Proof by going the reverse way

Question

Proving a mathematical statement usually involves coming to the required result from a known/given result(s). Am I allowed to do this the other way round i.e. coming to a known result from the required statement? Here is a simple example:

Proof of AM-GM inequality for two numbers:

$ (\sqrt x - \sqrt y)^2 ≥ 0 $

$\implies x + y - 2\sqrt{xy} ≥ 0 $

$\implies \dfrac{x + y}2 ≥ \sqrt{xy} $

My presentation:

If $ \dfrac{x + y}2 ≥ \sqrt xy $

$\iff x + y ≥ 2\sqrt{xy} $

$\iff x + y - 2\sqrt{xy} ≥ 0$

$ \iff (\sqrt x - \sqrt y)^2 ≥ 0 $ , which is true.

$\therefore (\sqrt x - \sqrt y)^2 ≥ 0 \implies \dfrac{x + y}2 ≥ \sqrt xy$

Is the second method considered a valid method of proof?

I know the second proof is just the reverse of the former but in some cases, it seems easier to go from the required statement to the given.

Answer 1

Your presentation is good, in fact almost perfect, however you should simply omit the initial word "If" which serves no purpose. The point of your presentation is that you are starting from the statement to be proved, then applying laws of algebra to find a sequence of equivalent statements, and then observing that the final statement of this sequence is true, hence all of them are true.

Answer 2

Doing the proof in reverse is perfectly fine and you can even write your proof in the following way:

$$ \dfrac{x + y}2 ≥ \sqrt xy ~~\Longleftarrow~~ x + y ≥ 2\sqrt{xy} ~~\Longleftarrow~~ x + y - 2\sqrt{xy} ≥ 0 ~~\Longleftarrow~~ (\sqrt x - \sqrt y)^2 ≥ 0 $$

where every "$\Longleftarrow$" should then be read as "is implied by". By noting that the last statement in the row is true, the first one must be too.

I have rarely seen this kind of backwards reasoning in proofs on paper, however it is the standard way proofs are done in most proof assistants.

Answer 3

The second method is what everyone writes down on a bit of scrap paper. Students often write it down as a proof as well, and often forget that a double-arrow is needed, which leads to proofs like this:

To prove $2 = 1$:

$2 = 1$

hence $1 = 2$ as well.

Add these to get $3 = 3$, which is true.

so we're done and it's true!

That pattern has led to a general disdain for such proofs, even when they're valid (like yours, aside from the unnecessary initial "If".). It helps the reader a lot if you start with a preliminary declaration that you're going to prove equivalence to some known thing, so I'd write your proof like this:

I'll show that AM-GM is logically equivalent to the true statement $(\sqrt{x} - \sqrt{y})^2 \ge 0$ for any real numbers $x,y$ $u$, through a sequence of equivalences.

\begin{align} \dfrac{x + y}2 &≥ \sqrt {xy} && \text{equivalent, by multiplying/dividing by $2$, to}\\ x + y &≥ 2\sqrt{xy}&& \text{equivalent, by subtracting/adding $2\sqrt{xy}$ from both sides, to}\\ x - 2\sqrt{xy} + y & >= 0 && \text{equivalent, by factoring/expanding, to}\\ (\sqrt x - \sqrt y)^2 &≥ 0 \end{align}

Because this final statement is true for all nonnegative $x,y$, so is the AM-GM inequality.

But by the time you add all that stuff, including the reason for each double-implication, it's actually harder to read than the "first form" proof. Furthermore, it's generally interesting to prove equivalence between two familiar things, but the second thing, in this case, isn't nearly as interesting in its own right as AM-GM. So a proof that goes from "known but not particularly inspiring basic thing" to "new and widely-useful statement" will make more sense to your reader.

Answer 4

The second argument is (almost) OK because you carefully used "if and only if" ($\iff$) rather than just "implies" ($\implies$).

You don't want to start that argument with "if".

I would recommend the usual straightforward implication even if the argument the other way is how you discovered your proof. The other way is prone to logical error.