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Math Help - Square Roots

  1. #1
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    Square Roots

    This was insipired by Treadstone71.
    Proof that for n\in Z^+
    Then, \sqrt{n} is rational or irrational.
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  2. #2
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    Mmm... There's something odd about this question. Suppose n is rational, then we are done. Suppose n is not rational, then by definition it is irrational. QED?
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  3. #3
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    Quote Originally Posted by Treadstone 71
    Mmm... There's something odd about this question. Suppose n is rational, then we are done. Suppose n is not rational, then by definition it is irrational. QED?
    Probably should read \sqrt  n,\ n \epsilon \mathbb{Z}_+ is either an integer or irrational.

    RonL
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  4. #4
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    Ha ha ha I did not realize what I said. CaptainBlack is right either an integer or irrational.
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  5. #5
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    It suffices to prove that the square of a non-integer rational number cannot be an integer.

    n,m\in \mathbb{Z}

    Let \frac{n}{m} non-integer and let it be in its lowest terms. That is, the intersection of the set of prime factors of n and m is empty. Let n/m be squared. However, since the set of prime factors of n and m did not change, the interesction is still empty, and therefore no cancellation can be made. \frac{n^2}{m^2} is therefore a rational number in its lowest terms with its denominator not equal to 1, and is therefore not an integer. QED.
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  6. #6
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    Here I thought of a proof,

    If,
    \sqrt{x}=\frac{p}{q}
    then,
    x=\frac{p^2}{q^2}
    Let, d=\gcd(p,q) then \gcd(\frac{p}{d},\frac{q}{d})=1
    Express x as,
    x=\frac{d^2\left(\frac{p}{d}\right)^2}{d^2\left( \frac{q}{d}\right)^2}
    Thus,
    \left(\frac{q}{d}\right)^2x=\left(\frac{p}{d} \right)^2
    But,
    \gcd(\left(\frac{p}{d}\right)^2,\left(\frac{q}{d} \right)^2)=1
    and,
    \left(\frac{p}{d} \right)^2\left|\left(\frac{q}{d} \right)^2x \right
    Thus,
    \left(\frac{p}{d} \right)^2 \left|x\right
    Thus,
    k\left(\frac{p}{d} \right)^2=x for some k\in\mathbb{Z}
    Substituting that into the equation we have,
    k\left(\frac{q}{d}\right)^2 \left(\frac{p}{d} \right)^2=\left(\frac{p}{d} \right)^2
    Thus,
    k=1
    Thus,
    \left(\frac{p}{d} \right)^2=x
    Thus, x is a square of some integer.
    Q.E.D.
    Last edited by MathMan; May 21st 2007 at 04:29 PM.
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