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Thread: [SOLVED] Please check proof

  1. #1
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    [SOLVED] Please check proof

    So I need to prove that$\displaystyle n^2 -1 $ is divisible by 8 iff n is odd
    I wrote n= 2P+1 where P = 2k

    then $\displaystyle (2P+1)^2 - 1 = 4P^2+4P + 1-1 $ subing 2k back in I get

    16k^2+8k = 8(2k^2+k)

    Is this okay?
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  2. #2
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    Quote Originally Posted by meg0529 View Post
    So I need to prove that$\displaystyle n^2 -1 $ is divisible by 8 iff n is odd
    I wrote n= 2P+1 where P = 2k

    then $\displaystyle (2P+1)^2 - 1 = 4P^2+4P + 1-1 $ subing 2k back in I get

    16k^2+8k = 8(2k^2+k)

    Is this okay?
    No. By setting $\displaystyle n=4k+1,$ you have only proven the statement for every other odd number: $\displaystyle \dots,\,-7,\,-3,\,1,\,5,\,9,\,13,\dots$.

    Fortunately, you can fix it by handling the case for odd $\displaystyle P$ separately. Then you get $\displaystyle n=4k+3,$ and

    $\displaystyle n^2-1=16k^2+24k+9-1=8(2k^2+3k+1).$
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  3. #3
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    Hello, meg0529!

    I need to prove that $\displaystyle X \:=\:n^2 -1 $ is divisible by 8 iff $\displaystyle n$ is odd
    I wrote $\displaystyle n\:=\: 2p+1$, where $\displaystyle p = 2k$ . . . . no

    $\displaystyle p$ does not have to be even.
    $\displaystyle p$ can be any integer, and $\displaystyle n$ will be odd.


    So we have: .$\displaystyle X \:=\:n^2-1 \:=\:(2p+1)^2-1 \:=\:4p^2 + 4p \:=\:4p(p+1)$

    We see that $\displaystyle X$ is divisible by 4.


    $\displaystyle \text{Note that: }\:X \:=\:4\cdot\!\!\!\!\!\!\!\!\underbrace{p(p+1)}_{\t ext{consecutive integers}} $

    Since $\displaystyle p(p+1)$ is the product of two consecutive integers,
    . . one of them is even, the other is odd.
    Hence: .$\displaystyle p(p+1)$ is divisible by 2.

    Therefore, $\displaystyle X$ is divisible by 8.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    It said if and only if, so we must prove it "the other way."

    . . If $\displaystyle n^2-1$ is divisible by 8, then $\displaystyle n$ is odd.


    We have: .$\displaystyle n^2-1$ is divisible by 8.

    . . Then: .$\displaystyle n^2-1 \:=\:8k$, for some integer $\displaystyle k.$

    And we have: .$\displaystyle n^2 \:=\:8k + 1$


    We see that $\displaystyle 8k$ is even . . . hence: $\displaystyle 8k+1$ is odd.

    . . So we have: .$\displaystyle n^2 \:=\:\text{odd}$


    It can be shown that:
    . . If the square of an integer is odd, the integer is odd.

    The proof is quite simple, but omitted here.

    Therefore: .$\displaystyle n$ is odd.

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  4. #4
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    Reckoner,

    Ok I see your point, So you are saying I need to do this is two different steps? Is there a way for me to do this in one shot?

    Thanks
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    Quote Originally Posted by meg0529 View Post
    Reckoner,

    Ok I see your point, So you are saying I need to do this is two different steps? Is there a way for me to do this in one shot?
    Certainly there is, and Soroban has provided it.

    Take care.

    Edit: Oh, and be sure you understand what is meant by "if and only if" (abbreviated iff). I misread your post initially and thought it only went one direction; there are basically two proofs that need to be made. Again, Soroban has this covered.
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  6. #6
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    Got it. Thanks you both very much
    Last edited by meg0529; Mar 22nd 2009 at 04:14 PM.
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  7. #7
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    I see you have edited your post, but I will post my response anyway since it is already written.

    Quote Originally Posted by meg0529 View Post
    So If my sentence contains an if and only if in it, I should set it up as an If then statement?
    No. An if-then statement means the implication goes in only one direction.

    The statement "If $\displaystyle p,$ then $\displaystyle q,$" which can also be written as $\displaystyle \text{``}p$ only if $\displaystyle q,\text{''}$ means that if $\displaystyle p$ is true, then $\displaystyle q$ must also be true (but it might be possible for $\displaystyle q$ to be true even though $\displaystyle p$ is false).

    The statement $\displaystyle \text{``}p$ if $\displaystyle q,\text{''}$ means that if $\displaystyle q$ is true, then $\displaystyle p$ must also be true (but it might be possible for $\displaystyle p$ to be true even though $\displaystyle q$ is false).

    The statement $\displaystyle \text{``}p,$ if and only if $\displaystyle q\text{''}$ means that if $\displaystyle p$ is true, then $\displaystyle q$ must also be true, and if $\displaystyle q$ is true, then $\displaystyle p$ must also be true. Do you see how the implication goes both ways?

    When proving a statement like $\displaystyle p\text{ if{f} }q,$ you need to prove both of these statements:
    1. If $\displaystyle p,$ then $\displaystyle q$
    2. If $\displaystyle q,$ then $\displaystyle p$
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  8. #8
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    Ya haha I re-read the chapter, thanks!
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