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Math Help - Explaination needed for Irrational Proof method

  1. #1
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    Explaination needed for Irrational Proof method

    Question proof √3 is irrational and show reason.
    ================================================== =====
    We prove that √3 is irrational. Assume to the contrary that √3 is rational,
    that is
    √3=p/q,

    where p and q are integers and q≠0. Morever, let p and q have no common divisor > 1.Then

    3=(p/q) => 3q=p (1)

    Since 3q is odd,it follows that q is odd. Then p is also odd.
    This means that there exists k ∈ Z such that

    p=3k. (2)

    Substituting (2) into (1), we get

    3q = (3k) => 3q=9k => q = 3k.

    Since 3k is even it follows that q is even. Then q is also even This is a contradiction.

    Note. I'm stuck with this solution and unable to understand the meaning of it. Any more simple alternative solution to reason irrational ?
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  2. #2
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    Re: Explaination needed for Irrational Proof method

    Yes, this proof is strange.

    Quote Originally Posted by vincor View Post
    Then

    3=(p/q) => 3q=p (1)

    Since 3q is odd
    It is not clear immediately why q is odd.
    Quote Originally Posted by vincor View Post
    it follows that q is odd. Then p is also odd.
    This means that there exists k ∈ Z such that

    p=3k. (2)
    This is clearly non sequitur. Not every odd number is a multiple of 3.

    One proof that \sqrt{3} is irrational parallels a similar proof for \sqrt{2}. Suppose 3q^2 = p^2 where GCD(p, q) = 1. Then 3 | p^2, so by Euclid's lemma, 3 | p, i.e., p = 3k for some integer k. Then 3q^2 = 9k^2, so 3 | q^2 and 3 | q, which is a contradiction with GCD(p, q) = 1.
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  3. #3
    Member agentmulder's Avatar
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    Re: Explaination needed for Irrational Proof method

    For me, when I look at 3q^2 = p^2 it is obvious this is impossible for integer p,q by fundamental theorem of arithmetic and exponent law (ab)^2 = (a^2)(b^2)

    No matter what values I pick for p,q I can never get that 3 'into the square' it doesn't fit.

    For non-square integer h, hq^2 = p^2 has no solutions in integers and therefore h is irrational.

    Now, if h is a square like 4, 9, 16, etc. then I can put h 'into the square' because it fits.

    Examples:hq^2 = p^2

    h = 4

    4q^2 = p^2, (2^2)(q^2) = p^2, (2q)^2 = p^2 it fits

    h = 225

    225q^2 = p^2, (15)^2(q^2) = p^2, (15q)^2 = p^2 it fits

    If h is not a square, for instance 72 how am i going to make it fit?
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