Thunder wrote:MarbitChow wrote:Infinity is not a number, so standard arithmetical operations cannot be performed on it. Having said that, I'd like to see what 'proof' you offer; I assume it's similar to the proof that 2=1:

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` a = x [true for some a's and x's]`

a+a = a+x [add a to both sides]

2a = a+x [a+a = 2a]

2a-2x = a+x-2x [subtract 2x from both sides]

2(a-x) = a+x-2x [2a-2x = 2(a-x)]

2(a-x) = a-x [x-2x = -x]

2 = 1 [divide both sides by a-x]

i hate that proof so much, your not allowed to divide by 0 people "trick" or no

Yep, this function's domain is all values where a =/= x. Since that was the original function, the whole proof is moot.

As for "Infinity - Infinity = 0", such subtraction is incorrect (it is called an indeterminant form) because it is unclear which "Infinity" has a higher value. While we can subtract ordinary numerals from infinity (any real number), we cannot subtract Infinity from Infinity since it is unclear which number 'reaches' Infinity first.

For example:

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`Limit_x->Infinity [2x - x] -> 2*Infinity - Infinity -> Infinity - Infinity (Indeterminant)`

BUT! Technically 2x reaches infinity sooner than x does. So Infinity - Infinity in this case would be Infinity. You can evade these problems by simplifying before evaluating the limit.

Limit_x->Infinity [2x - x] -> Limit_x->Infinity [x] -> Infinity

It is the same type of error caused by attempting to divide 0 by 0 (0/0 should be 1, but it is also undefined, and it is also 0, and it is also infinity, so the whole thing is called "indeterminant")