# How do you work backwards when solving word problems?

Oct 24, 2014

I am not certain the following is a fully satisfactory answer to this question, but it explains the logic behind logical transformations from something we want to prove to an obviously true statement and back.

"Working backwards" is the typical final part of a proof when from a statement we want to prove, for instance ${x}^{2} - 4 x + 6 > 1$, we derive an obviously true statement using some invariant (equivalent) transformations like these:
${x}^{2} - 4 x + 6 = {\left(x - 2\right)}^{2} + 2$, which is greater or equal to 2 (since a square of any algebraic expression is greater or equal to $0$), which, in turn, is greater than $1$.

The problem with the above "proof" is that, if the initial statement was false, using seemingly correct transformations, we can come up to an obviously true statement. So, the fact that from our original statement we have derived the obviously true final statement does not necessarily prove that our initial statement was true.

But, if all transformations we made are not only "correct", but invariant (or equivalent), which, in short, means reversible, then after we have derived a true statement we can conclude:
since all transformations are invariant (that is reversible), from the final true statement we can derive the initial, and that is the actual proof.
This is actually the "working back" part of a proof.

For the example above the real proof is the following sequence:
${\left(x - 2\right)}^{2} \ge 0$ - add 2 to both sides -
${\left(x - 2\right)}^{2} + 2 \ge 2 > 1$ - open parenthesis -
${x}^{2} - 4 x + 4 + 2 > 1$ - simplify -
${x}^{2} - 4 x + 6 > 1$ - which is what we had to proof.