Tag Archives: proof by contradiction

Just Beatty It

Algebra, Proofs, , , , , ,

[This is the 6th post in the current series about Wythoff’s game: see posts #1, #2, #3, #4, and #5. Caveat lector: this post is a bit more difficult than usual. Let me know what you think in the comments!]

Our only remaining task from last week was to prove the mysterious Covering Theorem: we must show that there is exactly one dot in each row and column of the grid (we already covered the diagonal case). Since the rows and columns are symmetric, let’s focus on columns.

The columns really only care about the x-coordinates of the points, so let’s draw just these x-coordinates on the number-line. We’ve drawn \(\phi,2\phi,3\phi,\ldots\) with small dots and \(\phi^2,2\phi^2,3\phi^2,\ldots\) with large dots. We need to show that there’s exactly one dot between 1 and 2, precisely one dot between 2 and 3, just one between 3 and 4, and so on down the line. For terminology’s sake, break the number line into length-1 intervals [1,2], [2,3], [3,4], etc., so we must show that each interval has one and only one dot:

Multiples of phi and phi^2 on a number line
Multiples of \(\phi\) (small blue dots) and of \(\phi^2\) (large green dots) perfectly interleave the integers on the number line. This is proved below.

Why is this true? One explanation hinges on a nice geometric observation: Take any small dot s and large dot t on our number-line above, and cut segment st into two parts in the ratio \(1:\phi\) (with s on the shorter side). Then the point where we cut is always an integer! For example, the upper-left segment in the diagram below has endpoints at \(s=2\cdot\phi\) and \(t=1\cdot\phi^2\), and its cutting point is the integer 3:

Splitting blue and green dots at integers
When a segment formed by a small and a large dot is cut into a \(1:\phi\) ratio (closer to the small dot), the cutting point is always an integer. In fact, the segment between \(j\cdot \phi\) and \(k\cdot \phi^2\) is cut at the integer j+k.

In general, if s is the jth small dot—i.e., \(s=j\cdot\phi\)—and \(t=k\cdot\phi^2\) is the kth large dot, then the cutting point between s and t is \(\frac{1}{\phi}\cdot s+\frac{1}{\phi^2}\cdot t = j+k\) (Why?![1]). But more importantly, this observation shows that no interval has two or more dots: a small dot and a large dot can’t be in the same interval because they always have an integer between them![2]

So all we have to do now is prove that no interval is empty: for each integer n, some dot lies in the interval [n,n+1]. We will prove this by contradiction. What happens if no dot hits this interval? Then the sequence \(\phi,2\phi,3\phi,\ldots\) jumps over the interval, i.e., for some j, the jth dot in the sequence is less than n but the (j+1)st is greater than n+1. Likewise, the sequence \(\phi^2,2\phi^2,3\phi^2,\ldots\) jumps over the interval: its kth dot is less than n while its (k+1)st dot is greater than n+1:

Showing no interval is empty
Illustrating the hypothetical situation where interval [n, n+1] contains no dot. This is used in a proof by contradiction to show that the interval in fact cannot be empty.

By our observation above on segment \(s=j\phi\) and \(t=k\phi^2\), we find that the integer j+k is less than n, so \(j+k\le n-1\). Similarly, \(j+k+2 > n+1\), so \(j+k+2 \ge n+2\). But together these inequalities say that \(n\le j+k\le n-1\), which is clearly absurd! This is the contradiction we were hoping for, so the interval [n,n+1] is in fact not empty. This completes our proof of the Covering Theorem and the Wythoff formula!

It was a long journey, but we’ve finally seen exactly why the Wythoff losing positions are arranged as they are. Thank you for following me through this!

A Few Words on the Column Covering Theorem

Using the floor function \(\lfloor x\rfloor\) that rounds x down to the nearest integer, we can restate the Column Covering Theorem in perhaps a more natural context. The sequence of integers $$\lfloor\phi\rfloor = 1, \lfloor 2\phi\rfloor = 3, \lfloor 3\phi\rfloor = 4, \lfloor 4\phi\rfloor = 6, \ldots$$ is called the Beatty sequence for the number \(\phi\), and similarly, $$\lfloor\phi^2\rfloor = 2, \lfloor 2\phi^2\rfloor = 5, \lfloor 3\phi^2\rfloor = 7, \lfloor 4\phi^2\rfloor = 8,\ldots$$ is the Beatty sequence for \(\phi^2\). Today we proved that these two sequence are complementary, i.e., together they contain each positive integer exactly once. We seemed to use very specific properties of the numbers \(\phi\) and \(\phi^2\), but in fact, a much more general theorem is true:

Beatty’s Theorem: If \(\alpha\) and \(\beta\) are any positive irrational numbers with \(\frac{1}{\alpha}+\frac{1}{\beta}=1\), then their Beatty sequences \(\lfloor\alpha\rfloor, \lfloor 2\alpha\rfloor, \lfloor 3\alpha\rfloor,\ldots\) and \(\lfloor\beta\rfloor, \lfloor 2\beta\rfloor, \lfloor 3\beta\rfloor,\ldots\) are complementary sequences.

Furthermore, our same argument—using \(\alpha\) and \(\beta\) instead of \(\phi\) and \(\phi^2\)—can be used to prove the more general Beatty’s Theorem!

Notes

  1. Hint: use the identity \(\frac{1}{\phi}+\frac{1}{\phi^2}=1\). [↩]
  2. To be thorough, we should also check that no interval has two small or two large dots. Why can’t this happen? [↩]

Logic Under Construction

Algebra, Math Foundations, Proofs, ,

In last week’s discussion of proofs by contradiction and nonconstructive proofs, we showed:

Theorem: There exist irrational numbers \(x\) and \(y\) with the property that \(x^y\) is rational.

However, our proof was nonconstructive: it did not pinpoint explicit values for \(x\) and \(y\) that satisfy the condition, instead proving only that such numbers must exist. Would a more constructive proof be more satisfying? Let’s see! I claim \(x=\sqrt{2}\) and \(y=\log_2 9\) work, because \(\sqrt{2}\) we already know to be irrational, \(y=\log_2 9\) can be similarly proved to be irrational (try this!), and $$x^y = \sqrt{2}^{\log_2 9} = \sqrt{2}^{\log_{\sqrt{2}}3}=3,$$ which is rational.

Let’s further discuss why last week’s proof was less satisfying. The following rephrasing of this proof may help shed some light on the situation:

Proof: Assume the theorem were false, so that any time \(x\) and \(y\) were irrational, \(x^y\) would also be irrational. This would imply that \(\sqrt{2}^{\sqrt{2}}\) would be irrational, and by applying our assumption again, \(\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}\) would also be irrational. But this last number equals 2, which is rational. This contradiction disproves our assumption and thereby proves the theorem, QED.

So perhaps this argument seems less satisfactory simply because it is, at its core, a proof by contradiction. It does not give us evidence for the positive statement “\(x\) and \(y\) exist”, but instead only for the negative statement “\(x\) and \(y\) don’t not exist.” (Note the double negative.) This distinction is subtle, but a similar phenomenon can be found in the English language: the double negative “not bad” does not mean “good” but instead occupies a hazy middle-ground between the two extremes. And even though we don’t usually think of such a middle-ground existing between logic’s “true” and “false”, proofs by contradiction fit naturally into this haze. In fact, these ideas motivate a whole branch of mathematical logic called Constructive logic that disallows double negatives and proofs by contradiction, instead requiring concrete, constructive justifications for all statements.

But wait; last week’s proof that \(\sqrt{2}\) is irrational used contradiction, and therefore is not acceptable in constructive logic. Can we prove this statement constructively? We must show that \(\sqrt{2}\) is not equal to any rational number; what does it even mean to do this constructively? First, we turn it into a positive statement: we must show that \(\sqrt{2}\) is unequal to every rational number. And how do we constructively prove that two numbers are unequal? By showing that they are measurably far apart. So, here is a sketch of a constructive proof: \(\sqrt{2}\) is unequal to every rational number \(a/b\) because $$\left|\sqrt{2} – \frac{a}{b}\right| \ge \frac{1}{3b^2}.$$ See if you can verify this inequality![1]

PS. In case you are still wondering whether \(\sqrt{2}^{\sqrt{2}}\) is rational or irrational: It is irrational (moreover, transcendental), but the only proof that I know uses a very difficult theorem of Gelfond and Schneider.

Notes

  1. This inequality would also have to be proven in a constructive manner. See these Wikipedia articles for more information: Intuitionistic logic (another name for Constructive logic) and Square root of 2: Constructive proof. [↩]

Methods of Irrationality

Algebra, Math Foundations, Proofs, ,

Being a mathematician requires you to think in strange ways.

For starters, you might think that mathematicians spend all day pondering purely theoretical things that only exist in Mathematical abstraction, right? Well, it can get even stranger than that: sometimes we have to think about things that don’t exist, even in the abstraction! It’s called a proof by contradiction, and here’s an example:

Theorem: The number \(\sqrt{2}\) is irrational.

Proof: To prove that \(\sqrt{2}\) is irrational, we have to show that we can never write it as a ratio of integers, \(\sqrt{2}=a/b\). So let’s assume we can find such a fraction and see where it leads us.

Let’s cancel common factors in the numerator and denominator so that \(a/b\) is in lowest terms. Squaring our equation shows that \(a^2=2b^2\), so \(a^2\) is even and so \(a\) itself is even. Since \(a/b\) is in lowest terms, \(b\) must be odd.

Since \(a\) is even, \(a^2\) is in fact divisible by 4. On the other hand, since \(b\) is odd, \(2b^2\) is not divisible by 4. But then the integer \(a^2=2b^2\) is both divisible by 4 and not divisible by 4, which is absurd! The only explanation for this contradiction is that our original assumption—that \(\sqrt{2}\) is rational—is false. So we’re done with the proof! Notice that we spent most of our effort reasoning about integers that never existed (specifically \(a\) and \(b\)).

But that’s not the only twisted thing about Mathematical thinking. Here’s a delightfully short yet aggravatingly unsatisfying proof:

Theorem: There exist irrational numbers \(x\) and \(y\) such that \(x^y\) is rational.

Proof: We already know that \(\sqrt{2}\) is irrational, so maybe we can use \(\sqrt{2}^{\sqrt{2}}\). Is this number rational? If it is, then we’re done: \(x=\sqrt{2},y=\sqrt{2}\) solves the problem. But what if \(\sqrt{2}^{\sqrt{2}}\) is irrational? In this case, I claim \(x=\sqrt{2}^{\sqrt{2}},y=\sqrt{2}\) works: indeed, $$\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot\sqrt{2}} = \sqrt{2}^2 = 2,$$ which is rational. In either case the required numbers \(x\) and \(y\) exist, so this completes the proof.

But wait; which is it? The proof shows us that one of the pairs \(x=\sqrt{2},y=\sqrt{2}\) or \(x=\sqrt{2}^{\sqrt{2}},y=\sqrt{2}\) works, but it doesn’t tell us which one! So, have we proven the theorem? Yes, technically, but only nonconstructively.

Frustrating, isn’t it?