Question

Prove that if m is composite then the set \( \mathbb{Z_m}-\{0\}\) is not closed under multiplication

Answer 1

if it is composite, then it factors right?

Answer 2

yup

Answer 3

say \(m=pq\)

Answer 4

and since \(pq=m=0\) you are done

Answer 5

it has to do with modulus a lil bit

Answer 6

\(\mathbb{Z_m}=\{0,1,2,...,m-1\}\)

Answer 7

yuppppp

Answer 8

with all operations mod m

Answer 9

so if \(m\) is composite, there are \(p<m, q<m\) with \(pq=m\) but modulo \(m\) it is \(pq=0\)