Not trying to be a fanboy. Of course, it's hard when you own the best console eve... dang it
Proof that 2 = 1
Let a and b be equal non-zero quantities
a = b
Multiply through by a
a^2 = ab
Subtract b^2
a^2 - b^2 = ab - b^2
Factor both sides
(a - b)(a + b) = b(a - b)
Divide out (a - b)
a + b = b
Observing that a = b
b + b = b
Combine like terms on the left
2b = b
Divide by the non-zero b
2 = 1
Q.E.D.
| wfz said: Eab, the problem with the "infinity times zero" question is that infinity isn't even a number. Also most of the time you're looking at that, 0 isn't a number either. They're both behaviors.
EDIT: Shameless, infinity times zero is not zero. Don't criticize people for butchering math and then do it yourself in the same post! |
1/3 means 1 divided by 3 like all fractions.
Zero is always zero, no matter what you choose to multiply it by.
| superchunk said: Proof that 2 = 1 Let a and b be equal non-zero quantities a = b Multiply through by a a^2 = ab Subtract b^2 a^2 - b^2 = ab - b^2 Factor both sides (a - b)(a + b) = b(a - b) Divide out (a - b) a + b = b Observing that a = b b + b = b Combine like terms on the left 2b = b Divide by the non-zero b 2 = 1 Q.E.D. |
Wrong.
If a = b then a - b = 0. You have divided by (a - b) which is dividing by 0.
Shameless said:
1/3 means 1 divided by 3 like all fractions. Zero is always zero, no matter what you choose to multiply it by.
|
Zero is always zero, no matter what real number you multiply it by. Infinity is not a number, so that doesn't work.
Infinity is a behavior. Also, 0 can be a behavior as well, so using L'hopital's rule is great when you have infinities and zeroes.
Yes.
Also 3/5 =/= 6/10.
We'll miss you George.
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My friends call me Hadoken because I'm down-right fierce
Shameless said:
Wrong. If a = b then a - b = 0. You have divided by (a - b) which is dividing by 0.
|
nice, you win. I was hoping it would take longer for someone to catch that. :)
I always liked this proof when I was shown it in my logic classes.
| superchunk said: Proof that 2 = 1 Let a and b be equal non-zero quantities a = b Multiply through by a a^2 = ab Subtract b^2 a^2 - b^2 = ab - b^2 Factor both sides (a - b)(a + b) = b(a - b) Divide out (a - b) a + b = b Observing that a = b b + b = b Combine like terms on the left 2b = b Divide by the non-zero b 2 = 1 Q.E.D. |
edited: someone got it
^_^
wfz said:
Zero is always zero, no matter what real number you multiply it by. Infinity is not a number, so that doesn't work.
Infinity is a behavior. Also, 0 can be a behavior as well, so using L'hopital's rule is great when you have infinities and zeroes.
|
Fair enough. I guess I'm getting confused when people define 1/0 = infinity when it should be written implicitly as "1/x = y; when x tends to 0, y tends to infinity", I'm getting tired ;).
wfz said:
Zero is always zero, no matter what real number you multiply it by. Infinity is not a number, so that doesn't work.
Infinity is a behavior. Also, 0 can be a behavior as well, so using L'hopital's rule is great when you have infinities and zeroes.
|
I was under the impression that L'hopital's rule only worked when it was 0/0 or Infinity/Infinity. It might change in later maths though; I don't know. In Calculuas AB, that's how it is, at least.
Also, math isn't messed up. .999... is the SAME number as 1. A lot of people have trouble with this conceptually because we can't fathom infinity. You can't look at this as approaching an infinite number of 9's (which would be less than 1) or anything like that. The number of 9's in .999... has actually reached infinity.
Here's another way to prove it.
c = .999...
10c = 9.999...
10c - c = 9.999... - .999...
9c = 9
c = 1
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