Thread: Divisibility of binomial coefficients, countability of N^k.

1. Divisibility of binomial coefficients, countability of N^k.

If anyone could shed some light on this problem, especially the second part to part a, I would be most grateful!

2. Originally Posted by LHS
If anyone could shed some light on this problem, especially the second part to part a, I would be most grateful!

Hmmm...what's the question? How to take the frog out without first melting the ice?

Tonio

3. Sorry? I don't understand, can you not see the inbeded image?

If not, here it is
http://img217.imageshack.us/img217/921/48698386.jpg

4. Originally Posted by LHS
Sorry? I don't understand, can you not see the inbeded image?

If not, here it is
http://img217.imageshack.us/img217/921/48698386.jpg

I'm afraid somebody's hacked the MHF: all over the place appear those annoying frogs inside ice cubes, and

Anyway, following the link above I get...a frog inside an ice cube again! I don't know what's going on.

Perhaps somebody hacked the imageshack site and now we've been infected...

Tonio

5. Haha.. right, ok, that certainly explains it! No worries!

6. Here's the question:

(a) Let $\displaystyle p\in\mathbb{Z}$ be a prime number. Prove that for any natural number $\displaystyle r$ such that $\displaystyle 0<r<p$ the binomial coefficient $\displaystyle p\choose r$ is divisible by $\displaystyle p.$ Hence prove that, for any positive integer $\displaystyle n,$ the integer $\displaystyle n^{p}-n$ is divisible by $\displaystyle p.$

(b) Let $\displaystyle k\in\mathbb{N}$ with $\displaystyle k\ge 2$ and let $\displaystyle \varphi:\mathbb{N}^{k}\to\mathbb{N}$ be given by

$\displaystyle \varphi((a_{1},\dots,a_{k}))=2^{a_{1}+1}\,3^{a_{2} +1}\dots p_{k}^{a_{k}+1},$

where $\displaystyle p_{k}$ is the $\displaystyle k$th prime. Deduce from the Fundamental Theorem of Arithmetic that $\displaystyle \varphi$ is injective and hence that $\displaystyle \mathbb{N}^{k}$ is a countable set for each $\displaystyle k\ge 2.$

[EDIT]: Both links and image are fine for me.