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Math Help - [SOLVED] Please check proof

  1. #1
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    [SOLVED] Please check proof

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

    then (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 n^2 -1 is divisible by 8 iff n is odd
    I wrote n= 2P+1 where P = 2k

    then (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 n=4k+1, you have only proven the statement for every other odd number: \dots,\,-7,\,-3,\,1,\,5,\,9,\,13,\dots.

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

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

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


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

    We see that X is divisible by 4.


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

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

    Therefore, X is divisible by 8.


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


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

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


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

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

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


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

    . . So we have: . 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: . 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; March 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 p, then q," which can also be written as \text{``}p only if q,\text{''} means that if p is true, then q must also be true (but it might be possible for q to be true even though p is false).

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

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

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