Let, $Z$ = $\mathbb{N}$ $*$ $\mathbb{N}$ and ~ be the relation defined as follows:
$(a,b)$ ~ $(c,d)$ $\Longleftrightarrow$ $a + d = b + c$
I have been asked to prove this relation is reflexive and symmetric. My attempt:
proof (by direct proof)
Reflexivity $\rightarrow$ $(a,b)$ ~ $(a,b)$ $\rightarrow$$ a+b = b+a$.
$\therefore$ by the commutivity of $\mathbb{N}$ this is clearly true.
Thus, ~ is reflexive.
Symmetric $\rightarrow$ $(c,d)$ ~ $(a,b)$ $\rightarrow$$ c+b= d+a$.
$\therefore$ the commutivity of $\mathbb{N}$ $\rightarrow$ $ b+c= a+d$ $\rightarrow$ $(a,b)$ ~ $(c,d)$.
Thus, ~ is symmetric.
$\blacksquare$
Is this sufficient? I understand I was vague in parts of the proof but this is simply since I was short on time. Are all my assumptions/reasoning logical?
It looks good to me.
There is one small detail you can use to make it go faster (barring concerns about circular logic*), namely that $$ a + d = b + c $$ is equivalent to $$ a-b = c-d $$ which makes things easier.
*Your relation is, technically, how the integers and subtraction are constructed from the naturals and addition (with $(n, 0)$ and all pairs related to it being the integer $n$ for $n\in \Bbb N$, and then $-n$ is given by $(0,n)$ and all pairs related to that). So depending on your level it would be circular to approach it like this.