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Math Help - Mathematical proofs

  1. #1
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    Mathematical proofs

    If a mod would like to move this to it's assigned section please do so, I don't know which one it belongs into.

    sorry i'm not good with latex

    Q1. Prove the following biconditional statement

    1. For any a in Z(set of integers) a not logically equivalent to 0(mod 3) if and only if a^2 is logically equivalent with 1(mod 3)

    2.Prove that if a is an odd integer, then for any x in Z(set of integers), x^2 - x - a does not equal 0

    3. Use mathematical induction to prove that for any natural number n, 3 divides (2^(2n) - 1 ). Specify each step in the process clearly.

    Thank you soo much (:
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by treetheta View Post
    If a mod would like to move this to it's assigned section please do so, I don't know which one it belongs into.

    sorry i'm not good with latex

    Q1. Prove the following biconditional statement

    1. For any a in Z(set of integers) a not logically equivalent to 0(mod 3) if and only if a^2 is logically equivalent with 1(mod 3)
    do you know how to change modular equations into algebraic ones? if so, i suggest you do that. for instance, a \equiv 1 \mod{3} means a - 1 = 3k for some k \in \mathbb{Z}.

    after rewriting the statements that way, see if you can figure out what to do.


    2.Prove that if a is an odd integer, then for any x in Z(set of integers), x^2 - x - a does not equal 0
    consider two cases: (1) x is even, (2) x is odd. show that the statement holds in either case. good ol' fashion algebra should do the trick. recall how to define even and odd integers.


    3. Use mathematical induction to prove that for any natural number n, 3 divides (2^(2n) - 1 ). Specify each step in the process clearly.

    Thank you soo much (:
    do you know the method of mathematical induction? what have you tried?

    Let P(n): 3 divides 2^{2n} - 1 for all n \in \mathbb{N} = \{ 0,1,2,3, \cdots \}

    Clearly P(0) is true. (check this!)

    Assume P(n) is true. Now we need to show that this implies P(n + 1) is true.

    Note that since P(n) is true, we have

    2^{2n} - 1 = 3k for some k \in \mathbb{Z}.

    So, P(n + 1) = 2^{2n + 2} - 1

    = 4 \cdot 2^{2n} ~{\color{red} + 4 - 4}~ - 1

    = \cdots
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  3. #3
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    okay i think i got the #2

    for the first 1 do i assume 3 cases for 0,1,2

    and the third one how did u get

    <br /> <br />
P(n + 1) = 2^{2n + 2} - 1<br />

    or did u multiply everything by 2^2 in which case u get

    <br /> <br />
P(n + 1) = 2^{2n + 2} - 4 = 12p<br />
2^(2n + 2) - 1 = 3(4n + 1)<br />

    and does 2n + 2 = 2k

    o.o i did these steps but i ended up getting 9/10 i don't know where i went wrong..
    [/FONT][/COLOR]
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by treetheta View Post
    okay i think i got the #2
    ok

    for the first 1 do i assume 3 cases for 0,1,2
    for (=>) direction you need only 2 cases: 1 (mod 3) and 2 (mod 3)

    for (<=) direction, use the contrapositive


    and the third one how did u get

    <br /> <br />
P(n + 1) = 2^{2n + 2} - 1<br />
    replace n with (n + 1)

    or did u multiply everything by 2^2 in which case u get

    <br /> <br />
P(n + 1) = 2^{2n + 2} - 4 = 12p<br />
2^(2n + 2) - 1 = 3(4n + 1)<br />

    and does 2n + 2 = 2k
    that's not what i did

    o.o i did these steps but i ended up getting 9/10 i don't know where i went wrong..
    [/font][/color]
    the answer here is not a number...
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  5. #5
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    replace n with (n + 1)
    i see how u got 2k + 2 now


    that's not what i did
    and sorry the latex came out wrong
     P(n + 1) = 2^{2n + 2} - 4 = 12p
    2^(2n + 2) - 1 = 3(4n + 1)


    so n = 2k

    the answer here is not a number...
    and that wasn't the answer that is what i got on a marking scale
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by treetheta View Post
    i see how u got 2k + 2 now




    and sorry the latex came out wrong
     P(n + 1) = 2^{2n + 2} - 4
    no, this is not P(n + 1)

    = 12p
    how??

    2^(2n + 2) - 1 = 3(4n + 1)


    so n = 2k
    no, and you are not looking for n in the first place. are you sure you know what the method of mathematical induction is?

    look up how to prove something via mathematical induction, then re-read my post and make sure you understand what is going on. then just pick up where i left off.
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  7. #7
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    okay, I've given up on what I did,
    and now I'm working on what you did

    Why did you multiply and add and subtract 4 on both sides

    can you explain it to me or link me to something that will help me understand it
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  8. #8
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    Omg, thanks jhevon i gett it ! : D
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by treetheta View Post
    Omg, thanks jhevon i gett it ! : D
    you got it? really? great!

    that adding and subtracting 4 (hence adding zero) is a neat trick, right?
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  10. #10
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    yea it was hard for me to follow why u did it
    but now it all makes sense : D
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