Dave Richeson at Division By Zero wrote recently about a “proof technique” for proving equalities or inequalities that is far too common: Starting with the equality to be proven and working backwards to end at a true statement. This is a technique that is almost a valid way to prove things, but it contains — and engenders — serious flaws in logic and the concept of proof that can really get students into trouble later on.

I left a comment there that spells out my feelings about why this technique is bad. What I wanted to focus on here is something I also mentioned in the comments, which was that it’s so easy to take a “backwards” proof and turn it into a “forwards” one that there’s no reason not to do it.

Take the following problem: Prove that, for all natural numbers \(n\),

\(1 + 2 + 2^2 + \cdots + 2^n = 2^{n+1} – 1\)

This is a standard exercise using mathematical induction. The induction step is trivial; focus on the induction step. Here we assume that \(1 + 2 + 2^2 + \cdots + 2^k = 2^{k+1} – 1\) for all natural numbers less than or equal to \(k\) and then prove:

\(1 + 2 + 2^2 + \cdots + 2^{k+1} = 2^{k+2} – 1\)

Here we have to prove two expressions are equal. Here’s what the typical “backwards” proof would look like (click to enlarge):

A student may well come up with this as his/her proof. It’s not a bad *initial draft* of a proof. Everything we need to make a totally correct proof is here. But the backwards-ness of it — all stemming from the first line, where we have assumed what we are trying to prove — needs fixing. Here’s how.

First, note that all the important and correct mathematical steps are taking place on the left-hand sides of the equations, and the right-hand sides are the problem here. So** delete all the right-hand sides of the equals signs and the final equals sign**.

Next, since the problem with the original proof was that we started with an “equation” that was not known to be true, **eliminate any step that involved doing something to both sides**. That would be line 4 in this proof. This might involve some re-working of the steps, in this case the trivial task of re-introducing a -1 in the final steps:

You could reverse these first two steps — eliminate all “both sides” actions and then get rid of the left-hand sides.

Then, we need to make it look nice. So for n = 1 to the end, **move the \((n+1)^\mathrm{st}\) left-hand side and justification to the \(n^\mathrm{th}\) right-hand side**:

Now we have a correct proof that does not start by assuming the conclusion. It’s shorter, too. Really the main thing wrong with the “backwards” proof is the repeated — and, notice, unnecessary — assertion that everything is equal to the final expression. Remove that assertion and the correct “forwards” proof is basically right there looking at you.