$ (a,b), (c,d) \in \mathbb{R} \times \mathbb{R}\ $ let us define a relation by $(a,b) \sim (c,d)$ if and only if $\ a + 2d = c+2b$

Question

Question:

For $ (a,b), (c,d) \in \mathbb{R} \times \mathbb{R}\ $ let us define a relation by $(a,b) \sim (c,d)$ if and only if $\ a + 2d = c+2b$

Is this an equivalence relation on $\mathbb{R} \times \mathbb{R}$?

My attempt:

Reflexive?

Notice that $ \forall (a,b) \in \mathbb{R} \times \mathbb{R}, a + 2b = a + 2b \implies ((a,b),(a,b)) \in R$.

Hence the relation is reflexive.

Symmetric?

If $ \ ((a,b), (c,d)) \in R \implies a + 2d = c + 2b \implies c + 2b = a + 2d \implies ((c,d), (a,b)) \in R $.

Hence the relation is symmetric.

Transitive?

If $ \ ((a,b), (c,d)) \in R$ and $ \ ((c,d), (e,f)) \in R \implies \ a + 2d = c+2b\ $ and $\ c + 2f = e + 2d \implies a + 2d + c + 2f = c+2b + e + 2d \implies a + 2f = 2b + e \implies a + 2f = e + 2b \implies ((a,b),(e,f)) \in R$

Hence the relation is transitive.

Therefore, the relation is an equivalence relation.

I am not quite sure if I have proved it correctly and if my approach is correct.

Answer 1

That's a correct direct approach. More generally note that $\,(a,b)\sim (c,d)\iff f(a,b) = f(c,d)\,$ for $\,f(u,v) = u-2v\,$ so we can apply the following very general equivalence kernel criterion.

---

It is straightforward to prove relations of form $\rm\, x\sim y {\overset{\ def}{\color{#c00}\iff}} f(x) = f(y)\, $ are equivalence relations.

More generally, suppose $\rm\ u\sim v\ \smash[t]{\overset{\ def}{\color{#c00}\iff}}\, f(u) \approx f(v)\ $ for a function $\rm\,f\,$ and equivalence relation $\,\approx.\, \ $ Then the equivalence relation $\rm\color{#0a0}{properties\ (E)}\,$ of $\,\approx\,$ transport (pullback) to $\,\sim\,$ along $\rm\,f$ as follows

• reflexive $\rm\quad\ \color{#0a0}{\overset{(E)}\Rightarrow}\, f(v) \approx f(v)\:\color{#c00}\Rightarrow\:v\sim v$

• symmetric $\rm\,\ u\sim v\:\color{#c00}\Rightarrow\ f(u) \approx f(v)\:\color{#0a0}{\overset{(E)}\Rightarrow}\:f(v)\approx f(u)\:\color{#c00}\Rightarrow\:v\sim u$

• transitive $\rm\ \ \ u\sim v,\, v\sim w\:\color{#c00}\Rightarrow\: f(u)\approx f(v),\,f(v)\approx f(w)\:\color{#0a0}{\overset{(E)}\Rightarrow}\:f(u)\approx f(w)\:\color{#c00}\Rightarrow u\sim w$

Such relations are called (equivalence) kernels. One calls $\, \sim\,$ the $\,(\approx)\,$ kernel of $\rm\,f.\,$ The equivalence classes $\,f_c = f^{-1}(c)\,$ are called the fibers or preimages, or level sets / curves of $f.$

Yours is the special case when $\,\approx\,$ is the equivalence relation of equality.

You can find many other examples of equivalence kernels in prior answers.

See also the more general notions of difference kernels and equalizers,

Answer 2

Yes, your proof is perfectly correct. We can simplify the proof a little bit, especially the transitivity part, if we observe that $a+2d=c+2b$ is equivalent to $a-2b=c-2d$, as this way each side of the condition uses entries from the same pair.

Answer 3

If you rearrange the defining equation as $a-c=2b-2d$, or $\frac{b-d}{a-c}=\frac12$, you can find a geometric interpretation. We have $(a,b)\sim (c,d)$ if and only if $(a,b)$ and $(c,d)$ lie on the same line with a slope of $\frac12$. Geometrically, it is clear that this is an equivalence: any point is on the same line as itself; if point $P$ is on the same line as point $Q$, then point $Q$ is on the same line as $P$; if $P$ and $Q$ are on the same line, and so are $Q$ and $R$, then so are $P$ and $R$.

The equivalence classes under this relation are all lines with $\frac12$ slope.