Question

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

Answer 1

\(n^3-n=n(n^2-1)=n(n-1)(n+1)\)

that should help.

Answer 2

thank you

Answer 3

No further questions?

Answer 4

not yet... im working it thru

Answer 5

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

Answer 6

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

Answer 7

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?

Answer 8

yes

Answer 9

Ok, good.

Answer 10

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

Answer 11

Ok, lets consider 3 possible cases (for possible integer k)

3k, 3k+1, and 3k+2

Answer 12

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.

Answer 13

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.

Answer 14

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

Answer 15

Do I sound rediculous, or is it understandable.

Answer 16

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

Answer 17

my brain is a bit mathed out

Answer 18

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

Answer 19

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

Answer 20

thank you