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-   -   could the numbering system be a loop? (http://tt.tennis-warehouse.com/showthread.php?t=444685)

pushing_wins 11-02-2012 01:41 PM

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

sureshs 11-02-2012 01:55 PM

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

Polaris 11-03-2012 12:37 PM

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

SystemicAnomaly 11-03-2012 01:01 PM

Quote:

Originally Posted by pushing_wins (Post 6990846)
+ infinity becomes - infinity

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

BHiC 11-03-2012 03:02 PM

Quote:

Originally Posted by pushing_wins (Post 6990846)
+ 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. :lol:

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

Quote:

Originally Posted by SystemicAnomaly (Post 6992780)
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 :twisted:

Squall Leonheart 11-03-2012 06:23 PM

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 11-03-2012 07:06 PM

best explanation here

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

pushing_wins 11-04-2012 08:18 PM

Quote:

Originally Posted by Squall Leonheart (Post 6993231)
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 11-05-2012 01:57 AM

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 11-05-2012 04:04 AM

Quote:

Originally Posted by pushing_wins (Post 6990846)
+ infinity becomes - infinity

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

sureshs 11-05-2012 05:35 AM

Quote:

Originally Posted by ninman (Post 6995414)
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 11-05-2012 05:36 AM

Quote:

Originally Posted by Sentinel (Post 6995515)
After a while, you get a numeric overflow and start at zero again. So, yes.

Though only for limited precision arithmetic.

Claudius 11-06-2012 10:26 AM

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

sureshs 11-06-2012 10:39 AM

Quote:

Originally Posted by BHiC (Post 6992950)
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 11-06-2012 10:40 AM

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

db10s 11-06-2012 10:47 AM

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

Claudius 11-06-2012 10:51 AM

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 11-06-2012 10:51 AM

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

sureshs 11-06-2012 10:54 AM

Quote:

Originally Posted by Claudius (Post 6998603)
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.


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