In class we gave the “classic” proof that \(\sqrt{2}\) is irrational, but we also gave a different proof using analysis. For each pair of positive integers \([a,b]\) with \(a/b \in [0,1]\) with \(b > 1\), we considered the interval
\[ I(a,b):= \left[ \frac{a}{b} – \frac{1}{b^{42}},\frac{a}{b} + \frac{1}{b^{42}} \right]\]
around the point \(a/b\). One possible guess is that the union ofthese intervals entirely covers \((0,1)\) (maybe with some overlap). There certainly will be some overlap since \(I(a,b)\) and \(I(ma,mb)\) are both centered on \(a/b = ma/mb\). But if these intervals do cover the entire interval \([0,1]\) then the total length of all these lines should be at least \(1\). But we can compute the length:
- The intervals \(I(1,2)\) and \(I(2,2)\) both have length \(2/2^{42}\).
- The intervals \(I(1,3)\), \(I(2,3)\), and \(I(3,3)\) both have length \(2/3^{42}\).
- The intervals \(I(1,4)\), \(I(2,4)\), \(I(3,4\), and \(I(4,4)\) both have length \(2/4^{42}\).
and so on, so the total length is at least
\[ \begin{aligned} S &:=
\frac{2}{2^{42}} + \frac{2}{2^{42}} + \frac{2}{3^{42}}+ \frac{2}{3^{42}}+ \frac{2}{3^{42}} + \frac{2}{4^{42}} + \ldots \\
& = 2 \left(\underbrace{\frac{1}{2^{42}} + \frac{1}{2^{42}}}_{2} + \underbrace{\frac{1}{3^{42}} + \frac{1}{3^{42}} + \frac{1}{3^{42}}}_{3} + \underbrace{\frac{1}{4^{42}} + \ldots + \frac{1}{4^{42}}}_{4} + \ldots \right) \\
& = 2 \left(\frac{2}{2^{42}} + \frac{3}{3^{42}} + \frac{4}{4^{42}}+ \frac{5}{5^{42}}+ \ldots \right) \\
& = 2 \left(\frac{1}{2^{41}} + \frac{1}{3^{41}} + \frac{1}{4^{41}}+ \frac{1}{5^{41}}+ \ldots \right) \\
& = 0.00000000094947\ldots\end{aligned} \]
(We did not and will not justify the last line) which is less than one. So irrational numbers must exist, and lots of them! This proof even shows that if you take a “random” real number \(\alpha\) then unless are really really unlucky, you must have
\[ \left| \alpha – \frac{p}{q} \right| \ge \frac{1}{q^{42}} \]
for all integers \(p\) and \(q\) with \(q > 1\). But unlike the first proof, this proof doesn’t show that any particular number is irrational. It’s a great feat of humanity that \(\pi\) is irrational, but does it also have this property?
Things to think about:
- Check that
\[ \left|\ \pi \ – \ \frac{22}{7} \right| = \frac{1}{7^{3.429288\ldots}} \] - Would the argument we gave in class work if one replaced \(42\) by another number? Which numbers?
- What is \(\sqrt{2}\) and \(\pi\) anyway?
We know (because we have been told) that \(\pi\) is irrational. In fact this was first proven by Lambert in 1760. It is a much harder (and more recent) theorem of Kurt Mahler from the 1950s that
\[ \left| \pi- \frac{p}{q} \right| \ge \frac{1}{q^{42}} \]
for all \(q > 1\). The optimal value of the exponent is not known!