# Prove that sqrt(2) is irrational using a specific technique

## 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 $\left(\frac{2b-a}{a-b}\right)^2=2$.
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.

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!

PeroK
Homework Helper
Gold Member
2020 Award
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!
First, there is no way to prove that a and b are positive. The question is why can you assume a and b are positive. You've done most of the work. All you have to do is say what to do if a, b are both negative.

HallsofIvy
Homework Helper
To amplify Peroks' response- you are not asked to show that a and b must be positive, only that they can be positive.

Thanks for the help PeroK and Hallsoflvy. So to establish that a and b can be positive can I simply comment that if (a/b)2=2, then it is also true that (-a/b)2=2 and that (-a/(-b))2=2, so it doesn't matter whether we consider the case in which a and b are both positive, both negative, or in which one is positive and one is negative since all three cases are equivalent?

PeroK
Homework Helper
Gold Member
2020 Award
Thanks for the help PeroK and Hallsoflvy. So to establish that a and b can be positive can I simply comment that if (a/b)2=2, then it is also true that (-a/b)2=2 and that (-a/(-b))2=2, so it doesn't matter whether we consider the case in which a and b are both positive, both negative, or in which one is positive and one is negative since all three cases are equivalent?

That's nearly it, although I still think you may be missing the key point. If there were such integers a' and b', then there would be positive integers a and b with the same property.

If you show that no such positive integers can exist, then no such integers can exist.

In this case, you could take a = |a'| and b = |b'|.

But, it's enough to say that if there are integers with the property a/b = √2, then there are positive integers a/b with this property. That's essentially the point.

Ah, that makes sense! I think I've got it now and can safely work on my solution, but if I run into any more questions I'll make another post. Thanks again for your help!

PeroK