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- anonymous

mathematical induction to prove that n 3 − n is divisible by 3, for every positive integer n

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- anonymous

mathematical induction to prove that n 3 − n is divisible by 3, for every positive integer n

- jamiebookeater

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- SolomonZelman

\(n^3-n=n(n^2-1)=n(n-1)(n+1)\)
that should help.

- anonymous

thank you

- SolomonZelman

No further questions?

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- anonymous

not yet... im working it thru

- anonymous

ok this is where im at
P(x+1)=(x+1)((x+1)−1)((x+1)+1) = 3m
do i divide

- SolomonZelman

you don't need anything, except for a little logic.

- SolomonZelman

here, tell me what happens if you have a product that consists of integers and one of these integers are divisible by 3?
Do you agree that the result is divisible by 3?

- anonymous

yes

- SolomonZelman

Ok, good.

- SolomonZelman

Now, lets come back to the fact that you want to prove that:
*n(n-1)(n+1)* is divislbe by 3, \(\forall {\bf n \in \mathbb Z}\)

- SolomonZelman

Ok, lets consider 3 possible cases (for possible integer k)
3k, 3k+1, and 3k+2

- SolomonZelman

If your number n falls under the category 3k
(Such that n --> 3k)
then the *n* component of *n(n-1)(n+1)*, is divisible by 3,
and thus the entire product *n(n-1)(n+1)* is divislbe by 3.

- SolomonZelman

If your number n falls under the category 3k+1
(Such that n --> 3k+1)
then the *n-1* component of *n(n-1)(n+1)*, is divisible by 3,
and thus the entire product *n(n-1)(n+1)* is divislbe by 3.

- SolomonZelman

And then if:
n --> 3k+2
then the *n+1* makes it divisible by 3.

- SolomonZelman

Do I sound rediculous, or is it understandable.

- anonymous

no u dont.. im just trying to absorb the concept

- anonymous

my brain is a bit mathed out

- SolomonZelman

\(\displaystyle\int\)\(\theta\) \(f(u)+nn=y\)

- SolomonZelman

If you have more questions to ask, I wil be back if I am online.

- anonymous

thank you

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