could the numbering system be a loop?
+ infinity becomes  infinity

No.........................

It could, if you are in the projective plane :) .

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On a serious note, no, it cannot form a loop, because a negative number cannot equal a positive number. Quote:

No... The easiest example off the top of my head is the following: the limit of e^x as x→∞ = ∞. The limit of e^x as x→∞ = 0. If +∞ = ∞, then the two limits would be equal, and no one is arguing the possibility of 0 = ∞.
If you then want to ask if 0 does indeed equal infinity, we could just use a similar argument: the the limit of e^x as x→0 = 1. By definition, 1 (a finite number) cannot equal infinity. 

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The real question is, is infinity really the largest number.
For example if we call infinity  omega, then omega + 1 is bigger than omega, but still infinitely large. If we keep going we can get omega + omega = 2omega. Continuing we get to omega^omega, but then omega^omega+1>omega^omega. If we keep going, we can get omega^omega^omega...  omega times. Then if we take that number +1, we get something bigger still, and start all over again. Meaning there are infinitely many, infinitely large numbers. By this definition omegaomega=0 (I think). 
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You need to define "infinity" first, which could mean a lot of things.

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Can anyone prove that i is not a real number? I can.

Public edu doesn't prepare me for TT math.... And I'm in Honors classes...

It has do with the fact that the complex numbers isn't an ordered field. The real numbers is either constructed from the rational numbers as dedekind cuts (look this up), or as equivalence classes of Cauchy sequences of rational numbers. You make R into an ordered field by saying a < b for two dedekind cuts a and b, if a is contained in b. (dedekind cuts are sets).
If follows by the axioms of an ordered field, that for any nonzero element x in the field x^2 > 0. Now , since i^2 = 1, you see why it can't be a real number. 
These are not topics usually taught in HS so don't worry

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