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Topic: Induction proof-need help. Thanks!
Replies: 1   Last Post: Oct 23, 2011 12:30 PM

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Alexander Kelmans

Posts: 3
Registered: 10/23/11
Re: Induction proof-need help. Thanks!
Posted: Oct 23, 2011 12:30 PM
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Dear Ben,

There is no need to know much to see that

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

Now by the induction hypothesis, (n^3 - n) is divisible by 3 and
obviously, 3n(n+1) is divisible by 3.
Therefore (n+1)^3 - (n+1) is also divisible by 3.

So it is not clear what you mean by yours:

``Your identity is far superior to what I was doing.''

Alexander




On Oct 23, 2011, at 4:47 AM, Esther Brink wrote:

> Alexander,
> Your identity is far superior to what I was doing. Thanks.
> Ben
>
> CC: discretemath@mathforum.org; benb@wcjc.edu
> From: kelmans@rutcor.rutgers.edu
> Subject: Re: Induction proof-need help. Thanks!
> Date: Sat, 22 Oct 2011 23:56:15 -0400
> To: bestes32@hotmail.com
>
> Obviously, (n+1)^3 - (n+1) = (n+1)(n^2 - n) + 3n(n+1). The proof by
> induction follows.
>
> On Oct 21, 2011, at 7:46 PM, Esther Brink wrote:
>
> Meo,
> For n = 1, note n^3 - n = 1^3 - 1 = 0, which is divisible by 3.
> Next suppose that k is an integer and the proposition "works".
> Then k^3 - k is divisible by 3.
> Notice that k^3 - k = k(k^2 - 1) = k(k-1)(k+1) = (k-1)k(k+1).
> Therefore 3 must divide either k-1, k or k+1.
> Now can we similarly divide (k+1)^3 - (k+1) into a product of
> three expressions? If 3 divides either k or k+1, will it divide (k
> +1)^3 - (k+1)? If not, what must 3 divide? Can we still conclude
> that 3 divides (k+1)^3 - (k+1)? Why?
> Thanks for providing a great question!
> Ben
>

> > Date: Fri, 21 Oct 2011 16:46:04 -0400
> > From: discussions@mathforum.org
> > To: discretemath@mathforum.org
> > Subject: Induction proof-need help. Thanks!
> >
> > Induction proof problem: write an induction proof for all natural

> numbers for n that n^3-n is diviable with 3.
> >
> > Thanks for any help.

>
>





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