Can I use proof where I assume the statement to be true and then arrive at a known fact.

Question

Prove x/y + y/x ≥ 2.

Now the book gives (x-y)² ≥0 and from there derived the above, hence proving it. I understand that, because you start at a known fact and end up at the thing you are trying to prove. I have omitted the algebra steps for convenience.

I also proved it by contradiction, so I stated that x/y + y/x < 2 and this then led to the contradiction (x-y)² <0.

 

My question is this; is it acceptable to start the proof as follows.

Assume the statement is true.

x/y + y/x ≥ 2

⇔x² + y² -2xy ≥ 0

⇔(x-y)² ≥0

The final statement is known to be always true, therefore the initial assumption must have also been true, hence proven.

Answer 1

First of all, let's look at the "proof".

 

(x-y)² ≥ 0

 

x² + y² ≥ 2xy

 

Now comes the critical step: divide by xy (assume xy≠0)

 

(x/y) + y/x ≥ 2

 

Now the problem: suppose x = 1 and y = -1

When you do the division, you change the direction of the inequality and the original statement is no longer true!

So when you do the division you must have xy>0!

 

Then on to your final question.  This is an opinion.

I do not think that you can assume something to be true and derive something true based on that assumption and then claim that the assumption is true.   However, if you can derive something known to be true from an assumption, I would expect that you could go the other way around, i.e. start with something you know to be true and derive the thing which you wish to prove.