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  1. #1
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    prove

    prove that

    squar root x^2+y^2 is a irrational if x and y is odd , positiv integer .

    thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by hoger View Post
    prove that

    squar root x^2+y^2 is a irrational if x and y is odd , positiv integer .

    thanks
    Suppose that:

    x, y \in \mathbb{Z} and that the square root of the sum of their squraes is rational, then there exist a, b \in \mathbb{N}_+ with \gcd(a,b)=1 such that:

     \sqrt{x^2+y^2} =\frac{a}{b}

    Then:

    b^2(x^2+y^2)=a^2

    Now consider any prime factor of b and the divisibility of a^2 by this prime, and in consequence the divisibility of a by the same prime.

    This will show that if \sqrt{x^2+y^2} is rational it is an integer and so x^2+y^2 ia a square.

    CB
    Last edited by CaptainBlack; November 1st 2008 at 09:52 AM.
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  3. #3
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    You can also consider quadratic residues mod 4. It is easy to show that all squares leave a remainder of 1 or 0 when divided by 4.
    x^2 + y^2 leaves a reminder of 2 if both x and y are odd so it cannot be a perfect square.

    Bobak
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  4. #4
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    Quote Originally Posted by bobak View Post
    You can also consider quadratic residues mod 4. It is easy to show that all squares leave a remainder of 1 or 0 when divided by 4.
    x^2 + y^2 leaves a reminder of 2 if both x and y are odd so it cannot be a perfect square.

    Bobak


    hi
    can you explain more for me .

    thank you
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by hoger View Post
    hi
    can you explain more for me .

    thank you
    You need both of the responses to follow what is going on. Suppose that x and y are integers and that \sqrt{x^2+y^2} is rational, then the first post shows that this is of necessity an integer and that x^2+y^2 is a square.

    Then bobak shows that any square leaves remainder 0 or 1 when divided by 4, but if x and y are odd integers x^2 leaves remainder 1 when divided by 4 as does y^2, so x^2+y^2 leaves a remainder of 2 and so cannot be a square. but this contradicts our assumption that \sqrt{x^2+y^2} is rational, and si it must be irrational.

    CB
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