Thread: 2 modular arithmetic proofs

Prove there is an integer e such that
(i) [a]modn + [e]modn = [e]modn + [a]modn = [a]modn for all integers a

(ii) for all integers a, there exists an integer b such that
[a]modn +[b]modn = [b]modn + [a]modn = [e]modn

Originally Posted by jsteel2

Prove there is an integer e such that
(i) [a]modn + [e]modn = [e]modn + [a]modn = [a]modn for all integers a

(ii) for all integers a, there exists an integer b such that
[a]modn +[b]modn = [b]modn + [a]modn = [e]modn

I'm not used to the [a] notation, but this is simply (i) additive identity and (ii) additive inverse. That is, 0 and -a (mod n).

If e isn't restricted to the set {0,1,...,n-1} then there in fact exist an infinite number of integers e, of the form {0, n, -n, 2n, -2n, ...}. Similar with integers b in part (ii), integers of the form -a + kn.