Showing $(a,b) \sim (c,d)$ iff $a+d=b+c$ is transitive on $\mathbb{N}$

Question

I am given:

$a,b,c,d \in \mathbb{N}$, $a \neq c$, and $b \neq d$. The relation $\sim$ on $\mathbb{N}\times\mathbb{N}$ is defined by $(a, b) \sim (c, d)$ iff $a+d=b+c$ for all $(a,b), (c,d) \in \mathbb{N}\times\mathbb{N}$.

I need to prove transitivity, but can't figure out where to start. I understand transitivity is basically $a=b$, $b=c$, then $a=c$, but with an even pair of elements, I'm not sure where to go from here.

Answer 1

Let $(a,b),(c,d)$ and $(e,f)$ be such that $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$. To show $(a,b)\sim(e,f)$, observe that $ a+d=b+c $ is equivalent to $a-b=c-d$. So by assumption you have both $a-b=c-d$ and $c-d=e-f$, which implies $a-b=e-f$. But this means $a+f=b+e$, i.e. $(a,b)\sim(e,f)$.

Answer 2

In this case it's not $a=b, b=c$ then $a=c$. That is the statement of transitivity for $=$. But you need to show not that $=$, but that $\sim$ is transitive. So you need to show that if $(a_L, a_R)\sim (b_L, b_R)$ and $(b_L, b_R) \sim (c_L, c_R)$, then $(a_L, a_R)\sim(c_L, c_R)$.

Using the definition of $\sim$, You can translate the conditions $(a_L, a_R)\sim (b_L, b_R)$ and and $(b_L, b_R) \sim (c_L, c_R)$ into the more primitive conditions on the numbers $a_L, a_R, b_L, b_R, c_L, c_R$. Then you can use those primitive conditions and the definition of $\sim$ to show that $(a_L, a_R)\sim(c_L, c_R)$.