist with the algorithms or procedures that one might use to find them.

Lemma (Euclidean division algorithm). For any two positive integers n and d, there are unique integers q and r for which n = qd + r and 0 ≤ r < d.

Note how we have chosen variable names so as to aid our understanding, since the lemma is fundamentally concerned with the operation of dividing n by d, so n stands for numerator, d for denominator or divisor, q for quotient, and r for remainder.

Proof. To prove uniqueness, suppose that qd + r = q'd + r', where 0 ≤ r,r' < d. It follows that r - r' = (q' - q)d, and so r - r' is a multiple of d. Since also |r - r'| < d, it follows that r - r' = 0 and so r = r', which implies q = q'. So the representation is unique when it exists. For existence, we shall prove that there can be no smallest failing instance of the lemma. Specifically, we shall prove that if the lemma holds up to a number n, then it also holds at n. So suppose that n and d are numbers and that the statement of the lemma holds with d and any number smaller than n. If n < d, then we can write n = 0 · d + n, and this fulfills the requirement that 0 ≤ r < d. Similarly, if n = d, then we may write n = 1 · d + 0, which also verifies the desired property. So we may assume that d < n. In this case, n - d is a positive integer smaller than n. By the assumption on n, the lemma holds for d and n - d, and so there are numbers q and r with n - d = qd + r and 0 ≤ r < d. By adding d to both sides, we see that n = (q + 1)d + r, which fulfills the desired statement for n and d. So there can be no minimal counterexample to the lemma, and consequently, there can be no counterexample at all. So the lemma holds for all n and d. ☐

Next, we prove Bézout's identity. Recall from chapter 1 that integers are relatively prime if they have no common factor larger than 1.

Lemma (Bézout's identity). If integers a and b are relatively prime, then there are integers x and y for which 1 = ax + by.

Proof. Assume that integers a and b are relatively prime. Let d be the smallest positive integer that is expressible as an integer linear combination of a and b, that is, as d = ax + by for some choice of integers x and y. Certainly d ≤ a, since we can write a = a · 1 + b · 0. I claim that d divides both a and b. To see this, apply the Euclidean algorithm to express a = kd + r for some integer k and remainder r, with 0 ≤ r < d.

Putting our equations together, observe that

We have therefore expressed r as an integer linear combination of a and b. Since r < d and d was the smallest positive such combination, it follows that r must be 0. In other words, a = kd is a multiple of d, as claimed. A similar argument shows that b also is a multiple of d, and so d is a common factor of a and b. Since these numbers are relatively prime, it must be that d = 1, and so we have achieved 1 = ax + by, as desired. ☐

Euclid's lemma. If p is prime and p divides ab in the integers, then p divides a or p divides b.

Proof. Assume that p is prime and that p divides ab. If p does not divide a, then a and p must be relatively prime, since there are no other nontrivial factors of p. By Bézout's lemma, it follows that 1 = ax + py for some integers x and y. Multiplying both sides by b, we see that

Since p divides ab, it follows that p divides the right-hand side of this equation, and so p divides b. So we have proved that if p does not divide a, then it divides b. And so p must divide one of them. ☐

We can generalize Euclid’s lemma to the situation of many primes:

Lemma. If a prime p divides a product of integers n₁n₂ ··· n_k, then p divides some n_i.

Proof. We know by Euclid’s lemma that this lemma is true when there are only two factors. Suppose that this lemma holds when there are fewer than k factors, and that we have a prime number p that divides a product n₁n₂ ··· n_k with k factors. The trick is to look upon the product n₁n₂ ··· n_k as a product of just two things, like this: n₁ · (n₂ ··· n_k). Since p divides this product of two things, we may conclude by Euclid’s lemma that either p divides n₁ or p divides the rest of the product n₂ ··· n_k. In the first case, we are done immediately, and in the second case, since there are now fewer than k factors in the product, we conclude by our assumption on k that p must divide one of the n_i for 2 ≤ i ≤ k. So in any case, p divides some n_i, and the lemma is proved. ☐

Fundamental theorem of arithmetic, uniqueness.

Finally, we can prove the uniqueness part of the fundamental theorem of arithmetic.

Fundamental theorem of arithmetic (uniqueness). Every positive integer has at most one prime factorization, in the sense that any two factorizations are simply rearranging the order of the prime factors appearing in them.

Proof. We have already proved the existence claim. What remains is the uniqueness claim. Suppose that all the numbers smaller than a number n have at most one representation as a product of primes (unique up to rearranging the order in which the prime factors appear in the product). Suppose that

are two representations of n as a product of primes. Since p₁ divides n, it follows by the generalization of Euclid’s lemma that p₁ must divide one of the q_j, and since these are all prime, it must be equal to one of the q_j. It might as well be q₁, by rearranging the qs. But in this case, we have

since these are both n / p₁, and by our assumption on n, these two products are a simple rearrangement of each other. So the original products also are obtained by rearranging, and we are done. ☐

The fundamental theorem of arithmetic amounts to the combination of the existence claim and the uniqueness claim, so it is now proved.

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Proof and the Art of Mathematics teaches the art of proof-writing in the diverse contexts of a variety of mathematical subjects—number theory, finite combinatorics, discrete mathematics, order theory, game theory, geometry, set theory, real analysis, and others. The proof-writing ideas are thus presented in the context of different mathematical topics, but always with mathematical theorems and statements that carry their own inherent interest and fascination—compelling mathematical results with interesting, elementary proofs.

So welcome to this first installment—I shall be regularly featuring additional excerpts from the book here on Infinitely More, and so look for them in the Proof and the Art of Mathematics section.

I particularly recommend this book to instructors who are teaching the art of proof-writing. But the book also serves well for any interested reader.

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A Classical Beginning

One of the classical gems of mathematics—and to my way of thinking, a pinnacle of human achievement—is the ancient discovery of incommensurable numbers, quantities that cannot be expressed as the ratio of integers.

The Pythagoreans discovered in the fifth century BC that the side and diagonal of a square have no common unit of measure; there is no smaller unit length of which they are both integral multiples; the quantities are incommensurable. If you divide the side of a square into ten units, then the diagonal will be a little more than fourteen of them. If you divide the side into one hundred units, then the diagonal will be a little more than 141; if one thousand, then a little more than 1414. It will never come out exactly. One sees those approximation numbers as the initial digits of the decimal expansion:

The discovery shocked the Pythagoreans. It was downright heretical, in light of their quasi-religious number-mysticism beliefs, which aimed to comprehend all through proportion and ratio, taking numbers as a foundational substance. According to legend, the man who made the discovery was drowned at sea, perhaps punished by the gods for impiously divulging the irrational.

The number √2 is irrational

In modern terminology, the claim is that √2 is irrational, since this is the ratio of diagonal to side in any square, and the rational numbers are those expressible as a fraction of integers p / q, with q ≠ 0. So we may state the claim as follows.

Theorem. The number √2 is irrational.

In plainer words, the theorem asserts that √2 cannot be expressed as a fraction. Let us prove the theorem.

Proof. Suppose toward contradiction that √2 is rational. This means that

for some integers p and q, with q ≠ 0. We may assume that p / q is in lowest terms, which means that p and q have no nontrivial common factors. By squaring both sides, we conclude that

and consequently

This tells us that p² must be an even number, which implies that p also is even, since the square of an odd number is odd. So p = 2k for some integer k. From this, it follows that

and therefore

So q² also is even, and therefore q must be even. Thus, both p and q are even, which contradicts our assumption that p / q was in lowest terms. So √2 cannot be rational. ☐

That was a traditional argument, and my impression is that if you were to knock on doors down the hall of a university mathematics department asking for a proof that √2 is irrational, then first of all, every single mathematician you find will be able to prove it and second, the odds are good for you to get a version of that argument. To my way of thinking, just as every self-respecting educated person should know something of the works of Shakespeare and of the heliocentric theory of planetary motion, similarly everyone should be able to prove the irrationality of the square root of 2. Mathematicians now have dozens of proofs of this beautiful classical result—the reader can find various collections online—and by the end of this chapter, we shall see several.

Notice that the proof proceeded by contradiction. In such a proof, one assumes (“toward contradiction”) that the desired conclusion is not true, aiming to derive from this assumption a contradiction, an impossible consequence. Since that consequence cannot be the true state of affairs, the assumption leading to it must have been wrong, and so the desired conclusion must have been true.

In the proof we gave, we had appealed to some familiar mathematical facts, such as the fact that the product of two odd numbers is odd and the fact that every rational number can be expressed as a fraction in lowest terms. These facts, of course, require their own proof, and indeed we shall provide proofs presently.

If √2 is not rational, then what kind of number is it? One can view certain developments in the history of mathematics as a process of recognizing problematic issues in our number systems and addressing them with new and better number systems. In ancient days, one had the whole numbers 1, 2, 3, and so on. This number system, however, lacks an additive identity, and by including such an identity, the number zero, we thereby form the natural numbers ℕ with 0, 1, 2, 3, and so on. Although one can always add two natural numbers together and still have a natural number, this is not true for subtraction, and so we are led to the integers ℤ, which include negative numbers ...,-2, -1, 0, 1, 2,... and thereby become closed under subtraction. While one can always multiply two integers and still have an integer, this is not true for division, and so we form the rational numbers ℚ to accommodate ratios of integers. The fact that √2 is irrational shows the need for us to extend our number systems beyond the rational numbers, to the real numbers ℝ and eventually to the complex numbers ℂ.

In truth, the actual historical development of the number systems was not like that; it was confused and incoherent. The ancients had the whole numbers and their ratios, and the positive rational numbers, but the number zero was strangely delayed and complicated, appearing early on in some places merely as a placeholder but taken seriously as an actual number only much later. The actual historical development of mathematical ideas is often messy and confounding, riddled with confusion and misunderstanding, yet afterward one can reinvent an imaginary but more logical progression. One can gain mathematical insight from the imaginary history; not every mistake in the actual historical development of mathematics is interesting.

The number √2, while irrational, is nevertheless still an algebraic number, which means it is the root of a nontrivial integer polynomial, in this case a solution of the equation x² - 2 = 0. The existence of transcendental real numbers, or nonalgebraic numbers, was proved in 1844 by Joseph Liouville. I view Liouville's result as an analogue of the classical √2 result, because it shows once again the need for us to extend our number system: the algebraic numbers do not exhaust the numbers we find important. Indeed, we now know that numbers such as e and π are transcendental, and in 1874 Georg Cantor proved in a precise sense that most real numbers are transcendental.

Meanwhile, the complex numbers extend the real numbers with the imaginary numbers, starting with the imaginary unit

Since the number i solves the equation

it is an algebraic complex number, but there are also transcendental complex numbers. Mathematicians continue to extend our various number systems, and we now have number systems with various kinds of points at infinity, such as the extended real numbers, which are the real numbers plus ±∞, as well as other number systems, such as the quaternions, the ordinal numbers, the hyperreal numbers, the surreal numbers, and more.

Returning to the integers and the rationals, let us consider our proof above more closely. We had used some fundamental properties of the integers and the rational numbers, such as the fact that every fraction can be put into lowest terms. That may seem familiar, but can we prove it? Actually, we can completely avoid the issue of lowest terms by arguing as follows.

Proof. (Alternative slightly revised proof) Suppose toward contradiction that √2 is rational. So √2 = p / q for some integers p and q, and we may assume that the numerator p is chosen as small as possible for such a representation. It follows as before that 2q² = p², and so p² and hence also p is even. So p = 2k for some k, which implies that q² = 2k² as before, and so q² and hence also q is even. So q = 2r for some r, and consequently

We have therefore found a rational representation of √2 using a smaller numerator, contradicting our earlier assumption. So √2 is not rational. ☐

This way of arguing, although very similar to the original argument, does not require putting fractions in lowest terms. Furthermore, an essentially similar idea can be used to prove that indeed every fraction can be put in lowest terms.

Lowest terms

What does it mean for a fraction p / q to be in lowest terms? It means that p and q are relatively prime, that is, that they have no common factor, a number k > 1 that divides both of them. I find it interesting that the property of being in lowest terms is not a property of the rational number itself but rather a property of the fractional expression used to represent the number. For example, 3 / 6 is not in lowest terms and 1 / 2 is, yet we say that they are equal:

But how can two things be identical if they have different properties? These two expressions are equal in that they describe the same rational number; the values of the expressions are the same, even though the expressions themselves are different. Thus, we distinguish between the description of a number and the number itself, between our talk about a number and what the number actually is. It is a form of the use/mention distinction, the distinction between syntax and semantics at the core of the subject of mathematical logic. How pleasant to see it arise in the familiar elementary topic of lowest terms.

Lemma. Every fraction can be put in lowest terms.

Proof. Consider any fraction p / q, where p and q are integers and q ≠ 0. Let p' be the smallest nonnegative integer for which there is an integer q' with

That is, we consider a representation p' / q' of the original fraction p / q whose numerator p' is as small as possible. I claim that it follows that p' and q' are relatively prime, since if they had a common factor, we could divide it out and thereby make an instance of a fraction equal to p / q with a smaller numerator. But p' was chosen to be smallest, and so there is no such common factor. Therefore, p' / q' is in lowest terms. ☐

This proof and the previous proof of the theorem relied on a more fundamental principle, the least-number principle, which asserts that if there is a natural number with a certain property, then there is a smallest such number with that property. In other words, every nonempty set of natural numbers has a least element. This principle is closely connected with the principle of mathematical induction, which we shall discuss in a later chapter. For now, let us simply take it as a basic principle that if there is a natural number with a property, then there is a smallest such number with that property.

A geometric proof

Let us now give a second proof of the irrationality of √2, one with geometric character, due to Stanley Tennenbaum. Mathematicians have found dozens of different proofs of this classic result, many of them exhibiting a fundamentally different character from what we saw above.

A geometric proof of the theorem. If √2 is rational p / q, then as before, we see that

which means that some integer square has the same area as two copies of another smaller integer square.

We may choose these squares to have the smallest possible integer sides so as to realize this feature. Let us arrange the two medium squares overlapping inside the larger square, as shown here.

Since the large blue square had the same area as the two medium gold squares, it follows that the small orange central square of overlap must be exactly balanced by the two smaller uncovered blue squares in the corners. That is, the area of overlap is exactly the same as the area of the two uncovered blue corner spaces. Let us pull these smaller squares out of the figure to illustrate this relation as follows.

Notice that the squares in this smaller instance also have integer-length sides, since their lengths arise as differences in the side lengths of previous squares. So we have found a strictly smaller integer square that is the sum of another integer square with itself, contradicting our assumption that the original square was the smallest such instance. ☐

Generalizations to other roots

Let us generalize the theorem in a few ways. For example, an essentially similar argument to our original proof works also for the cube root of 2.

Theorem. The number ∛2 is irrational.

Proof. We can basically follow our first argument for √2. Namely, suppose that

in lowest terms. By cubing both sides, we conclude that

So p³ is even and therefore p is even, since odd × odd × odd is odd. So p = 2k for some k, and therefore

Dividing by 2, we conclude that q³ = 4k³, and so q³ is even and therefore q is even, contrary to our assumption that p / q is in lowest terms. ☐

This argument works essentially the same with

and so on, as the reader will verify in the exercises. Another way to generalize the result is to consider the square roots of numbers other than 2. For example:

Theorem. The number √3 is irrational.

Proof. Suppose that √3 = p / q in lowest terms. So 3q² = p². So p² is a multiple of 3. This implies that p is a multiple of 3, since otherwise it would not arise in the prime factorization of p². So p = 3k for some integer k; therefore,

and so q² = 3k². Thus, q² is a multiple of 3, and so q is a multiple of 3. This contradicts our assumption that p / q was in lowest terms. ☐

This proof relied on the existence and uniqueness of prime factorizations, a familiar fact, which will also be covered in detail in a later chapter. In the exercises, the reader is asked to prove further generalizations. Some generalizations can be handled as a consequence of what we have already done. A corollary is a theorem-like statement having an easy proof as a consequence of an earlier theorem.

Corollary. The number √18 is irrational.

Proof. Notice that

If 3√2 = p / q were rational, then √2 = p / (3q), and so √2 would also be rational, which it is not. So √18 cannot be rational. ☐

Let us also prove this corollary directly, without relying on the earlier theorem.

Alternative direct proof of the corollary. Assume toward contradiction that √18 is rational. So √18 = p / q for some integers p and q, and we may assume that p / q is in lowest terms. Squaring both sides and simplifying, we see that

and so p² is even. So p = 2k for some integer k; consequently,

and so

So 9q² is even, but since 9 is odd, it must be that q² is even and hence also that q is even. So both p and q are even, contrary to our assumption that p / q was in lowest terms. So √18 cannot be rational. ☐

All the theorems and corollaries in this chapter are unified by a grand universal theorem, asserting that

is irrational unless n is itself a perfect integer kth power, meaning that n = r^k for some integer r. This is equivalent to saying that all the exponents in the prime factorization of n are multiples of k.

Meanwhile, see my musical proof of the classic theorem on √2, in collaboration with Hannah Hoffman:

Mathematical Habits

• State claims explicitly. Do not allow ambiguity in your mathematical claims and theorem statements. Distinguish between similar but inequivalent statements. Formulate your claim to state exactly what you intend.

• Know exactly what you are trying to prove. Before embarking on an argument or proof, get completely clear about the meaning and content of the claim that is to be proved.

• Insist on proof. Be prepared to prove essentially every mathematical statement you make. When challenged, be prepared to give further and more detailed justification. Do not make assertions that you cannot back up. Instead, make weaker statements that you can prove. Some of the exercises ask you to “prove your answer,” but this is redundant, because, of course, you should always prove your answers in mathematical exercises.

• Try proof by contradiction. When trying to prove a statement, imagine what it would be like if the statement were false. Write, “Suppose toward contradiction that the statement is false, ...” and then try to derive a contradiction from your assumption. If you succeed, then you will have proved that the statement must be true.

• Try to prove a stronger result. Sometimes a difficult or confusing theorem can be proved by aiming at the outset to prove a much stronger result. One overcomes a distracting hypothesis simply by dispensing with it, by realizing it as a distraction, an irrelevant restriction. In the fully general case, one sometimes finds one's way through with a general argument.

• Generalize. After proving a statement, seek to prove a more general statement. Weaken the hypothesis or strengthen the conclusion. Apply the idea of the argument in another similar-enough circumstance. Unify our understanding of diverse situations. Seek the essence of a phenomenon.

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Exercises

1. Prove that the square of any odd number is odd. (Assume that a positive integer is odd if and only if it has the form 2k+1.)

2. Master several alternative proofs of the irrationality of √2, such as those available at http://www.cut-the-knot.org/proofs/sq_root.shtml, and present one of the proofs to the class.

3. Prove that

   is irrational. Give a direct argument, but kindly also deduce it as a corollary of theorem on √2.

4. Prove that

   is irrational for every integer m ≥ 2.

5. Prove that √5 and √7 are irrational. Prove that √p is irrational, whenever p is prime.

6. Prove that √20 is irrational as a corollary of the fact that √5 is irrational.

7. Prove that √(2m) is irrational, whenever m is odd.

8. In the geometric proof that √2 is irrational, what are the side lengths of the smaller squares that arise in the proof? Using those expressions and some elementary algebra, construct a new algebraic proof that √2 is irrational. [Hint: Assume that p² = q² + q² and that this is the smallest instance in the positive integers. Now consider 2q - p and p - q.]

9. Criticize this “proof.” Claim. √n is irrational for every natural number n. Proof. Suppose toward contradiction that √n = p / q in lowest terms. Square both sides to conclude that nq² = p². So p² is a multiple of n, and therefore p is a multiple of n. So p = nk for some k. So nq² = (nk)² = n²k², and therefore q² = nk². So q² is a multiple of n, and therefore q is a multiple of n, contrary to the assumption that p / q is in lowest terms. ☐

10. Criticize this “proof.” Claim. √14 is irrational. Proof. We know that √14 = √2·√7, and we also know that √2 and √7 are each irrational, since 2 and 7 are prime. Thus, √14 is the product of two irrational numbers and therefore irrational. ☐

11. For which natural numbers n is √n irrational? Prove your answer. [Hint: Consider the prime factorization of n, and consider especially the exponents of the primes in that prime factorization.]

12. Prove the unifying theorem mentioned at the end the chapter, namely, that

    is irrational unless n is itself an integer kth power.

13. Prove that the irrational real numbers are exactly those real numbers that are a different distance from every rational number. Is it also true if you swap “rational” and “irrational”?

A full discussion of the exercises, their solutions (often by multiple distinct methods), and the further ideas to which they lead can be found the supplemental volume: