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Prove sqrt(6) is irrational

  1. Nov 23, 2003 #1
    Alright, heading says it all. This is a nice problem heh.. I can see how to prove sqrt(5) is irrational. I think this method works up to the points where the fact 5 is a prime is used, (ie prime lemma) on 5 which doesn't work so well on 6! hehe

    Was thinking of maybe using product of primes somehow but.. hmm dunno

    Anyway for sqrt(5) went like this (proof by contradiction)

    Prove sqrt(5) is not rational.

    Suppose sqrt(5) = a/b, where a,b E Z+ (suppose sqrt(5) is rational)
    we'll also assume gcd(a,b) = 1 (otherwise just divide a,b by gcd)
    sqrt(5) = a/b <=> 5 = a^2/b^2 or 5*b^2=a^2
    so 5|a^2, but 5 is a prime so 5|a Ea' 5*a'=a so 5b^2=(5a)^2 or 5b^2=5^2*a^2 or b^2 = 5*a'^2 so 5|b^2 by prime lemma 5|b 5|a ^ b|b so 5|gcd(a,b)
    so 5|1 which is nonsense therefore our initial assumption is incorrect and 5 is in fact irrational.

  2. jcsd
  3. Nov 23, 2003 #2
    Suppose sqrt(6) is rational

    gcd(a,b) = 1

    [tex] \sqrt{6} = \frac{a}{b} [/tex]

    [tex] 6b^2 = a^2 [/tex]

    Now there are 4 possibilities.

    1) both a and b are even.
    It cannot be true since gcd(a,b)=1

    2) a is odd and b is even
    It cannot be true too. If a is odd, then left hand side will be even while right hand side will be odd.

    3) both a and b are odd.
    It cannot be true too and the reason is the same as that in (2)

    4) a is even and b is odd.

    Let a = 2m and b = 2n + 1

    [tex] 6b^2 = a^2 [/tex]

    [tex] 6*{(2n+1)}^2 = {2m}^2 [/tex]

    [tex] 12m^2 + 12m +3 = 2n^2 [/tex]

    left hand side is odd while right hand side is even


    Therefore sqrt(6) is irrational
  4. Nov 23, 2003 #3


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    "Let a = 2m and b =2n + 1
    6b2= a2

    6(2n+1)2= 2m2"

    No, it should be 6(2n+1)2= (2m)2

    "12m2+ 12m+ 3= 2n2"

    Now this completely baffles me. Even assuming you accidently switched m and n, it doesn't follow.
    From 6(2n+1)2= (2m)2 you get
    6(4n2+ 4n+ 1)= 4m2 or
    24n2+ 24n+ 6= 4m2 and both sides are clearly even.
  5. Nov 23, 2003 #4
    How about dividing both sides by 2, then we get
    12n2+ 12n+ 3= 2m2
    where left hand side is odd while right hand side is even.

    If we don't divide both sides by 2, left hand side cannot be divisible by 4 while right hand side can.
  6. Nov 23, 2003 #5


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    How about the following (for any n not a square of an integer).

    sqrt(n)=a/b, where a and b are integers with no common factors, and b >1.


    The numerator and denominator still have no common factors, so n is not an integer.
  7. Nov 25, 2003 #6
    any n in what? is n any nonnegative rational number, then? if it's any nonnegative integer n not the square of an integer, it seems odd to end by saying "n is not an integer."
  8. Nov 25, 2003 #7


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    Sorry for the confusion. I assumed we were discussing integers in general. My discussion is about (positive) integers which are not squares of integers. In n is a square of an integer, then b=1.
  9. Nov 25, 2003 #8
    then wouldn't "n is not an integer" prove it?
  10. Nov 28, 2003 #9


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    Hi phoenixthoth, Mathman is using proof by contradiction . In the first line of his proof what he means is : Assume that n is a positive integer that is not a perfect square.

    He goes on to show that assuming sqrt(n) to be rational, but non integer (see b>1), leads directly to a contradiction of the assumption that n was an integer. Hence the result that the square root of all integers, other than the perfects squares, are irrational. It's a perfectly valid proof.
    Last edited: Nov 28, 2003
  11. Nov 28, 2003 #10
    it seems fishy because it was easier to prove it in general for nonsquare nonnegative integers n than for n=6 and it's usually the case that it's easier to do it for n=6 and harder in general. i'm at some point if i feel like it going to check where the assumption that n is not a square was used and if this proof applies to square integers as well. i suppose that's what enables one to say b>1.
  12. Nov 30, 2003 #11
    and 4 = 2*2 both of which are prime
  13. Nov 30, 2003 #12


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    What has 4 = 2 X 2 got to do with that? Some squares are squares of primes and some aren't, like 6 X 6 = 36. How does that affect the proof?
  14. Dec 1, 2003 #13

    (Repost of 'disappeared' Post with Refinement)


    1) Each SQRT(Prime Number) is irrational &

    2) SQRT(Prime Number 1) * SQRT(Prime Number 2) is irrational if Prime Number 1 is not equal to Prime Number 2.

    3) SQRT(6) = SQRT(2) * SQRT(3)

    4) 2 & 3 are two different Prime Numbers.


    5) SQRT(6) is irrational.

    This completes the proof...


    Last edited by a moderator: Dec 1, 2003
  15. Dec 1, 2003 #14
    this was in reference to a deleted post.
  16. Dec 1, 2003 #15

    assumption #1 is as clear as conclusion #5. some people would not take #1 without proof.

    selfadjoint, this post in its original form was about six words long and the point in bringing up 4=2x2 is that it should have been stated that the two primes must be different; it was stated in the second form. this argument would have to be messaged before it would work for all n. the main thing there would be to prove/say a nonsquare has not all even powers in its prime factorization.
    Last edited: Dec 1, 2003
  17. Dec 1, 2003 #16


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    The problem is that you never really establish the truth of either point 1 or point 2 anywhere in the proof kx21

    I still think that the best proof is as follows.

    Let [tex]\sqrt{n} = \frac{a}{b}[/tex]

    Then [tex] a^2 = b^2 n[/tex] <- Eqn1

    From the fundamental therom of arithmetic (that is the uniqueness of prime factorization) it is clear that any integer squared contains only even powered prime factors.

    Thus the LHS of Eqn1 can contain only even powered prime factors whereas the RHS of Eqn1 will contain odd powered prime factors whenever n contains any odd powered prime factors.

    So LHS = RHS is possible if (and only if) n contains only even powered prime factors. This is of course equivalent to the requirement that n be a perfect square.

    In other words, either n is a perfect square or [tex]\sqrt{n}[/tex] is irrational.
    Last edited: Dec 1, 2003
  18. Dec 1, 2003 #17
    *** a 'M' Generalization with a bonus Link Phi, e & i... ***

    Theorem- The irrationality of SQRT(N)

    SQRT(N) is irrational if

    N = p1*p2...*pk, where

    pi i =1 to k are different Prime Numbers.

    For instances,

    1) Is SQRT(6) irrational?


    It's irrational as 6 = 2 * 3


    2) Is SQRT(30) irrational?

    A simple Number Test,

    with a Bonus Link about Phi (Golden Ratio), e, pi for all:-

    http://superstringtheory.com/forum/extraboard/messages12/376.html [Broken]

    Happy exploring & have a wonderful memory & time...


    Last edited by a moderator: May 1, 2017
  19. Dec 1, 2003 #18


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    While correct that theorem is only "half baked" as there are very many integers n for which sqrt(n) is irrational but yet n cannot be expressed as such a simple prime product as stated by the therom (that is, with all prime factors to unity power only).

    Why not do it properly as per either mathmans or my own post, and establish irrationality for sqrt of all integers which are not perfect squares ?

    Also, if you're going to just quote a therom that has already established primeness of certain square roots (as opposed to establishing it from something more fundamental like the uniqueness of prime factorization) then why pick such a "half baked" theorem in the first place? Why not use a better theorem to begin with.

    Theorem : Either n is a perfect square or sqrt(n) is irrational.

    Then all you need to do is to say. "Since 6 in not a perfect then by the above theorem then the sqrt of 6 is irrational." QED.

    Really, if you are going to be lame then at least do it properly. ;-)
  20. Dec 1, 2003 #19

    Let me think...

    Are there any 'Topological Defects' hidden in your Theorem.


    Your Theorem is invalid if...

    n = -1.


    Given SQRT(-1)= i * i

    This Completes the proof...


    Last edited by a moderator: Dec 1, 2003
  21. Dec 1, 2003 #20
    it was clarified earlier that n is a nonnegative nonsquare integer. i is irrational if irrational means not rational. but irrational usually refers to R\Q. technically, this doesn't violate an or statement though i think what was mean by adding the word "either" was exclusive or, Xor.
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