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

  • #1

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!
 

Answers and Replies

  • #2
PeroK
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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.
 
  • #3
HallsofIvy
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To amplify Peroks' response- you are not asked to show that a and b must be positive, only that they can be positive.
 
  • #4
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?
 
  • #5
PeroK
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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.
 
  • #6
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!
 
  • #7
PeroK
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I just realised that the term you're looking for is wlog, "without loss of generality". A very useful concept!
 

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