Find a function $f$ with $\ker(f) ={\sim}$ for equivalence relation $\sim$

Question

Find a function $f: \mathbb{R} \setminus \{0\} \rightarrow ??$, which satisfies

$$\ker(f) = ~ \sim$$

for the equivalence relation

$$x \sim y \iff \exists ~ p,q \in \mathbb{Q}: \frac{x}{y} = p+q \sqrt7 \text{ for }x,y \in \mathbb{R} \setminus \{0\},$$

where $\ker(f)$ is defined as $\ker(f):= \{(x,y) \in A \times A: f(x) = f(y) \}$

Answer 1

Recall that equivalence relations on a set $X$ correspond to partitions of $X$:

1. Given an equivalence relation $\sim$, the corresponding partition of $X$ is the set of equivalence classes of $X$ modulo $\sim$.

2. Given a partition $\mathcal{P}$ of $X$, the corresponding equivalence relation is the relation $x\sim y$ if and only if there exists $A\in\mathcal{P}$ such that $x,y\in P$.

In particular, given an equivalence relation $\sim$ on a set $X$, we can let $X/\sim$ be the seet of equivalence classes on $X$. Then we always have a map $$\pi\colon X\to X/\sim$$ given by $\pi(x) = [x]_{\sim}$, where $[x]_{\sim}$ is the equivalence class of $x$ under $\sim$. Note that by definition we will have $$\pi(x)=\pi(y)\iff [x]_{\sim}=[y]_{\sim}\iff x\sim y.$$ This is the "canonical projection" onto the set of equivalence classes.

In addition, one often defines the kernel of a function $f\colon A\to B$ to be the set $$\ker f= \{ (x,y)\in X\mid f(x)=f(y)\}.$$ This is the set that defines the equivalence relation "have the same image under $f$".

Given these facts, it should be straighforward to find a set and function that satisfy the desired properties.

Answer 2

One can say $x \sim y$ iff $\frac{x}{y}\in\mathbb{Q(\sqrt{7})}(:=\{a +b\sqrt{7}~|~a,b\in\mathbb{Q} \})$.
So if we want $ker(f)=~\sim ~$, we need: $f(x)=f(y)$ iff $\frac{x}{y}\in\mathbb{Q(\sqrt{7})}$.

Intuition:
If $f(1)=a$, we get $f(x)=a$ iff $x=\frac{x}{1}\in\mathbb{Q(\sqrt{7})}$.
So we get all elements from $\mathbb{Q}(\sqrt{7})$ go to $a$.
In the same manner, we get for some $r\in\mathbb{R}$:
if $f(r)=k$ than $f(x)=k$ iff $\frac{x}{r}\in\mathbb{Q(\sqrt{7})}\iff x\in r\cdot\mathbb{Q(\sqrt{7})}$
By this logic, we see that each equivalence class is of the type:
$r\cdot\mathbb{Q(\sqrt{7})}$.

We can indeed see that:$$r\cdot\mathbb{Q(\sqrt{7})} = k\cdot\mathbb{Q(\sqrt{7})}\iff r(a +b\sqrt{7})=k(c +d\sqrt{7})\iff\frac{r}{k}=\frac{a +b\sqrt{7}}{c +d\sqrt{7}}\in\mathbb{Q}(\sqrt{7})$$

And the same in the opposite way if $\frac{r}{k}\in\mathbb{Q}(\sqrt{7})$ we can get $r\cdot\mathbb{Q(\sqrt{7})} = k\cdot\mathbb{Q(\sqrt{7})}$.

Define $A:=\{r\cdot\mathbb{Q(\sqrt{7})}~|~r\in\mathbb{R}\backslash\{0\}\}$.

We can define $f$ as follows: $$f:\mathbb{R}\backslash \{0\}\rightarrow A\\x\mapsto x\cdot\mathbb{Q}(\sqrt{7})$$

We get $f(x)=f(y) \iff x\cdot\mathbb{Q(\sqrt{7})} = y\cdot\mathbb{Q(\sqrt{7})}\iff\frac{x}{y}\in\mathbb{Q}(\sqrt{7})\iff x \sim y$