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Math Help - proof by contradiction

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    proof by contradiction

    If x is an element of the integers and x^2 is even, then x is even.

    assume x is odd when x satisfies the hypothesis. When you square an odd you get an odd because odd times odd is odd. so x^2 is odd and x^2 is even which is falser because a number cannot be even and odd

    im unsure if this is right
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    sorry i forgot i had one more question.

    if a,b are elements of the integers and ab is odd, then a and b are both odd.

    assume for sake of contradiction a and b are not both odd when the hypothesis is satisfied. so a could be even and b odd. an even number times an odd is even so ab is even. ab cannot be even and odd so there cannot be a counterexmple
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    Quote Originally Posted by leinadwerdna View Post
    sorry i forgot i had one more question.

    if a,b are elements of the integers and ab is odd, then a and b are both odd.

    assume for sake of contradiction a and b are not both odd when the hypothesis is satisfied. so a could be even and b odd. an even number times an odd is even so ab is even. ab cannot be even and odd so there cannot be a counterexmple
    Claim 1. Let a,b be odd, then a*b is odd.
    Proof of Claim 1. Assume the contrary that a,b are not both odd but a*b is odd. Then there are two possible cases as follows.
    (i) a is odd and b is even. Then for some n,m\in\mathbb{Z}, we may write a=2*n-1 and b=2*m, which yields a*b=(2*n-1)*(2*m)=2*(2*m*n)-2*m=2*(m*(2*n-1)) (even). Hence this case is not possible.
    (ii) a and b are both even. Then for some n,m\in\mathbb{Z}, we may write a=2*n and b=2*m, which yields a*b=(2*n)*(2*m)=2*(2*m*n) (even). Hence this case is not possible either.
    Therefore, if a,b are odd, then a*b is odd. \rule{0.2cm}{0.2cm}

    Quote Originally Posted by leinadwerdna View Post
    If x is an element of the integers and x^2 is even, then x is even.

    assume x is odd when x satisfies the hypothesis. When you square an odd you get an odd because odd times odd is odd. so x^2 is odd and x^2 is even which is falser because a number cannot be even and odd

    im unsure if this is right
    Claim 2. If x^{2} is even and x is an integer, then x is even.
    Proof of Claim 2. Again assume the contrary that x^{2} is even but x is odd. Then, we have from Claim 1 that x^{2}=x*x is odd.
    This contradiction proves Claim 2. \rule{0.2cm}{0.2cm}

    Probably you are requested to prove Claim 2 by using Claim 1.
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