Talk:Congruence relation/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 "$${\scriptstyle *}$$" congruence be called "$${\scriptstyle \dagger}$$" congrunece for greater clarity. That is, I'm under the impression that $${\scriptstyle P^* = P^{\dagger}}$$.

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 <>


 * 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)