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Destrus 04-30-2009 08:25 AM

why is math so painful
 
:cries:
i used to like it, then i got to college

Hessah 04-30-2009 08:34 AM

-pat pat-

Destrus 04-30-2009 08:53 AM

i hate howwhile studying i look at the first exam and see that hey, i can do a lot ofthis pretty easily and i feel like i've actually gotten something done cause i looked it over so quickly but in reality i've barely scratched the surface of my problems

Destrus 04-30-2009 09:11 AM

:cries:

how do i prove that for any natural number n n^4 - n^2 is divisible by 3:cries:

Vasu 04-30-2009 09:20 AM

Use mathematical induction!

lamchopz 04-30-2009 09:22 AM

n^4 - n^2 = n^2(n^2 - 1) = n*n*(n-1)*(n+1)

Since, n -1 ,n , n + 1 are three consecutive natural numbers, one of them has to be divisible by 3, hence your expression is divisible by 3.

Vasu 04-30-2009 09:26 AM

LOL, the method I was thinking of was, prove it for n=1, then assume it is true for n=k, and then prove it for n=k+1. A bit more elaborate. :urweird:

Destrus 04-30-2009 09:26 AM

so that was a test question from the first exam i couldn't remember how to do (even then T_T). if he actually put it on the final i'd be so happy i might have to kiss you lam (but settle for a hug)


also, vasu, die a horrible death D:<
cause that's on this too and i hate it. i hate it soooo much.

Destrus 04-30-2009 09:34 AM

Quote:

Originally Posted by Vasu (Post 330242)
LOL, the method I was thinking of was, prove it for n=1, then assume it is true for n=k, and then prove it for n=k+1. A bit more elaborate. :urweird:

o_o oh you posted again, didn't notice.

freaking induction

lamchopz 04-30-2009 09:35 AM

I'm unsure if he'd accept that but it's something used liberally in arithmetics.

It follows from this more general statement: For any n consecutive integers, at least one of them is divisible by n.

The proof is relatively painless:

Each number can be represented as nq + r where r < n (due to Euler's integer theorem, or whatever it's called. can't remember now xD)

r is therefore any integer from 0 to n -1

since u have n numbers and they're consecutive, they are essentially:

nq + 0
nq +1
...
nq + (n - 1)

or any valid combination (we don't illustrate this in the formal answer, however).

Normally, you just say since the remainders include all values from 0 to n -1, and there are n consecutive numbers, one of them has to have a remainder of 0, which means the same as one of them is divisible by n.

Again check with your professor if he'll accept this without the burden of proof.

and LOL @ your random rage, Des.


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