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Thread: 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 $\displaystyle a=\sqrt{n-1}+\sqrt{n+1}$ is rational.
    Then
    $\displaystyle a^2=2n+2\sqrt{(n+1)(n-1)}$
    $\displaystyle \frac{a^2-2n}2=\sqrt{(n+1)(n-1)}$
    Hence $\displaystyle \sqrt{(n+1)(n-1)}$ is rational, which implies that $\displaystyle (n+1)(n-1)=n^2-1$ is a perfect square.
    We get $\displaystyle n^2-1=m^2$ for some integer m.
    But the only solutions in integere are $\displaystyle n=\pm1$ and $\displaystyle m=0$.
    But for n=1 we have $\displaystyle a=\sqrt2$, 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 $\displaystyle \sqrt{n-1}+\sqrt{n}+1$?
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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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    $\displaystyle a=\sqrt{n-1}+\sqrt{n}+1$ is rational if and only $\displaystyle b=a-1=\sqrt{n-1}+\sqrt{n}$ is rational.

    You get:
    $\displaystyle b=\sqrt{n-1}+\sqrt{n}$
    $\displaystyle b^2=2n-1+2\sqrt{(n-1)n}$
    $\displaystyle \frac{b^2-2n+1}2=\sqrt{(n-1)n}$
    meaning that $\displaystyle \sqrt{(n-1)n}$ 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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    $\displaystyle
    a=\sqrt{n}-1+\sqrt{n}+1
    $


    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
    $\displaystyle
    a=\sqrt{n}-1+\sqrt{n}+1
    $


    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:
    $\displaystyle a=2\sqrt{n}$
    $\displaystyle \sqrt{n}=\frac a2$
    Hence $\displaystyle \sqrt{n}$ is rational if and only if n is a perfect square.

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