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Math Help - a few simple proofs

  1. #1
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    a few simple proofs

    hey guys
    having a problem with a couple of calculus proofs...please help!

    (1) prove that n is odd iff n^2 is odd
    i have done the <--- part of this, im just having a problem proving that n is odd if n^2 is odd (the ---> part)

    (2) Show that there exists no rational number x such that x^2 = 3.
    i have attempted to use proof by contradiction but im getting nowhere!

    thanks
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  2. #2
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    Quote Originally Posted by wik_chick88 View Post
    hey guys
    having a problem with a couple of calculus proofs...please help!

    (1) prove that n is odd iff n^2 is odd
    i have done the <--- part of this, im just having a problem proving that n is odd if n^2 is odd (the ---> part)

    (2) Show that there exists no rational number x such that x^2 = 3.
    i have attempted to use proof by contradiction but im getting nowhere!

    thanks
    1
    (a) if n is odd, then n^2 is odd
    Let n = 2p + 1.
    n^2 = (2p + 1)^2 = 4p^2 + 4p + 1 = 2(2p^2 + 2p) + 1 = 2m + 1 which is odd

    Now show if n^2 is odd, then n is odd.
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  3. #3
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    i have already proven that, im having problem proving that if n^2 is odd then n is odd!
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  4. #4
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    Quote Originally Posted by wik_chick88 View Post
    i have already proven that, im having problem proving that if n^2 is odd then n is odd!
    Contradiction. Assume n is even.
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  5. #5
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    but then aren't you just proving that if n is odd, then n^2 is odd? not the other way around?
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    Quote Originally Posted by wik_chick88 View Post
    (2) Show that there exists no rational number x such that x^2 = 3.
    i have attempted to use proof by contradiction but im getting nowhere!
    Try the rational root theorem. x^2 = 3 \implies x^2 - 3 = 0

    The only rational roots of this equation are going to be \pm1,~\pm3. Since none of these work then the solutions of x^2 - 3 = 0 are not rational. Therefore they are irrational.

    -Dan
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  7. #7
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    Quote Originally Posted by wik_chick88 View Post
    but then aren't you just proving that if n is odd, then n^2 is odd? not the other way around?
    We have already shown that if n is odd, then n^2 is odd. Next, we have to show if n^2 is odd, then n is odd.

    Contradiction: n^2 is odd and n is even.

    Since n is even, 2|n, and since n^2 is odd, 2\nmid n^2

    2\nmid (n\cdot n)\Rightarrow 2\nmid n \ \text{or} \ 2\nmid n.

    But, we have 2|n. Therefore, we have reached a contradiction, and if n^2 is odd, then n is odd.
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  8. #8
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    To show that \displaystyle n^2\textrm{ is odd} \implies n\textrm{ is odd}, you can use a proof by exhaustion. You need to consider the cases \displaystyle n\textrm{ is odd}, n\textrm{ is }0, n\textrm{ is even} and show what you get for each when you get \displaystyle n^2.
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