# Ask Uncle Colin: A Bogus Proof

Dear Uncle Colin,

I’ve been shown a proof that goes like this:

To show: $a^n = 1$ for all non-negative integers $n$ and for all non-zero real numbers $a$.

Proceed by induction.

Base case: $a^0 = 1$ by definition, so the base case holds.

Inductive step: Suppose $a^j = 1$ for all integers $j$ such that $0 \le j \le k$. Consider $a^{k+1} = \frac{\left(a^k\right)\left(a^k\right)}{a^{k-1}} = \frac{1 \times 1}{1}$ So the inductive step holds.

Conclusion: Since $a^0 = 1$ and if $a^j = 1$ for all $j \le k$, then $a^{k+1} = 1$, $a^n = 1$ for all non-negative integers $n$ $\blacksquare$That’s clearly not true, but it looks fine. What am I missing?

Believable, Obviously Gravely Unsound Semantics

Hi, BOGUS, and thanks for your message! I’m a big fan of bogus proofs – they’re an excellent way to test one’s understanding of The Rules.

One of the ways to check where a proof has gone wrong is to try it with specific numbers rather than general ones - so we might take $a=2$ and see whether the arguments hold up.

The base case holds just fine; it’s the induction step that’s the problem.

We can suppose $a^j = 1$ for all non-negative integers $j \le k$ - in fact, we know that’s true when $k=0$, so let’s examine that case specifically - it’s a case that satisfies the premises, but we think ought to break the conclusion.

We’re to consider $a^{k+1}$, which in this case is $2^1$. (We know it’s 2! The argument says it’s 1.) We’re told we can write that as $2^1 = \frac{\left(2^0\right)\left(2^0\right)}{2^{-1}}$.

Aha! Here’s the problem: we’ve got $2^{-1}$ on the bottom, and -1 is *not* a non-negative integer. In terms of the proof, we haven’t said anything about $2^{-1}$ being 1 - or (going back to the more general case) $a^{-1}$ being 1.

The first inductive step, which is supposed to take us from $k=0$ to $k=1$, is flawed, because it involves $a^{-1}$, which we’ve made no assumption about – and in particular, we haven’t assumed that it’s 1.

Hope that helps!

- Uncle Colin