Integer proof equivalence class

Question

I've been searching online but I couldn't find help on this matter. How can I prove that $[(a,b)]+[(c,d)]=[(a+c,b+d)]$ is independent of the choice I make of representatives of the equivalence classes? Thanks!

Answer 1

You start with the assumption $$ (a, b) \sim (a', b'), \quad (c, d) \sim (c', d'), $$ and you have to prove $$ (a + c, b + d) \sim (a' + c', b' + d'). $$ Just appeal to the definition of $\sim$. It is just a matter of some simple calculations then.

Answer 2

$\rm\begin{eqnarray} {\bf Hint}\qquad\quad\ \ \ \rm (A,B) \sim (a,b) &\iff&\rm A+b\, =\, a+B\\ \rm (C,D) \sim (c,d) &\iff&\rm C+d\, =\, c+D\\ \rm(A\!+\!C,B\!+\!D)\sim (a\!+\!c,b\!+\!d)&\iff&\rm A\!+\!C\!+\!b\!+\!d\, =\, a\!+\!c\!+\!B\!+\!D \end{eqnarray}$

The definition of addition does not depend on choice of class rep's if the first two rhs equations imply the third rhs equation. That inference is, hopefully, clear from the above.