I have this problem I do not know how to solve.
Given the function $f:\mathbb{Q}\rightarrow \mathbb{Q}_{\geq > 0}, x \mapsto \left | 2x+3 \right |$. (1) Determine the kernel of $f$ and (2) the transversal of the quotient set of $f$.
Here are some definitions:
A function $f:X\to Y$ defines an equivalence relation $\sim$ on $X$ in the following way: for $x_0,x_1\in X$, $x_0\sim x_1$ if and only if $f(x_0)=f(x_1)$.
Definition (quotient set): Given a set X and an equivalence relation $c$ on $X$. The set $X/c:=\{\left [ x \right ]_c\ : x \in X \}$ is called the quotient set of $X$ with regard to $c$.
Definition (transversal): Given a set $X$ and an equivalence relation $c$ on $X$. A transversal of $c$ with regard to $X$ is a subset $T$ of $X$ such that for each $K \in X/c$ there is exactly one $t \in T$ with $K = \left [ t \right ]_c$.
How do I solve this?
For (1): If I plot $f$, I can see that for $x,y \in \mathbb{Q}$ we have $f(x)=f(y) \Leftrightarrow \left |x \right |=\left | y+3 \right |$. But I do not know if this has anything to do with this.
I think you are wrong about (1), and $$f(x) = f(y) \Longleftrightarrow | x + 3/2 | = | y + 3/2 |,$$ from which it follows that $$\ker(f) = \{ (x,y) \in \mathbb{Q}^2 : | x + 3/2 | = | y + 3/2 | \}.$$
In what concerns (2), if I understood your definition right, the transversal is a set of representatives of the equivalence classes, in this case, it must be of the kernel of $f$ (although that is not very clear).
One such set of representatives could be, for example, the set $$\{ x \in \mathbb{Q} : x \geq -3/2 \}.$$