ThePhysicist

linus-just-the-tip-tips

I am three days into folding but i only completed 9 WU...I can squeeze something like 5-10 hours a day of folding and I have a pretty old pc... i5-650 with a 6870

Your math teacher was wrong.
 
There are actually 4 proofs to this, two of which have been mentioned in this thread already, and I do not remember the 4th.  Here are the three I know, 1 & 2 already stated in this thread:
 
1) 1/3 = .3 repeating (I'm unable to draw the line over the .3, so we accept the word "repeating").
 
and 2/3 = .6 repeating (no problem here.  It exactly IS .6 repeating)
 
and 3/3 = .9 repeating = 1
 
 
2) Fact: If two numbers are NOT equal, then there is an infinite set of numbers BETWEEN those 2 numbers.  There are no numbers between .9 repeating and 1, so therefore we know they are equal.
 
3) From algebra days, if you do the same thing to an equation, and then subtract those two equations, you are left with the same thing.  Thus:
 
Let x = .9 repeating
 
Multiply both sides by 10, and we now have 10x = 9.9 repeating.
 
Subtract the original equation from the new equation:
 
10x = 9.9 repeating
-   x =   .9 repeating
________________
 
 9x = 9  (the .9 repeating has been subtracted leaving just 9)
 
Now, solve for x
 
x = 1

1/3 is 0.3 recurring, it is by definition a rational number, so we know what it is even if we're physically incapable of writing it somewhere. It's a common misconception that there is a missing 1 at the end of 0.9 recurring, but the fact is that there is no "end". The number goes on forever. And it is equal to 1. It's not an approximation, they are the same number. 0.9 recurring is also a rational number, meaning it is equal to a/b where a and b are integers. Can you find two integers different from 1 that give 0.9 recurring?

You mean problems which don't require difficult math to solve or problems that don't use complex numbers?
 
The former basically makes content beyond a highschool level impossible.