$ (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:

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.

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,

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.

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.