I've been tidying my flat today, which is a big deal for me. I can't remember the last time I did that. Anyway, I found a bag of money in my room. Its full of 10p and 20p pieces that I must have been stashing away a long time ago. There must be about £2 in there.

What should I spend it on?

Beer!

give it to a tramp...its christmas

give it to a tramp...its christmas

http://etext.lib.virginia.edu/etcbin/to ... ision=div1

give it to a tramp...its christmas

Yup, then the tramp can buy beer.

give it to a tramp...its christmas

Yup, then the tramp can buy beer.

Or smack.

Ahh, tramps and beer. NOW I'm in the holiday spirit!

You forgot about the smack!

mmmmm...beeeer...

Save it for kids.

Kids drink enough beer.

But they don't get enough smack.

Kids drink enough beer.

You can never drink enough beer. I actually have mathematical proof of that.

Are you kidding? Goat's avatar proves my point! *barf* looked at it AGAIN

I'm not kidding. I'll prove it, using the technique of induction.
I'll try to keep it simple. The comments between { } explain what rule I used.

To prove this, we assume P(n) to be true. If we can prove a certain basis to be true, and a step P(n+1) to be true, then P is valid.

P(n) is called the Induction Hypothesis.

First, I define a function P(n):
P(n) = "One has not had enough beer, for n beers" (n in N)

Then, I prove a certain basis: n=1
P(1)
= { definition of P }
"One has not had enough beer, for 1 beer."
= { theorem: "One beer is no beer" }
"One has not had enough beer, for 0 beer."
= { empty domain }
True

Next, I will prove that P is valid, for every step that follows another, by proving P(n+1).
P(n+1)
= { theorem: "One beer is no beer" }
P(n+0)
= { nil-element of addition )
P(n)
= { Induction Hypothesis }
true

Q.E.D.