# How to convert a “backwards” proof into a “forwards” proof

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.

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Filed under Abstract algebra, Geometry, Math, Problem Solving