:rolleyes: Algebra. He’s doing ALGEBRA?
Sure, I took Algebra I.
Like, in the EIGHTH grade! 
STUPID CANADIANS!
Hahahahahah!
<smack> :smack:
Ouch!!
I’m also in tenth grade…you think that’s bad? Try my math analysis (pre-calc) homework.
phraser, I can remember by differential calculus. 
Anyhoop, if anyone wants to know the error in the proof that A+B=B…
[spoiler]If A=B, than A-B=0.
Thus, in the next-to-last step, both sides are divided by zero. Any number divided by zero equals infinity, or does not exist, depending on your math persuasion. A nonzero devided by zero never equals a rational number.
It’s an insidious version of the following proof…
1 x 0 = 0
2 x 0 = 0
Therefore, 1 x 0 = 2 x 0
Devide each side by zero…
1 = 2
This, obviously, doesn’t wash.[/spoiler]
I used to annoy my math teachers by giving my variables odd names. I’d be solving for the value of Ringo or 
(yes, I had a sheet full of smiley-face stickers about the size of the eraser on a #2 pencil, and I placed a sticker onto my work paper every time I had to name the variable). My teacher eventually started to specify “USE X AS VARIABLE EXCEPT WHERE OTHERWISE NOTED” on the blackboard when he assigned the homework.
That’s no fun.
I could be wrong here (and I have been from time to time) but division by zero is undefined. There is no such number as “infinity” and I’m not sure what it means for a number to be divided by zero and cease to exist.
I’ll just divide my credit card balance by zero and…
MonkeyMensch is correct about the division by zero bit. It ain’t infinity, that’s for damn sure.
THEOREM 1: All positive integers are equal.
Proof: Given real numbers a,b define max{a,b} to be the larger of the two or either when they are equal. Thus max{3,7} = max{7,7} = 7. We shall prove
For every positive integer n, if a and b are positive integers such that max{a,b} = n, then a = b.
Clearly this will imply Thm 1.
(i) If max{a,b} = 1 then a = b = 1. Thus the result holds when n = 1.
(ii) In general, given a,b with max{a,b} = n let [symbol]a[/symbol] = a-1 and [symbol]b[/symbol] = b-1. Then
max{[symbol]a[/symbol],[symbol]b[/symbol]} = n-1.
By the inductive hypothesis it follows that [symbol]a[/symbol] = [symbol]b[/symbol] and so a = b. This completes the proof by induction.
In the same spirit we can also prove
THEOREM 2: In any exam all candidates achieve the same score.
Proof: Again by induction on the number n of candidates. The theorem is trivial when n = 1. In general split the set of candidates into two subsets A and B which are both smaller than the whole set and which have one candidate in common. By the inductive hypothesis all the candidates in set A achieve the same score. Similarly for the candidates in B. Since A and B have one person in common this score is the same for A and B. Thus all the candidates get the same score and we are done.
Does it count if your math teachers thought youd reached your math potential about the 6th grade and most of there time was spent making sure you didnt forget the basic stuff ?
Ah well, not all the ‘proofs’ can be interesting.
Hmm…I’ve got a pretty good idea where the error is in Jabba’s proof, but I don’t think I could explain it succinctly. More time…
It’s nice to know I can get work as a variable.
That’ll teach you to do vanity searches.
The errors in Jabba’s proof are pretty simple actually…
[spoiler]In the first proof, neither a-1 nor b-1 are guaranteed to be positive integers, but the inductive step assumes that they are.
In the second proof, it isn’t possible to construct “two subsets A and B which are both smaller than the whole set and which have one candidate in common” when n=2.
[/spoiler]
I knew the second one. Thought I’d use it as an interview question (for a tech interview).
The second one, I would’ve figured out soon. Thanks to Math Geek.