What luck!
Posted:
Wed Dec 14, 2005 12:46 pm
by ianlogan123
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?
Posted:
Wed Dec 14, 2005 1:05 pm
by FUBAR
give it to a tramp...its christmas
Posted:
Wed Dec 14, 2005 1:29 pm
by Biert
FUBAR wrote:give it to a tramp...its christmas
Yup, then the tramp can buy beer.
Posted:
Wed Dec 14, 2005 3:34 pm
by BlueRaja
Ahh, tramps and beer. NOW I'm in the holiday spirit!
Posted:
Wed Dec 14, 2005 5:21 pm
by psychotic
You forgot about the smack!
Posted:
Thu Dec 15, 2005 11:47 am
by Goat
But they don't get enough smack.
Posted:
Thu Dec 15, 2005 11:53 am
by Biert
BlueRaja wrote:Kids drink enough beer.
You can never drink enough beer. I actually have mathematical proof of that.
Posted:
Thu Dec 15, 2005 12:00 pm
by BlueRaja
Are you kidding? Goat's avatar proves my point! *barf* looked at it AGAIN
Posted:
Thu Dec 15, 2005 12:43 pm
by Biert
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.