1. ## proof

by definition, a~b means there exists a k in Z such that a-b=3k

now

if a~a' and b~b' show ab~a'b'

Note: S=Z=(0,+or-1,+or-2,.......) and
_ _ __
a*b = ab

2. Originally Posted by natester
by definition, a~b means there exists a k in Z such that a-b=3k

now

if a~a' and b~b' show ab~a'b'

Note: S=Z=(0,+or-1,+or-2,.......) and
_ _ __
a*b = ab

$a-a'=3n$ where n is an integer and

$b-b'=3m$ where m is an integer.

so $a=a'+3n$ and $b=b'+3m$

so

$ab=(a'+3n)(b'+3m)=a'b'+3a'm+3b'n+9mn=a'b'+3(a'm+b' n+3mn)$

the term $a'm+b'n+3mn$ is an integer by the closer property of the integers.

so lets call $a'm+b'n+3mn=q$ then

$ab=a'b'+3q$ so ab~a'b'