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 Message Boards » » You can't have inequalities with imaginary numbers Page [1]  
Walter
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7/25/2009 4:20:32 PM

Fareako
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haha

7/25/2009 4:22:11 PM

ScHpEnXeL
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ahahaha

7/25/2009 4:23:11 PM

d7freestyler
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lol

7/25/2009 4:25:19 PM

miska
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haha

7/25/2009 4:51:11 PM

ShawnaC123
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7/25/2009 8:35:47 PM

EuroTitToss
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7/25/2009 8:37:01 PM

ScubaSteve
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hahahahaha lmao

7/25/2009 8:37:07 PM

not dnl
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^^ha now thats good

[Edited on July 25, 2009 at 8:37 PM. Reason : .]

7/25/2009 8:37:13 PM

JeffreyBSG
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hahahahaha

7/25/2009 9:11:34 PM

ScubaSteve
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bttt for night

7/26/2009 12:40:48 AM

moron
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the real component would just be 0 wouldnt it...?

7/26/2009 12:52:02 AM

casummer
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^ it begins.

7/26/2009 12:56:08 AM

Tarun
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i <3 this thread

You just can't have inequalities with imaginary numbers

7/26/2009 1:43:51 AM

lewisje
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Here's why: https://en.wikipedia.org/wiki/Ordered_field#Which_fields_can_be_ordered.3F

Basically, in every ordered field, every nonzero number is either negative (less than 0) or positive (greater than 0), and multiplying both sides of an inequality by a negative number changes its sense; this means that all nonzero square numbers are less than 0, and 0 cannot be expressed as a sum of nonzero squares, but because it is possible to express 0 as the sum of nonzero squares of complex numbers (i^2+1^2=0), the complex numbers cannot be ordered.

9/29/2013 6:19:18 PM

 Message Boards » Chit Chat » You can't have inequalities with imaginary numbers Page [1]  
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