Equivalences and bijections

Question

I have to show that the following are an equivalence relation on $A$ and find a bijection between $A/\sim$ and $B$.

I know that to show something is an equivalence relation it needs to satisfy the reflective, symmetry and transitive property, but I don't know how to prove that. I also don't know anything about bijections.

a) $A=\mathbb{Z}$ (integers); $m\sim n$ means that $m^2=n^2$;$B=\mathbb{N}$ (Natural numbers)

b) $A=\mathbb{R}\times \mathbb{R}; (x,y)$ is equivalent to $(x_1, y_1)$ means that $x^2+y^2=x_1^2+y_1^2$; $B=\{x \in \mathbb{R}|x \geq 0\}$

Any help would be appreciated!

Answer 1

a) You need to prove:

• $\color{red}{\text{Reflexive property. }} n\sim n$ for all $n\in \mathbb{Z}$, what means $n^2=n^2$, which is clear.
• $\color{red}{\text{Symmetry property. }}$ If $m \sim n$ for $n,m\in \mathbb{Z}$, then $n\sim m$. Since equality satisfies the symmetry property we have: $$m^2=n^2 \iff n^2=m^2;\,\,\,\,\text{i.e.}\,\,\,\,m\sim n \iff n\sim m$$
• $\color{red}{\text{Transitive property. }}$ If $\{m,n,p\}\subset\mathbb{Z}$ s.t. $m\sim n$ and $n\sim p$, then $m^2=n^2$ and $n^2=p^2$, so, by transitive property of the equality it follows $m^2=p^2$, which means $m\sim p$.

Therefore, $\sim$ is an equivalence relation on $\mathbb{Z}$. It is no hard to show that a equivalence class for $n\in\mathbb{Z}$ under $\sim$ is: $[n]=\{n,-n\}$. Now, we need to prove that there is an injective and surjective mapping $f:\mathbb{Z}/\sim \longrightarrow \mathbb{N}$. So, what about $x\mapsto |x|$?