Modulos i

[Martin]

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?

[Tony]

yes, where did you get i * i ~ 9 from?

[Martin]

http://mathworld.wolfram.com/Congruence.html

[Tony]

eh, too much reading...

in short - you're wrong

[Andy]

couldnt u two just fite at home? instead of here? *sigh* its kinda getting old...

[Martin]

Ignore Tony. Does anyone have an answer to my question? More generically:

i ~ k (mod k^2 + 1) for all k in N.

[Martin]

*** 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

[md]

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...

[Tony]

Cornflake wrote:

therefore, martin, your wrong.

tony wrote:

in short - you're wrong

Just what I said in the beginning

[bugzpodder]

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...