1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations


    3. 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.

  2. jcsd
  3. Sep 11, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
  4. Sep 11, 2014 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    To amplify Peroks' response- you are not asked to show that a and b must be positive, only that they can be positive.
  5. Sep 11, 2014 #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?
  6. Sep 11, 2014 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
  7. Sep 11, 2014 #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!
  8. Sep 11, 2014 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I just realised that the term you're looking for is wlog, "without loss of generality". A very useful concept!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

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