Cornflake wrote:
therefore, martin, your wrong.
tony wrote:
in short - you're wrong
Just what I said in the beginning
Computer Science Canada Modulos i |
| Author: | Martin [ Mon Nov 08, 2004 9:29 pm ] |
| Post subject: | Modulos i |
Let i be the number such that i * i = -1 Let ~ mean 'is congruent to' i * i ~ -1 (mod 10) i * i ~ 9 (mod 10) .: i ~ 3 (mod 10) Is there any flaw to this logic? |
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| Author: | Tony [ Mon Nov 08, 2004 10:03 pm ] |
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yes, where did you get i * i ~ 9 from? |
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| Author: | Martin [ Mon Nov 08, 2004 10:23 pm ] |
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http://mathworld.wolfram.com/Congruence.html |
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| Author: | Tony [ Mon Nov 08, 2004 10:32 pm ] |
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eh, too much reading... in short - you're wrong |
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| Author: | Andy [ Mon Nov 08, 2004 10:34 pm ] |
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couldnt u two just fite at home? instead of here? *sigh* its kinda getting old... |
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| Author: | Martin [ Mon Nov 08, 2004 10:51 pm ] |
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Ignore Tony. Does anyone have an answer to my question? More generically: i ~ k (mod k^2 + 1) for all k in N. |
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| Author: | Martin [ Tue Nov 09, 2004 12:49 am ] |
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*** not posted by martin*** since i^2 is -1 (as a complex number), using your logic the root of -1 = 3 mod 10. me thinks you have an error in your logic -- cornflake http://www.nxor.org |
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| Author: | md [ Tue Nov 09, 2004 1:05 am ] |
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the above was me in case anyone is confused... anyway, after further consideration, the proof is definitly wrong since any two numbers multiplyed together are always greater than or equal to 0 (in the reals), since the assumption is wrong, it follows that the rest of the solution is also wrong therefore, martin, your wrong. [edit] martin seems to be unsure about wether or not we're talking about the complex or real numbers... |
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| Author: | Tony [ Tue Nov 09, 2004 3:22 am ] |
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Cornflake wrote: therefore, martin, your wrong.
tony wrote: in short - you're wrong
Just what I said in the beginning |
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| Author: | bugzpodder [ Tue Nov 09, 2004 9:45 am ] |
| Post subject: | |
actually, those identities are not the point of finding a congruence eqn for i. Consider equations such as: x^2+y^2=0 (mod 25), x,y are positive integers then we can use first factor: (x+yi)(x-yi)=0 (mod 25), since i=7 (mod 25) then we get: (x+7y)(x-7y)=0 (mod 25) etc... |
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