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Math Help - Proof by Induction - Divisibility Proofs

  1. #1
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    Proof by Induction - Divisibility Proofs

    Prove by induction that 13^n - 6^(n - 2) is divisible by 7, n subset N.

    Step 1: n = 1...

    13^1 - 6^(1 - 2) = 13 - 6^-1 = 13 - 0.1666 = 12.833

    Can someone help me spot where I've gone wrong on the calcualtion to prove n = 1 yields a number divisible by 7?

    Thank you.
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  2. #2
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    Re: Proof by Induction - Divisibility Proofs

    Quote Originally Posted by GrigOrig99 View Post
    Prove by induction that 13^n - 6^(n - 2) is divisible by 7, n subset N.

    Step 1: n = 1...

    13^1 - 6^(1 - 2) = 13 - 6^-1 = 13 - 0.1666 = 12.833

    Can someone help me spot where I've gone wrong on the calcualtion to prove n = 1 yields a number divisible by 7?

    Thank you.
    I think you need to start from n=2.
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  3. #3
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    Re: Proof by Induction - Divisibility Proofs

    Thanks. I suppose it seems obvious, but all the examples in the book were operating from n = 1, so I figured I didn't have the choice.
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    Re: Proof by Induction - Divisibility Proofs

    Suppose 13^k-6^{k-2} is divisible by 7

    Is 13^{k+1}-6^{k-1} also divisible by 7 ?

    (13)13^k-(6)6^{k-2}=(14)13^k-(7)6^{k-2}-\left[13^k-6^{k-2}\right]

    so the inductive proof is verified.
    You just need to know which value of n it becomes valid for.
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    Re: Proof by Induction - Divisibility Proofs

    Got it. Thanks.
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    Re: Proof by Induction - Divisibility Proofs

    Proof By Induction tries to show the following...

    Proposition being true for n=a causes the proposition to be true for n=a+1,
    being true for n=a+1 in turn causes it to be true for n=a+2,
    being true for n=a+2 in turn causes it to be true for n=a+3,
    which in turn causes it to be true for n=a+4,
    which in turn causes it to be true for n=a+5,
    which in turn...... all the way to infinity.

    We show this for every pair of terms in general using n=k and n=k+1.

    It's like Japanese dominoes.

    The value "a" is most often 1, but may be otherwise.
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