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Math Help - proof that if a square is even then the root is too.

  1. #1
    No one in Particular VonNemo19's Avatar
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    proof that if a square is even then the root is too.

    Show that if m^2=m\cdot{m} is even, then m is even.

    I think I've got it:


    By definition, we can write m^2=2n for some integer n. This implies that m\cdot{m}=2n. But, this means that m\cdot{m} has a factor of 2. Therefore, m has a factor of 2. So, by definition, m is even.

    Now, my gut instinct is that I butchered this. Can someone clean it up?
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    Quote Originally Posted by VonNemo19 View Post
    Show that if m^2=m\cdot{m} is even, then m is even.

    I think I've got it:


    By definition, we can write m^2=2n for some integer n. This implies that m\cdot{m}=2n. But, this means that m\cdot{m} has a factor of 2. Therefore, m has a factor of 2. So, by definition, m is even.

    Now, my gut instinct is that I butchered this. Can someone clean it up?
    Since m = 2n, then m*m = 2n *2n = 4n^2 = 2(2n^2) which is of the form 2p
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by dwsmith View Post
    Since m = 2n, then m*m = 2n *2n = 4n^2 = 2(2n^2) which is of the form 2p
    Isn't this circular reasoning. You are assuming m is even, that is what we want to prove. I think a cleaner way is to prove the logically equivalent statement the contra-positive.

    Remember that if p then q is equivalent to not q implies not p. The reformulation is

    If m is not even then m^2 is not even

    Now just set m=2k+1 \implies m^2=2(2k^2+2k)+1 is odd. QED
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    Quote Originally Posted by TheEmptySet View Post
    Isn't this circular reasoning. You are assuming m is even, that is what we want to prove. I think a cleaner way is to prove the logically equivalent statement the contra-positive.

    Remember that if p then q is equivalent to not q implies not p. The reformulation is

    If m is not even then m^2 is not even

    Now just set m=2k+1 \implies m^2=2(2k^2+2k)+1 is odd. QED
    I thought m is even when I read it.
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  5. #5
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    The Contrapositive is the quickest way, but if you wanted to prove it directly, you'd have to consider the cases " \displaystyle m is even" and " \displaystyle m is odd" and see what happens when you square each.
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  6. #6
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    I think the original proof is fine, but could possibly use another sentence or two. You can always just say that if m is odd, then m and 2 are relatively prime. Since 2 divides the LHS but doesn't divide m, it must divide the other factor, which is m. Contradiction.

    The contrapositive still seems fastest and most basic though.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    Show that if m^2=m\cdot{m} is even, then m is even.

    I think I've got it:


    By definition, we can write m^2=2n for some integer n. This implies that m\cdot{m}=2n. But, this means that m\cdot{m} has a factor of 2. Therefore, m has a factor of 2. So, by definition, m is even.

    Now, my gut instinct is that I butchered this. Can someone clean it up?
    Another way to do this is that if 2\nmid m then by the FTA m=p_1^{\alpha_1}\cdots p_n^{\alpha_n} with p_k\ne 2\;\; k\in[n] and thus m^2=p_1^{2\alpha_1}\cdots p_n^{2\alpha_n} and since none of these primes are two we arrive at a contradiction.
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Isn't this circular reasoning. You are assuming m is even, that is what we want to prove.
    This is what I was thinking, but as I am new to proof, I thought maybe I was just not seeing something.
    Quote Originally Posted by TheEmptySet View Post
    Remember that if p then q is equivalent to not q implies not p. The reformulation is

    If m is not even then m^2 is not even

    Now just set m=2k+1 \implies m^2=2(2k^2+2k)+1 is odd. QED
    Thank you.
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