So I am given this:
"Assume (a1,b1)Z∼(a2,b2) and (c1,d1)Z∼(c2,d2), where each component is a natural number. Then (a1+c1,b1+d1)Z∼(a2+c2,b2+d2).
Prove dis shiznit."
The relation Z∼ is defined as (a,b) = a - b where that difference is an integer. So (a,b)Z∼(c,d) if a+c=b+d.
The thing is, I have no idea whatsoever on how to combine 4 ordered pairs into a binary operation. Any and all help would be much appreciated!
I try adding (a1,b1) with (c1,d1), but I'm not quite sure how to prove that it is related to the sum of (a2,b2) and (c2,d2).
If $$ (a1, b1)Z\sim(a2, b2) $$ and $$ (c1, d1)Z\sim(c2, d2) $$ Then $$ a1 + a2 = b1 + b2 $$ and $$ c1 + c2 = d1 + d2 $$ hence $$ (a1 + c1) + (a2 + c2) = (b1 + d1) + (b2 + d2) $$ Due to the ring properties of the natural numbers.
This means that $$ ((a1 + c1), (b1 + d1))Z\sim((a2 + c2), (b2 + d2)) $$