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Math Help - real number problem 3

  1. #1
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    real number problem 3

    prove that there is no positive integer n for which root n - 1 + root n + 1 is rational.
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  2. #2
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    Quote Originally Posted by saha.subham View Post
    prove that there is no positive integer n for which root n - 1 + root n + 1 is rational.
    Suppose a=\sqrt{n-1}+\sqrt{n+1} is rational.
    Then
    a^2=2n+2\sqrt{(n+1)(n-1)}
    \frac{a^2-2n}2=\sqrt{(n+1)(n-1)}
    Hence \sqrt{(n+1)(n-1)} is rational, which implies that (n+1)(n-1)=n^2-1 is a perfect square.
    We get n^2-1=m^2 for some integer m.
    But the only solutions in integere are n=\pm1 and m=0.
    But for n=1 we have a=\sqrt2, which is irrational.

    EDIT: Corrected typo n(n-1) to (n+1)(n-1) on one place.
    Last edited by kompik; May 19th 2010 at 12:09 AM.
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  3. #3
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    its root n + 1 not whole root n + 1
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  4. #4
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    Quote Originally Posted by saha.subham View Post
    its root n + 1 not whole root n + 1
    So you want to work with \sqrt{n-1}+\sqrt{n}+1?
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  5. #5
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    yeah exactly plzz solve it
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  6. #6
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    a=\sqrt{n-1}+\sqrt{n}+1 is rational if and only b=a-1=\sqrt{n-1}+\sqrt{n} is rational.

    You get:
    b=\sqrt{n-1}+\sqrt{n}
    b^2=2n-1+2\sqrt{(n-1)n}
    \frac{b^2-2n+1}2=\sqrt{(n-1)n}
    meaning that \sqrt{(n-1)n} is rational and therefore (n-1)n is a perfect square.

    Now since n-1 and n are coprime, both n-1 and n are perfect squares. So again you have two squares which differ by 1 and you can continue int he same way as I did in my first post. You get that the only solution is n=1.

    BTW do you know the result, that a square root of an integer is rational if and only if this integer is a perfect square? If you did not learn this, you were probably supposed to look for a completely different solution. (Perhaps there is a simpler one and I have overlooked it...)
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  7. #7
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    <br />
a=\sqrt{n}-1+\sqrt{n}+1<br />


    sorry i meant to said this one. extremely sorry ti disturb u again
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  8. #8
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    Quote Originally Posted by saha.subham View Post
    <br />
a=\sqrt{n}-1+\sqrt{n}+1<br />


    sorry i meant to said this one. extremely sorry ti disturb u again
    This one is extremely easy. But this expression is rational for many integers. You have:
    a=2\sqrt{n}
    \sqrt{n}=\frac a2
    Hence \sqrt{n} is rational if and only if n is a perfect square.

    You can try it yourself for 1=1^2, 4=2^2, 9=3^2, 16=4^2, etc.
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