if $R = \{(x,y) \in \Bbb{R}^2 : x^2 + y^2 = 1\}$ and $S = R^2$. I need write $R \circ S$

Question

I haven’t had any mathematics since high school (have been a while) and now i’m trying with the basics in logic and set theory but of course i don’t get it easy.

I have this operation: if $R = \{(x,y)\in {\Bbb R}^2 : x^2 + y^2 = 1\}$ and $S = \Bbb R^2$. I need write $R \circ S$, but I’m having troubles. I know it isn’t $R$. Using the definition: $$ R \circ S = \{(x,z) \in {\Bbb R}^2 : \exists y\ [y \in {\Bbb R} \text{ and } (x,y) \in S \text{ and } (y,z) \in R]\} $$ Now I replace $$ R \circ S = \{(x,z)\in {\Bbb R}^2 : \exists y\ [y \in {\Bbb R} \text{ and } (x,y) \in R^2 \text{ and }x^2 + y^2 = 1\} $$ I thought it ends $R \circ S = R$, but it doesn’t, and I don’t know how to continue from here.

Answer 1

$R = \{\langle x,y\rangle \in {\Bbb R}^2 : x^2 + y^2 = 1\}$ and $S = \Bbb R^2$

$$\begin{align}R\circ S & =\{\langle x,z\rangle:\exists y~.(\langle x,y\rangle\in S\wedge\langle y,z\rangle \in R)\}&&\text{def'n of }\circ\\[1ex]&=\{\langle x,z\rangle:\exists y~.(\langle x,y\rangle\in\Bbb R^2\wedge(\langle y,z\rangle\in\Bbb R^2\wedge y^2+z^2=1))\}&&\text{def'n }S, R\\[1ex]&\text{as you had (mostly); note that it was $\langle y,z\rangle\in R$ rather that $\langle x,y\rangle$.}\\[1ex]R\circ S&=\{\langle x,z\rangle:\exists y~.(x\in\Bbb R\wedge y\in\Bbb R\wedge z\in\Bbb R\wedge y^2+z^2=1)\}&&\text{def'n of }\Bbb R^2\\[1ex]&=\{\langle x, z\rangle\in\Bbb R^2:\exists y\in\Bbb R~. y^2=1-z^2\}&&\text{rearrange}\\[1ex]&=\{\langle x,z\rangle\in\Bbb R^2:z^2\leqslant 1\}&&\text{when does it exist?}\\[1ex]&=\{\langle x,z\rangle\in\Bbb R^2:-1\leq z\leq 1\}\\[1ex]&= \Bbb R\times[-1~..~1]\end{align}$$

Answer 2

Hint. First observe that $(x,z) \in R^2$ if and only if there exists $y \in {\Bbb R}$ such that $(x,y) \in R$ and $(y,z) \in R$, that is $x^2 + y^2 = 1$ and $y^2 + z^2 = 1$. I claim that $(x,z) \in R^2$ if and only if $x^2 = z^2 \leqslant 1$. Can you see why?

Next the composition of two relations is usually defined as follows (what you wrote is the definition of $S \circ R$, unless you have other conventions) $$ (x,z) \in R \circ S \iff \text{there exists $y$ such that $(x,y) \in R$ and $(y,z) \in S$}. $$ Thus, in your case $$ (x,z) \in R \circ S \iff \text{there exists $y$ such that $x^2 + y^2 = 1$ and $ y^2 = z^2 \leqslant 1$} $$ Can you conclude from there?