could the numbering system be a loop?

[pushing_wins]

+ infinity becomes - infinity

[sureshs]

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

[Polaris]

It could, if you are in the projective plane .

[SystemicAnomaly]

Originally Posted by pushing_wins

+ infinity becomes - infinity

Conversely, if I lose enough money at the blackjack tables I will eventually become very rich?

[BHiC]

Originally Posted by pushing_wins

+ infinity becomes - infinity

It will be on 12/21/12, when we move into a new dimension where a continuous loop will prove useful to mathematicians, so they can understand our new enlightenment.

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

Originally Posted by SystemicAnomaly

Conversely, if I lose enough money at the blackjack tables I will eventually become very rich?

Wait seriously?! I am going to go break Vegas now! I will lend you my private yacht during the winters for coming up with the idea

[Squall Leonheart]

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.

[sapient007]

best explanation here

http://www.youtube.com/watch?v=kIq5CZlg8Rg

[pushing_wins]

Originally Posted by Squall Leonheart

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.

it all boils down to "by defintion"

[ninman]

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 omega-omega=0 (I think).

[Sentinel]

Originally Posted by pushing_wins

+ infinity becomes - infinity

After a while, you get a numeric overflow and start at zero again. So, yes.

[sureshs]

Originally Posted by ninman

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 omega-omega=0 (I think).

+inf - (+ inf) is undefined.

[sureshs]

Originally Posted by Sentinel

After a while, you get a numeric overflow and start at zero again. So, yes.

Though only for limited precision arithmetic.

[Claudius]

You need to define "infinity" first, which could mean a lot of things.

[sureshs]

Originally Posted by BHiC

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

Yeah otherwise the number will be both greater than 0 and less than 0.

[sureshs]

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

[db10s]

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

[Claudius]

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.

[sureshs]

These are not topics usually taught in HS so don't worry

[sureshs]

Originally Posted by Claudius

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 element x in the field
x^2 > 0. Now , since i^2 = -1, you see why it can't be a real number.

The proof I know was less sophisticated but probably amounts to the same. If i is real, it must be >, =, or < than 0 (your ordered field). Since i^2 = -1, and using your axiom, none of the 3 possibilities can be true.