1. Not finding help here? Sign up for a free 30min 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 root 3 is irrational

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that there is no rational x such that x2=3

    2. The attempt at a solution

    Suppose that there is a rational x=[itex]\frac{a}{b}[/itex]=[itex]\sqrt{3}[/itex] and that the fraction is fully simplified. (ie. a and b have no common factor)

    Then a2/b2=3 which means a2=b2.3 and it follows that a is a factor of 3 and be written a=3.k (k is an integer). Therefore a2/b2 = 9.k2/b2=3. Therefore b2=3.k2.

    This means both a and b are multiples of 3 and this contradicts our original assumption that the fraction is fully simplified.

    3. Question

    As far as I can tell the proof is sound but if you replace 3 with 4 the same logic holds and this means there is no rational number x such that x2=4 which is obviously wrong since 2 does. So what is wrong with this proof.
     
  2. jcsd
  3. Aug 19, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, the "same logic" does not work with 4 because 3 is a prime number and 4 is not. 6 is not a multiple of 4 but [itex]6^2= 36= 4(9)[/itex] is. That's because 6(6)= (2(3))(2(3))= (2)(2)(3)(3)= 4(9). You can't do that with 3.

    More formally, every integer is of the form 3k or 3k+1 or 3k+ 2 for some integer k. [itex](3k)^2= 9k^2= 3(3k^2)[/itex] is a multiple of 3. [itex](3k+1)^2= 9k^2+ 6k+ 2= 3(3k^2+ 2k)+ 1[/itex] is not a multiple of 3. [itex](3k+2)^2= 9k^2+ 12k+ 4= 3(3k^2+ 4k+ 1)+ 1[/itex] is not a multiple of 3. That is, for any integer, n, [itex]n^2[/itex] is a multiple of 3 if and only if n is a multiple of 3.

    That does NOT work with 4. Every integer is of the form 4k, or 4k+1, or 4k+ 2, or 4k+ 3. BOTH [itex](4k)^2= 16k^2= 4(4k^2)[/itex] and [itex](4k+ 2)= 16k^2+ 16k+ 4= 4(4k^2+ 4k+ 1)[/itex] are multiples of 4.
     
    Last edited: Aug 19, 2012
  4. Aug 19, 2012 #3
    Your proof uses your previous result, that if [itex]p^2[/itex] is a multiple of 3, then [itex]p[/itex] also is. This is only true for prime numbers.
     
  5. Aug 19, 2012 #4
    Thanks for clearing it up that makes perfect sense
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Prove root 3 is irrational
Loading...