View Single Post
Old 01-04-2008, 05:16 PM   #10
AnimeSpirit
SBLive! Veteran
 
AnimeSpirit's Avatar
 
Join Date: Dec 2006
Location: Nowhere
Age: 37
Posts: 1,659
Rep Power: 208 AnimeSpirit has a brilliant future AnimeSpirit has a brilliant future AnimeSpirit has a brilliant future AnimeSpirit has a brilliant future AnimeSpirit has a brilliant future AnimeSpirit has a brilliant future AnimeSpirit has a brilliant future AnimeSpirit has a brilliant future AnimeSpirit has a brilliant future AnimeSpirit has a brilliant future AnimeSpirit has a brilliant future
Send a message via Yahoo to AnimeSpirit
Quote:
Originally Posted by j.nc View Post
yeah, I like this one though. As JP says...
if a+b=c then (a+b-c) = 0

So in the last step you are actually saying x divided by zero equals y divided by zero:
3/(a+b-c)=4/(a+b-c)
or
3/0=4/0

heh, tricky
Yes, that's another couple of rules of algebra broken.

(a+b-c) would equal zero. That, I can agree with, but if you are to add a 3 to one side of the equation, you must add a 3 to the other in order to keep the equation balanced, getting (3/0=3/0). This, in fact, would be a mathematical impossibility because, as Joe said, you can't divide by zero. However, discussing division is a side issue anyway, because no division took place in this equation nor should it have.

So let's see where the author started. He started with this, (a+b=c), which equals zero when you rearrange it as (a+b-c=0). Now, if you wanted to distribute 3 and 4 across (a+b-c), you can NOT put 3 on one side and 4 on the other as the author did. This imbalances the equation so no wonder he got such an impossible outcome.

When you add an element to an equation, you have to add equally to BOTH sides of the equation in order to keep it balanced, getting this (3*4(a+b-c)=3*4(0)). If (a+b-c) equals zero, then while following proper order of operations, this would work down to (0=0), a perfectly acceptable outcome.
__________________
Confuscious says~
AnimeSpirit is offline   Reply With Quote