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Math Help - proofs question

  1. #1
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    Question proofs question

    Could someone tell me if I proved this correctly?

    prove that for all integers m and n, if m and n are odd, then m + n is even.

    p: m and n is odd
    q: m+n is even(conclusion)

    There exists an integer, k[1], such that m = 2 k[1] + 1
    There exist an integer, k[2], such that n = 2 k[2] + 1

    m + n is even

    m + n = k[1]+1 + 2 k[2]+1 = 2(k[1]+k[2])+2
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  2. #2
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    Roughly speaking, yes.

    for all integers m and n, if m and n are odd, then m + n is even.
    A proof of a statement that starts with "for all", unless it is a proof by induction, must start with words like "fix arbitrary", "consider", "let," etc. In this case, you can say, "Fix arbitrary m and n" or "Let m and n be some integers." Otherwise, when you are talking about m and n later, it is not clear what those are.

    Further, a proof of a statement that starts with "if" must start with "assume." In this case, you say, "... and assume that m and n are odd."

    p: m and n is odd
    q: m+n is even(conclusion)
    This is not necessary since you are not using the letters p and q later. In general, one should not just write a proposition as a part of a proof but rather explain its status: whether the proposition is assumed, whether it's the next goal to prove, etc.

    m + n is even
    To show that m + n is even, consider...

    = 2(k[1]+k[2])+2
    = 2(k[1] + k[2] + 1); therefore, this number is even.
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