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Math Help - Contrapositive

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    Contrapositive

    Can someone please tell me what the contrapositive to this statement is: the square of an even number is an even number.

    I thought it was if a number is not even, then its square is not even.
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    Quote Originally Posted by sjenkins View Post
    Can someone please tell me what the contrapositive to this statement is: the square of an even number is an even number.

    I thought it was if a number is not even, then its square is not even.
    mmmm

    Let P ~:~ \text{n is an even number}
    Q ~:~ \text{its square is even}


    The first sentence is P \implies Q with some different syntax, but though the same meaning. I'll try to explain it... Grammatically, the "square of" comes after the "even number", in the sense that you have in a first time "an even number" and then, you get its "square". This is why you consider the square part as a consequence...

    The contraposive is \bar{Q} \implies \bar{P}, that is to say :
    \text{the square of a number is not even} \implies \text{the number is not even}.

    In other words : "If the square of a number is not even, then this number is not even".


    Is it clear ? Does it look correct to you ?
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    Yes, thanks so much Moo! I appreciate your constant willingness to help me!!!
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    Hello, sjenkins!

    Can someone please tell me what the contrapositive to this statement is:
    The square of an even number is an even number.
    We must carefully write the statement as an implication.

    "The square of an even number is an even number."

    \text{This says: }\;\text{If }\underbrace{x\text{ is even,}}_p\text{ then }\underbrace{x^2\text{ is even.}}_q


    We want: . \sim\: q \to \sim p

    \text{This is: \;If }\underbrace{x^2\text{ is not even,}}_{\sim q}\text{  then } \underbrace{x\text{ is not even.}}_{\sim p}

    In other words: If the square of a number is odd, then the number is odd.



    Too fast for me, Moo!
    .
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    More contrapositive help

    I just don't think I was thinking it through. Now, can anyone tell me if the following is a proof of this statement???

    Assume that x is not an even number
    Then x is not a multiple of 2
    In other words, x=2k+1 for some integer k
    So x^2=(2k+1)^2=4k^2+1=2(2k^2)+1=2n where n=2k^2+1
    We check that n is also an integer which it is since k is an integer and the integers are closed under multiplication
    So, x is not a multiple of 2
    Therefore, x^2 is not even
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    Moo
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    Quote Originally Posted by sjenkins View Post
    I just don't think I was thinking it through. Now, can anyone tell me if the following is a proof of this statement???

    Assume that x is not an even number
    Then x is not a multiple of 2
    In other words, x=2k+1 for some integer k
    So x^2=(2k+1)^2=4k^2+1=2(2k^2)+1=2n where n=2k^2+1
    We check that n is also an integer which it is since k is an integer and the integers are closed under multiplication
    So, x is not a multiple of 2
    Therefore, x^2 is not even
    (a+b)^2=a^2+2ab+b^2 \neq a^2+b^2 !!!

    By the way, what you proved here is :

    \text{x is not even} \implies \text{the square of x is not even}, but this is not the contraposive, nor the statement.
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    Why do I have such a difficult time understanding all of this. I am really trying here and nothing I do seems to work out.
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    Quote Originally Posted by Moo View Post
    The contraposive is \bar{Q} \implies \bar{P}
    That is not the contrapositive. The contrapositive of P\implies Q is \neg\,Q\implies\neg\,P.

    Q\implies P is the converse of P\implies Q.
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    Quote Originally Posted by algebraic topology View Post
    That is not the contrapositive. The contrapositive of P\implies Q is \neg\,Q\implies\neg\,P.

    Q\implies P is the converse of P\implies Q.
    Did you not see that Moo used \overline Q for \neg Q?
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  10. #10
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    I beg Mooís pardon, the bar didnít show up clearly on my browser (and itís also the first time Iíve ever seen this notation being used.)
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    Trying this proof again...

    Here's try number 2, does this work to prove if the square of a number is not even, then the number is not even.

    Assume that x^2 is not an even number
    Then x^2 is not a multiple of 2
    In other words, x^2=(2k+1)^2 for some integer k
    It follows that x=2k+1
    Therefore x is not even
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    Your proof is okay, except for one little thing:

    Quote Originally Posted by sjenkins View Post
    In other words, x^2=(2k+1)^2 for some integer k
    It follows that x=2k+1
    This does not work. Suppose x = -1. We can write x^2 = (-1)^2 = (1)^2 = (2\cdot0 + 1)^2 = (2k + 1)^2,\;k=0\in\mathbb{Z}, but x = -1\neq 2k + 1 = 1. In other words, when taking the square root of a square, you have to be aware that \sqrt{x^2} = \lvert x\rvert. Of course, this is easy to rectify.

    However, I think a direct proof of the original statement would be a lot simpler: Let n be an even integer, i.e., \exists p\in\mathbb{Z},\;n = 2p, and then show, through a chain of implications, that n^2 must also be even.
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    I agree that a direct proof of the original statement may have been easier, unfortunately the book said I had to create an indirect proof
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    Can I ask what you would write to fix this problem...I'm confused!
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    Quote Originally Posted by sjenkins View Post
    I agree that a direct proof of the original statement may have been easier, unfortunately the book said I had to create an indirect proof
    Okay, then. Your proof is almost there; just fix the flaw I pointed out. This may help you, however: if n^2 is not even, then 2 does not divide it. But if n^2 = n\cdot n is not divisible by 2, neither is any of its factors, so 2 does not divide n.
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