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Thread: Proof involving even integers...

  1. #1
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    Proof involving even integers...

    Question: Prove than an n ∈ ℤ is even if and only if n + 6 is even. Show the if part and the only if part.

    My attempt...

    Only if part

    Prove that an integer n is even only if n + 6 is even.

    Assume n is even; n = 2k.

    Now, n + 6 is equivalent to 2k + 6.

    2(k + 3) where k + 3 is obviously an integer.

    ∴ Integer n is even only if n + 6 is even.

    If part

    Prove that if n + 6 is even; n is an even integer.

    Assume n + 6 is even; n + 6 = 2k.

    n = 2k - 6.

    n = 2(k - 3) where k - 3 is obviously an integer.

    ∴ If n + 6 is even; integer n is even.

    Have I made any mistakes? I have properly read the "if" and "only if" parts out of the question? I want to be sure as I'm new to proofs.

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  2. #2
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    Re: Proof involving even integers...

    Quote Originally Posted by Induction View Post
    [FONT="]Question: Prove than an n ∈ ℤ is even if and only if n + 6 is even. Show the if part and the only if part.

    My attempt...

    Only if part

    Prove that an integer n is even only if n + 6 is even.

    Assume n is even; n = 2k.

    Now, n + 6 is equivalent to 2k + 6.

    2(k + 3) where k + 3 is obviously an integer.

    [/FONT]
    ∴ Integer n is even only if n + 6 i[FONT="]s even.

    If part

    Prove that if n + 6 is even; n is an even integer.

    Assume n + 6 is even; n + 6 = 2k.

    n = 2k - 6.

    n = 2(k - 3) where k - 3 is obviously an integer.

    [/FONT]
    ∴ If n + 6 is even; integer n is even.

    Have I made any mistakes? I have properly read the "if" and "only if" parts out of the question? I want to be sure as I'm new to proofs.

    Those are both correct. Well done!
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