Recent content by lynchu

Thanks a bunch. That's just about what I needed. :)

Sorry, forgot to include that. Pascal's Identity is: \binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k} for non-negative n's.

But \binom{n}{0} = \binom{n-1}{0} = 1 It's the number of 0-element subsets of an n-elements set. Which is 1, because there's only the empty set.

Homework Statement Use mathematical induction and Pascal's Identity to prove: \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - ... + (-1)^{k}\binom{n}{k} = (-1)^{k}\binom{n-1}{k} The Attempt at a Solution First, I guess this means something like: \sum_{i=0}^{k}(-1)^{i}\binom{n}{i} =...

Appreciate the help so far. I've spent way too much time on this and yet I just don't see it. :cry:

Out of x > y I got -y + x > 0. So that 0 < -y + x =< k But that's not quite a contradiction? Or is it?

Homework Statement Let x and y be real numbers. Prove that if x =< y + k for every positive real number k, then x =< y The Attempt at a Solution x =< y + k -y + x =< k since k is positive, the lowest value it can take doesn't include 0: -y + x < 0 x < y So I get x < y from x =< y...