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Archive 1

Modular arithmetic

The prototypical example is modular arithmetic: for n a positive integer, two integers a and b are called congruent modulo n if a − b is divisible by n."

I think that there is no need to a and b to be a integers. Definition works well if a and b are real numbers with feature "a-b is integer" --Čikić Dragan 13:43, 23 February 2006 (UTC)

Things are most interesting when the numbers are integers, see modular arithmetic. If you are a computer guy, see modulo operation for the real number case. Oleg Alexandrov (talk) 23:20, 23 February 2006 (UTC)

LinearAlgebra

I'm not familiar enough with the concept of congruence in complex matricies to go ahead and edit this myself, but could "" congruence be called "" congrunece for greater clarity. That is, I'm under the impression that .

Kevmitch 01:52, 1 March 2006 (UTC)

i am not good with wiki's math symbols so can someone plz add the following information
a = b(mod n) implies n|(a − b)
and if n X (a - b) then <<a is not congruent to b (mod n)>>
if a = b(mod n) and m|n then it can be proved that (for integer m)
a = b (mod m)
--164.58.59.64 03:11, 26 March 2006 (UTC)
That's in the article titled modular arithmetic, and applies to one, but not all, of the congruence relations considered here. Michael Hardy 03:45, 26 March 2006 (UTC)

Untitled

I'm not sure if this is correct, so please help me out. Say I have to integers, e, x and n, and e and x are congruent modulo n. Then iss this equation correct for calculating x:

((n(e-1))+1) / e = x

--90.224.122.164 (talk) 16:25, 19 June 2008 (UTC)

Clear this up?

The article states that numbers are congruent mod m if:

(a-b) mod m = 0

or

(a/m) = (b/m)

Are these equivalent? Is it possible for: ((a-b) mod m) to be non-zero while (a mod m) = (b mod m)? —Preceding unsigned comment added by 157.228.91.78 (talk) 20:40, 3 March 2009 (UTC)