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## Homework Statement

Prove that √2 is irrational as follows. Assume for a contradiction that there exist integers a, b with b nonzero such that (a/b)

^{2}=2.

1. Show that we may assume a, b>0.

2. Observe that if such an expression exists, then there must be one in which b is as small as possible.

3. Show that [itex]\left(\frac{2b-a}{a-b}\right)^2=2[/itex].

4. Show that 2b-a>0, a-b>0.

5. Show that a-b<b, a contradiction.

This problem is from Galois Theory, Third Edition by Ian Stewart.

My question is how can we show that a, b>0? Assuming that a and b are positive I've completed steps 2. through 5. Now all that I have left to do is show a and b are positive, which seems like it should be the simplest part. Nonetheless, I'm unclear how to do it.

## Homework Equations

None

## The Attempt at a Solution

Here's my line of thought so far. I'm not sure if it's correct.

First note that √2>1.

Now consider the possibilities for the signs of a and b. If exactly one of a or b is negative, then a/b<0. But we have (a/b)

^{2}=2, which would imply √2=a/b<0, which is false. If both a and b are positive or both a and b are negative then a/b>0, so (a/b)

^{2}=2 implies √2=a/b>0, which is true. It is therefore not valid to treat the case in which exactly one of a or b is negative, while it is valid to treat either the case where a, b<0 or when a, b>0 since they are equivalent.

I'm very unsure of what I have written up there. Any help or guidance is much appreciated.

Thanks!