2=0

[ChrisP]

So if you define 2 as the sum of an infinite amount of infinitesimally small numbers, or as a basic example, .5 + .5 + .5 + .5 = 2, or .25 + .25 +.25 +.25 + .25 +.25 +.25 + .25 = 2 etc etc until you have infinity multiplied by 0, which, as defined by mathematics, equals 0.
Therefor 2=0, thanks to calculus.

[JackG]

So you're saying that n*k = 2. As n approaches infinity, then k approaches zero. Also, the product of limits equals the limit of the products. The limit of n is infinity, and the limit of k equals zero. Infinity times zero is an indeterminate form. Therefore, you can't say what their product equals.

EDIT: Fixed wording.

[Molten]

Originally Posted by JackG

Infinity times zero is an indeterminate form.

Beat me to it.

[daltore]

You could actually think of it in terms of derivative calculus. Since you're multiplying 0 times infinity, you can also represent 0 as 1/infinity, which will result in infinity/infinity, therefore L'Hopital's rule can apply (and don't think that it equals 1...) Similarly, you can represent infinity as 1/0, which will give you 0/0, and the same thing is true.

[O'Sancheski]

to put it simple.... 2 is the new 0, or the other way around

[Brandon_L]

Using complex mathematics not yet known to man, I have come up with the following theory:

The bird is greater or equal to the word.

But seriously, I'll believe that 2=0 when I type 2 in my calculator and it says 0.

You guys think to hard!

[Molten]

Originally Posted by Brandon_L

But seriously, I'll believe that 2=0 when I type 2 in my calculator and it says 0.

Don't always believe your calculator. Sometimes they lie.

[Ian Curtis]

Originally Posted by Molten

Don't always believe your calculator. Sometimes they lie.

What does your TI-84 have to say about: 1782^12 + 1841^12 = 1922^12?

Fermat's Last Theorem

[JackG]

Nice trick. Of course, anyone looking at the results for 1782^12 + 1841^12 and 1922^12 individually will think that it breaks Fermat's theorem, but entering (1782^12) + (1841^12) - (1922^12) shows that the two sides are not in fact equal. If you could just find values of a, b, and c so that their difference is greater than 12 decimal points away, however, I think that the calculator would fall for the trick because of rounding problems due to memory issues. At least it would work in Matlab, and I'm guessing the calculator stores data the same way.

So, anyone know some a, b, and c such that (c^n)/((a^n)+(b^n)-(c^n)) > 1*(10^13) ? Did I even do that right?

[AndrewJS]

If you are adding like this,
.5 + .5 + .5 + .5 = 2
Etc, then you are really doing a Sigma(2/n) for n times as n approaches infinity. Using the distributive property this sigma expression can be written:
n * (2 / n ) = 2
For all nonzero real numbers, this expression equals 2. At n = 0, there's a hole in the graph, yet its limit is 2.

The key about infinity and zero is that they are not well-defined. It depends HOW you're reaching infinity or zero, not only the fact that you are.

[Molten]

Originally Posted by AndrewJS

The key about infinity and zero is that they are not well-defined.

Both are just as well-defined as anything else is. Zero is no less defined then one. Understood that infinity is a sort of hard concept to grasp, but it is pretty well set even if you can't see it yourself in the real world.

Or in wikiwords [citation needed]

[PAR_WIG1350]

Originally Posted by iCurtis

What does your TI-84 have to say about: 1782^12 + 1841^12 = 1922^12?

Fermat's Last Theorem

that proof was EPIC, and it still is