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Math Help - is this enough prove?

  1. #1
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    is this enough prove?

    hey I should perhaps have seminar presentation on this task and wonder if that evidence is good enough?


    show that the 5^(n +1) -1 is divisible by 4, for all n> _ = 0 (greater or equal to 0)

    my solve:
    5^(n+1)-1
    51 (mod 4)
    5^(n+1) = 5^n*5 1^n*1 (mod 4)
    1^n*1-1≡0 (mod 4)
    1^n = 1 n> _ = 0 (greater or equal to 0)
    1-1≡0 (mod 4)
    what i would prove
    Last edited by Petrus; September 20th 2012 at 09:43 AM.
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  2. #2
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    Re: is this enough prove?

    Hello, Petrus!

    It's hard to follow what you're doing.
    . . I hope you don't present it like that . . .
    And I don't like that n\!+\!1 power.


    \text{Show that }5^n -1\text{ is divisible by 4, for all }n \ge 1

    Are you trying an Inductive Proof?


    Verify S(1)\!:\:5^1 - 1 \:=\:4 . . . True!


    Assume S(k)\!:\;5^k - 1\text{ is divisible by 4.}

    That is: . 5^k - 1 \:=\:4a\text{ for some intger }a.


    Add 4\!\cdot\!5^k to both sides:

    . . 5^k - 1 + 4\!\cdot\!5^k \;=\;4a + 4\!\cdot\!5^k

    . . 5^k + 4\!\cdot\!5^k - 1 \;=\;4(a + 5^k)

    . . 5^k(4 + 1)-1 \;=\;4(a + 5^k)

    . . . . . 5^k\!\cdot\!5 - 1 \;=\;4(a + 5^k)

    . . . . . 5^{k+1} - 1 \;=\; \underbrace{4(a+5^k)}_{\text{divisible by 4}}


    We have proved S(k+1).
    The inductive proof is complete.
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  3. #3
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    Re: is this enough prove?

    Hello!
    That way u solved it is not how i would do but i got 2 question
    1. The way i solved it would i get wrong on a test and i did solve it like that
    2. I dont get it what happend with ^(n+1)?? On that i did 5^1-1 =4 do i mean that n=0?
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  4. #4
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    Re: is this enough prove?

    @Petrus, your proof seems correct. However you know that 5 \equiv 1 (\mod 4) so therefore

    5^{n+1} \equiv 1^{n+1} \equiv 1 (\mod 4) (this is a more direct method).
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  5. #5
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    Re: is this enough prove?

    Ohh never thought about it... Stupid of me sorry
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