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!

In proof of SQRT(2) is irrational, why can't a,b both be even

  1. Sep 14, 2014 #1
    In the classic proof of irrationality of SQRT(2) we assume that it can be represented by a rational number,a/b where a, b are integers. This assumption after a few mathematical steps leads to a contradiction, namely that both a, b are even numbers.
    Why is that a contradiction?
    Well you can say that the rational fraction has to be in its lowest terms;therefore either a or b or both must be odd.
    However, that wasn't in the assumption(lowest terms). The assumption was just two integers a,b.
    Why can't they both be even?
     
  2. jcsd
  3. Sep 14, 2014 #2
    In the correct version of the classic proof, it is assumed that a and b are not both even.
     
  4. Sep 14, 2014 #3

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    And the reason you want to make the assumption that both are even right at the start is...
     
  5. Sep 15, 2014 #4
    The proof I use starts by assuming a and b are integers with no common factors i.e. a/b is in its lowest terms. Then you end up with a loop in which a and b always have a common factor of 2 - but this is impossible as the numerator and denominator would get smaller without limit. An impossibility for positive integers.
     
  6. Sep 15, 2014 #5

    pwsnafu

    User Avatar
    Science Advisor

    If a and b are both even, then let ##a=2c## and ##b=2d##. Repeat the argument on c and d, so we conclude they too are even. Let ##c=2e## and ##d=2f##....which are also even. And so on. This argument should keep going forever.

    But here's the thing. The natural numbers have a minimum (namely 1). So we can't keep halving forever. At some point this process must terminate. This contradicts our original argument.
     
  7. Sep 15, 2014 #6
    Thanks for the help
     
  8. Sep 17, 2014 #7

    CKH

    User Avatar

    Another proof:

    In ##2 = (a/b)^2##, assume that ##a## and ##b## have been reduced to have no common factors.
    Then ##2b^2=a^2##, so ##a## must be even (contain a factor 2).
    Thus ##a=2c## for some ##c##
    Thus ##2b^2=4c^2##
    Thus ##b^2=2c^2##, so now ##b## must be even also.

    This contradicts the assumption that ##a## and ##b## have no common factor, so no such ##a## and ##b## can exist.
     
  9. Sep 21, 2014 #8

    lurflurf

    User Avatar
    Homework Helper

    Just do this

    Suppose $$\sqrt{2}=a/b$$
    where
    a=2^m c
    b=2^n d
    c and d are odd integers
    then
    $$
    \sqrt{2}=a/b \\
    \sqrt{2}b=a \\
    2b^2=a^2 \\
    2(2^n d)^2=(2^m c)^2 \\
    2^{2n+1} d^2=2^{2m} c^2 \\
    $$
    thus
    2n+1=2m
    2n+1 is odd while 2m is even
    an even number cannot equal an odd number
    contradiction
     
    Last edited: Sep 21, 2014
  10. Sep 21, 2014 #9

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    If you have never seen Vi Hart's videos, here is an amusing one on Pythagoras and the proof that 2 is irrational.

    P.S. While you are there, check out her Mobius strip story of "Wind and Mr Ug"
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: In proof of SQRT(2) is irrational, why can't a,b both be even
Loading...