Prove that $\sim$ is an equivalence relation Let $A = \{1, 2, 3,...,9\}$ and let $\sim$ be the relation on $A\times A$ defined by
   $(a,b) \sim(c,d)$ if $a+d = b+c$.
Prove that $\sim$ is an equivalence relation.
Really stuck on this question, maybe its the use of ~ that is throwing me off.  Any suggestions on how to do this?  Maybe I could swith ~ out for something else while solving then put it back in.
Thanks
 A: Clearly, the statement $a+d=b+c$ is just saying that $a-b=c-d$, i.e. two ordered pairs are similar if the difference of their terms is equal. To show it's an equivalence relation, we need reflexivity, transitivity and symmetry. The first and last are obvious. 
For transitivity, we need to show $x\sim y$ and $y\sim z$ implies $x\sim z$. Let $x=(x_1,x_2),y=(y_1,y_2),z=(z_1,z_2)$. The conditions tell you that $x_1-x_2=y_1-y_2$, but also that $y_1-y_2=z_1-z_2$. Clearly then, $x_1-x_2=z_1-z_2$. But this is exactly what it means for $x\sim z$!
A: Just use the definitions:
For reflexivity, you have to show that for any $(a,b)$, it is true that $(a,b)\sim (a,b)$, i.e. that $a+b=b+a$. Well, $a$ and $b$ are natural numbers, and those are commutative, so yes, this is true.
Symmetry: here you have to show that if $(a,b) \sim (c,d)$, then $(c,d) \sim (a,b)$, which is to say: show that if $a+d=b+c$, then $c+b=d+a$ ... can you show this?
Finally, now that I have shown what you need to prove to demonstrate reflexivity and symmetry, can you figure out what you need to show to demonstrate transitivity?
