Question: On the set $N\times N$ define $(m,n)\simeq(k,l)$ if $m+l=n+k$.

Draw a sketch of $N\times N$ that shows several equivalence classes. (hint: sketch points on graph paper).

I'm not quite sure how to sketch the equivalence classes. I know that the following is in the relation, but do I just graph the points? Also, how do I find out how many equivalence classes there are and how do I distinguish them? Are there $4$ different equivalence classes ($1$ for $m$, $1$ for $n$, $1$ for $k$, and $1$ for $l$)? I know that if my relation were $\mod p$, there would be $p$ equivalence classes.

Answer: 
The condition $x^2\ge0$, which implies $2+x^2>0$ just tells you that the function is defined on $\mathbb{R}$, but not that it is surjective. For instance, the function
$$
g(x)=\frac{1}{\sqrt{2+x^2}}
$$
is not surjective if considered as $g\colon\mathbb{R}\to\mathbb{R}$, because $0<g(x)\le\frac{1}{\sqrt{2}}$ for all $x$.

If you're allowed to use calculus, surjectivity and injectivity follow from considering these facts:

1. $\displaystyle\lim\_{x\to-\infty}f(x)=-\infty$
2. $\displaystyle\lim\_{x\to\infty}f(x)=\infty$
3. $\displaystyle f'(x)=\sqrt{2+x^2}+\frac{x^2}{\sqrt{2+x^2}}>0$

Therefore $f$ is increasing and assumes every real value.

Without calculus it's a bit more difficult. Let's try solving the equation
$$
y=x\sqrt{2+x^2}
$$
with respect to $x$. First we note that if $x<0$, then also $y<0$ and conversely. Now we can square:
$$
y^2=2x^2+x^4
$$
or
$$
x^4+2x^2-y^2=0
$$
that furnishes
$$
x^2=-1+\sqrt{1+y^2}
$$
(the other solution of the quadratic equation must be discarded, because it's negative). Thus we can write
$$
|x|=\sqrt{\sqrt{1+y^2}-1}
$$
and, remembering the consideration above, we have
$$
x=\begin{cases}
\sqrt{\sqrt{1+y^2}-1} & \text{if $y>0$}\\[2ex]
0 & \text{if $y=0$}\\[2ex]
-\sqrt{\sqrt{1+y^2}-1} & \text{if $y<0$}\\[1ex]
\end{cases}
$$
and this proves both surjectivity and injectivity, because the function has an inverse.
**Hint** Let $a=x^2 \geq 0$ and $b=y^2 \geq 0$. The equation reduces to $$2a+a^2=2b+b^2$$