The relation $T$ on $\mathbb{R}\times \mathbb{R}$ given by $(x,y)T(a,b)$ iff $x^2+y^2=a^2+b^2$. Sketch the equivalence class of $(1,2)$; of $(4,0)$.

Question

The relation $T$ on $\mathbb{R}\times \mathbb{R}$ given by $(x,y)T(a,b)$ iff $x^2+y^2=a^2+b^2$. Sketch the equivalence class of $(1,2)$; of $(4,0)$.

$T$ is an equivalence relation on $\mathbb{R}\times \mathbb{R}$, because:

$T$ is reflexive on $\mathbb{R}\times \mathbb{R}$, and we can proving this as the following:

for all $(x,y) \in \mathbb{R}\times \mathbb{R}$, we have

$(x,y)T(x,y) \Leftrightarrow x^2+y^2=x^2+y^2$

$T$ is symmetric on $\mathbb{R}\times \mathbb{R}$, and we can proving this as the following:

for all $(x,y),(a,b) \in \mathbb{R}\times \mathbb{R}$, we have

$(x,y)T(a,b) \Leftrightarrow x^2+y^2=a^2+b^2$

$\Leftrightarrow a^2+b^2=x^2+y^2$

$\Leftrightarrow (a,b)T(x,y)$

$T$ is transitive on $\mathbb{R}\times \mathbb{R}$, and we can proving this as the following:

for all $(x,y),(a,b),(p,q) \in \mathbb{R}\times \mathbb{R}$, we have:

$(x,y)T(a,b) \land (a,b)T(p,q) \Leftrightarrow x^2+y^2=a^2+b^2 \land a^2+b^2=p^2=q^2$

$ \Leftrightarrow x^2+y^2+=p^2=q^2$

$ \Leftrightarrow (x,y)T(p,q)$.

We have,

• $(1,2)/ T =[(1,2)]=\{(x,y)\in \mathbb{R}\times \mathbb{R} | (x,y) T (1,2)\}=\{(x,y)\in \mathbb{R}\times \mathbb{R} | x^2+y^2=5\}$

• $(4,0)/ T =[(4,0)]=\{(x,y)\in \mathbb{R}\times \mathbb{R} | (x,y) T (4,0)\}=\{(x,y)\in \mathbb{R}\times \mathbb{R} | x^2+y^2=16\}$

Is that true please?

Answer 1

Yes, every thing is true. Your solution is fine !