Complete Induction proof

**coobe** · October 21st 2009, 06:46 AM

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

**tonio** · October 21st 2009, 07:14 AM

Originally Posted by coobe

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

**CaptainBlack** · October 22nd 2009, 06:41 AM

Originally Posted by coobe

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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