Consider the following relation on $\mathbb{R}$: $x$ ~ $y$ if any only if, $x^2-y^2 \in \mathbb{Z}$

1) Prove this is an equivalence relation

My attempt:

An equivalence relation has to be reflexive, symmetric, and transitive.

Reflexivity: For any $x ∈$ set $S$, $(x, x) ∈ R$ (reflexive property)

• $x^2$ - $x^2$ = $0 \in \mathbb{R}$, hence it is reflexive.

Symmetry: For any $x, y ∈$ set $S$, if $(x, y) ∈ R$, then $(y, x) ∈ R$ (symmetric property)

• $x^2$ - $y^2$ = $-(y^2-z^2) \in \mathbb{Z}$, hence it is symmetric.

Transitivity: For any $x, y, z ∈ S$, if $(x, y) ∈ R$ and $(y, z) ∈ R$, then $(x, z) ∈ R$ (transitive property)

• Let $a, b, c \in R$ such that $x, y \in R$ and $y, z \in R$
• We have that $x^2 - y^2 \in R$ and $y^2 - z^2 \in R$
• Let $x^2 - y^2 = m \in \mathbb{Z}$ and $y^2 - z^2 = n \in \mathbb{Z}$
• $y^2 - z^2$ = $- z^2 + y^2 = n$
• $((x^2-y^2) = m)$ + $((- z^2 + y^2) = n)$ = $x^2 - z^2 = m+n \in \mathbb{Z}$, hence it is transitive.

2) Let $C$ be the equivalence class of $0$ and $I$ the closed interval$[5, 6]$ . How many elements are in $C ∩ I$? Explain.

I have no idea how to approach this... I think that the equivalence class of $[0]$ = All $\in \mathbb{Z}$ where $x^2 - y^2 = 0$. But I don't know how to show how many elements are there in $I$.

Can someone please provide some assistance?

Solution 1:

You want the set of all numbers between $5$ and $6$ whose squares are integers. The squares are between $25$ and $36$. So the answer is the set of positive square roots of $25, 26,27,28,29, 30,31, 32, 33, 34, 35, 36$.