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Math Help - Introduction

  1. #1
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    Introduction

    Hi,

    I came to MHF through google, looking for a *direct* proof that for all integers x it holds that if x^2 is even, then x is even. There's a post on this here: proof that if a square is even then the root is too., but it doesn't give a direct proof, unfortunately. Anyway, MHF seemed like a nice and helpful site, so I registered, and here I am.

    Tomas
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    Re: Introduction

    The contrapositive proof is more direct than the direct proof in this case...
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  3. #3
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    Re: Introduction

    Yes, it's quite neat, that's precisely the point: I'm looking at this as an example of a case where proving the contrapositive is a lot easier than giving a direct proof. Of course, I need the direct proof to justify this claim. (I'm talking about deriving that there exists an a s.t. x=2a from the given that x^2=2b for some integer b.
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    Re: Introduction

    See what happens when x is odd and you square it.
    See what happens when x is even and you square it.
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  5. #5
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    Re: Introduction

    Well, of course if x is odd, then x^2 is odd, and if x is even, then x^2 is even. But without wanting to sound pedantic, I'd say that's a contrapositive proof in disguise. I'd like to start reasoning about x^2 and end at x.

    I just discussed it with a colleague, and we came up with this.

    By the fundamental theorem of arithmetic, x^2 has a unique prime factorization, so x^2=p_1^{c_1}\cdot p_2^{c_2}\cdot\ldots\cdot p_n^{c_n}, for certain primes p_i, and positive whole numbers c_i. (Assume these primes are listed in ascending order.) Because x^2 is even, p_1=2 and because it is a square, we have that for all i, 2\mid c_i. (This shows why, if x^2 is even, 4\mid x^2.) So x=2^{c_1/2}\cdot p_2^{c_2/2}\cdot\ldots\cdot p_n^{c_n/2}, which means x is even. QED.
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    Re: Introduction

    Quote Originally Posted by tomasklos View Post
    Well, of course if x is odd, then x^2 is odd, and if x is even, then x^2 is even. But without wanting to sound pedantic, I'd say that's a contrapositive proof in disguise. I'd like to start reasoning about x^2 and end at x.

    I just discussed it with a colleague, and we came up with this.

    By the fundamental theorem of arithmetic, x^2 has a unique prime factorization, so x^2=p_1^{c_1}\cdot p_2^{c_2}\cdot\ldots\cdot p_n^{c_n}, for certain primes p_i, and positive whole numbers c_i. (Assume these primes are listed in ascending order.) Because x^2 is even, p_1=2 and because it is a square, we have that for all i, 2\mid c_i. (This shows why, if x^2 is even, 4\mid x^2.) So x=2^{c_1/2}\cdot p_2^{c_2/2}\cdot\ldots\cdot p_n^{c_n/2}, which means x is even. QED.
    It's not a contrapositive proof. You are seeing what the possible ways are to get even-valued x^2. It's a proof by exhaustion, which is direct.
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    Re: Introduction

    OK, thank you. What I meant was "direct" in the sense that I wanted to reason from x^2 being even to x being even. What I meant by "contrapositive in disguise" is that just one of those cases (the one of x being odd) already provides the contrapositive proof.

    I'm sorry, I'm new here; still learning the proper way of expressing myself on the forum. Thanks.
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    Re: Introduction

    Quote Originally Posted by tomasklos View Post
    OK, thank you. What I meant was "direct" in the sense that I wanted to reason from x^2 being even to x being even. What I meant by "contrapositive in disguise" is that just one of those cases (the one of x being odd) already provides the contrapositive proof.

    I'm sorry, I'm new here; still learning the proper way of expressing myself on the forum. Thanks.
    Logically speaking, a direct proof is an argument that can show that statement 1 implies statement 2. You can start wherever you like, as long as you get to an argument that shows this implication. Exhaustive proofs like what I've shown you are direct.
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    Re: Introduction

    @thomasklos that's a hard 1. i dont even know the answer. help please!
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    Re: Introduction

    Quote Originally Posted by EhtaChan View Post
    @thomasklos that's a hard 1. i dont even know the answer. help please!
    Hi, there are several (ideas for) proofs on this page. Can you try to explain what you don't understand?
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  11. #11
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    Re: Introduction

    Hi Friends!
    I am the new member of the forum ...........
    So first of all I would like say ".............HELLO........"
    to all of u Dear..................

    Thanks...........
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  12. #12
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    Re: Introduction

    Can u suggest me how to solve pie chart questions very quickly??????????????

    thanks
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  13. #13
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    Re: Introduction

    Hi, pssingh1001,

    Two things. First, start a new thread for a new issue (introduction, new question, etc.). Second, you are more likely to receive help if you post a concrete question. A forum is not a replacement for textbooks.
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  14. #14
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    Re: Introduction

    Thanks for your suggestion ...........
    I will remember in future
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