A Short Number Theory Joke

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A common exercise in introductory discrete math courses is the proof that \sqrt{2} is irrational. The general structure of the proof is to assume (for the sake of contradiction) that we can write \sqrt{2} as a fraction in reduced form. That is,

\displaystyle{\sqrt{2} = \frac{a}{b}}   where   a, b \in \mathbb{N}   and   \mathrm{gcd}(a, b) = 1

Rewrite as 2b^2 = a^2 and from here easily show that a and b must both be even, contradicting our assumption that they share no common factor.

Suppose that instead we were tasked with showing that \sqrt[3]{2} is irrational. Let’s see what happens if we try the same approach. Suppose

\displaystyle{\sqrt[3]{2} = \frac{a}{b}}     where     a, b \in \mathbb{N}     and     \mathrm{gcd}(a, b) = 1

Rewrite as 2b^3 = a^3. From here one can proceed in the same manner as above and show that a and b must both be even. However, a particularly cheeky student might rewrite the equation again as

    \[b^3 + b^3 = a^3\]

and simply apply Fermat’s Last Theorem to conclude that no positive integer solutions exist. 😛

This is of course circular reasoning and totally not a valid proof, but it is hilarious. Just for fun, lets try to write this argument with more generality. First, observe that for any k \ge 3, the same trick works to show that \sqrt[k]{2} is irrational since FLT applies for cubes, 4th powers, 5th powers, etc. What about for radicands other than 2? For 9, it also works with a just a little bit more effort:

    \begin{align*} \sqrt[3]{9} &= \frac{a}{b} \\ 9b^3 &= a^3 \\ b^3 + 8b^3 &= a^3 \\ b^3 + (2b)^3 &= a^3 \end{align*}

We can apply FLT here because 9 = 1^3 + 2^3 can be written as the sum of two cubes. In general, the trick works for \sqrt[k]{n} if we can write n as

    \[n = r^k + s^k\]

for positive integers r and s. For the first few values of n and k, this looks like

n\diagdownk3456789
2✓✓✓✓✓✓✓
9✓
16✓
17✓
32✓
33✓
35✓
54✓
64✓
65✓✓