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Thread: prove

  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:

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

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

    Then:

    $\displaystyle b^2(x^2+y^2)=a^2$

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

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

    CB
    Last edited by CaptainBlack; Nov 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.
    $\displaystyle x^2 + y^2$ 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.
    $\displaystyle x^2 + y^2$ 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 $\displaystyle x$ and $\displaystyle y$ are integers and that $\displaystyle \sqrt{x^2+y^2}$ is rational, then the first post shows that this is of necessity an integer and that $\displaystyle x^2+y^2$ is a square.

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

    CB
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