Revision as of 18:08, 14 September 2008 by Dfreidin (Talk)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

It's been a while since I've taken Discrete Math... How do you do i mod n Z?


I believe this is how you do it:

n = iq + r, where q is the quotient and r is the remainder.

i mod n Z = r



Let a = i mod nZ, then a - i = nZ. This shows that Z divides a - i by n. We can also look at the formula as a = nZ + i which tells us that a is a product of Z and the remainder of it.


I understand the whole i mod nZ, but I am wondering what is the difference when the operator is addition instead of multiplicaton.

The Example in class (Z mod 6Z, +). We said a=1 but is not the identity. Is the only difference with the operator that 0 is the identity in addition, and 1 the identity in multiplication? -Neely


Yes, just like in normal arithemetic, 1 is the multiplicative identity, and 0 is the additive identity. It works the same way in modular arithmetic.

--Dfreidin 22:08, 14 September 2008 (UTC)

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett