# Addition and multiplication in Z/nZ

1. Oct 16, 2011

### autre

I need to show that addition and multiplication are well-defined in Z/nZ. So far, I've figured out that I need to work within this framework:

Suppose a ~ a' and b ~ b'. Try to prove a + b ~ a' + b'.

If a ~ a'...something
If b ~ b'...something

Any ideas?

2. Oct 16, 2011

### Deveno

it sounds like you have a hazy notion of why we need to check that addition and multiplication are well-defined, and what it involves.

after all, setting

3 (mod 7) + 4 (mod 7) = 0 (mod 7)

makes perfect sense.

but here's the thing:

3 (mod 7) isn't just a single number: it's a whole slew of numbers

3 (mod 7) = {......,-11,-4,3,10,17,24,......}

so when we're using 3 (mod 7) in a sum, we might pick any one of this infinite set as a "representative". so if we use -11, say, instead of 3, as a representative, we want to make sure that doesn't get us into trouble. so, even though:

3 (mod 7) looks like a single number, it's actually a stand-in, for an entire set.

now, what does it mean for a number k to be in the set 3 (mod 7)?

it means k = 3 + 7*something, or, as its usually put:

3-k is a multiple of 7. so, if

a ~ a', that is a = a' (mod n), what this MEANS is a-a' is a multiple of n, say sn.

b ~ b', so that b = b' (mod n), means b and b' differ by a multiple of n, b-b' = tn.

what you need to show, is if a' is in a (mod n), and b' is in b (mod n) (remember, these are sets), then a'+b' is in a+b (mod n), which is to say:

a+b - (a'+b') = kn, for some k (this integer k is what you are really trying to find. if you can find it, all is good. if you can't...houston, we have a problem).

now you KNOW a - a' = sn, and b - b' = tn. time to do a lil algebra now.