# could the numbering system be a loop?

Discussion in 'Odds & Ends' started by pushing_wins, Nov 2, 2012.

1. ### pushing_winsHall of Fame

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+ infinity becomes - infinity

2. ### sureshsBionic Poster

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

3. ### PolarisHall of Fame

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It could, if you are in the projective plane .

4. ### SystemicAnomalyG.O.A.T.

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

5. ### BHiCRookie

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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. :lol:

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

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 :twisted:

6. ### Squall LeonheartRookie

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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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8. ### pushing_winsHall of Fame

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it all boils down to "by defintion"

9. ### ninmanHall of Fame

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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 omega-omega=0 (I think).

10. ### SentinelBionic Poster

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

11. ### sureshsBionic Poster

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+inf - (+ inf) is undefined.

12. ### sureshsBionic Poster

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Though only for limited precision arithmetic.

13. ### ClaudiusProfessional

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

14. ### sureshsBionic Poster

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

15. ### sureshsBionic Poster

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

16. ### db10sHall of Fame

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

17. ### ClaudiusProfessional

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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.

Last edited: Nov 6, 2012
18. ### sureshsBionic Poster

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

19. ### sureshsBionic Poster

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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.