Hi can anyone give a detailed solution to this?

If a and b are natural numbers, then is an integer

Trying to teach my self numbeer theory but hitting a few walls =p

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- August 20th 2010, 03:56 PMjames12division problem
Hi can anyone give a detailed solution to this?

If a and b are natural numbers, then is an integer

Trying to teach my self numbeer theory but hitting a few walls =p - August 20th 2010, 04:12 PMArchie Meade
- August 20th 2010, 04:18 PMjames12
Hi Archie Meade, thanks for your reply, but is there a way to show/prove it's an integer?? This question is from a problem set in divisibility. Sorry, should of probably included that in the question

- August 21st 2010, 05:08 AMHallsofIvy
One way to do it is to note that is the coefficient of in the binomial . Since the coefficients of x and y are 1, all such coefficients are positive integers.

And, of course, with n= a+ b and k= b.

If you don't like that, try proving it by induction on b with a fixed.

If b= 1, , an integer.

Assume true for given b and look at . - August 21st 2010, 09:20 PMjames12
Thanks for your reply, I may see what I can find with the induction path, cheers =)

- August 22nd 2010, 02:49 AMUnbeatable0
See here

- August 24th 2010, 06:02 PMArchie Meade
Hi James,

I'll try a "Proof By Induction" on this,

though it has not been as straightforward as the PBI proofs I've done up to now.

Aside from that, the following is an attempted proof using PBI.

and prove it must be a positive integer for all of the terms being positive integers,

then deduce the binomial from that by setting all but 2 terms to zero,

since

Or, we can simply work with the binomial exclusively.

So, taking the multinomial coefficient...

**P(k)**

This means that the integers summing to k**may be different sets of integers**, but they**do**sum to k.

It also means that is an**integer variable**, that is, it can vary but takes on only positive integer values.

To illustrate....

Therefore, the base case involves establishing this framework for initial values.

For example, suppose k=5 (though we would start with k=smallest practical value of interest)...

Then...

for positive integers, and n=2 or 3.

**P(k+1)**

The set of integers sum to k+1.

However, we utilise the fact that

Then...

Hence the inductive step is complete.