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Math Help - Complete Induction proof

  1. #1
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    Complete Induction proof

    can someone give me a hint on how to solve such questions in general ?

    "Proof with complete induction that every number with 2^{3n}-1 with n element of Natural numbers, is divisible by 7"

    thanks in advance
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  2. #2
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    Quote Originally Posted by coobe View Post
    can someone give me a hint on how to solve such questions in general ?

    "Proof with complete induction that every number with 2^{3n}-1 with n element of Natural numbers, is divisible by 7"

    thanks in advance

    \mbox{For}\,n=1\,\,\mbox{we get}\,2^{3\cdot1}-1=8-1=7\,\mbox{ so it is true}

    \mbox{Assume now for }\,\,n\,\mbox{ and we'll show for }\,\,n+1:

    2^{3(n+1)}-1=2^{3n+3}-1=8\cdot2^{3n}-1=2^{3n}-1+7\cdot2^{3n}

    Apply now the inductive hypothesis and we're through.

    Tonio
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  3. #3
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    Quote Originally Posted by coobe View Post
    can someone give me a hint on how to solve such questions in general ?

    "Proof with complete induction that every number with 2^{3n}-1 with n element of Natural numbers, is divisible by 7"

    thanks in advance
    Complete (or mathematical) induction is a method of proving an assertion P(n) is true for every natural number by first showing that the base case (that is P(1) or P(0) ) is true, then showing that the truth of P(k) for some k implies the truth of P(k+1). When we have done these two things we have proven P(n) for all n \in \mathbb{N}.

    Now look at Tonio's solution and you will see that it fits this proof method.

    Complete induction is often taken as an axiom of arithmetic.

    CB
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