# Complete Induction proof

• Oct 21st 2009, 06:46 AM
coobe
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 $\displaystyle 2^{3n}-1$ with n element of Natural numbers, is divisible by 7"

• Oct 21st 2009, 07:14 AM
tonio
Quote:

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 $\displaystyle 2^{3n}-1$ with n element of Natural numbers, is divisible by 7"

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

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

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

Apply now the inductive hypothesis and we're through.

Tonio
• Oct 22nd 2009, 06:41 AM
CaptainBlack
Quote:

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 $\displaystyle 2^{3n}-1$ with n element of Natural numbers, is divisible by 7"

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