Prove that the relation $R$ in set of integers $Z$ given by $R = \{(a,b) : (a-b) ext{ is divisible by number } 5\}$ is an equivalence relation.

**OR**

Find $g \circ f$ and $f \circ g$ if $f: R o R$ and $g: R o R$ are given by functions $f(x) = x^2 + 2$ and $g(x) = \dfrac{

Question

Prove that the relation $R$ in set of integers $Z$ given by $R = \{(a,b) : (a-b) 	ext{ is divisible by number } 5\}$ is an equivalence relation.

**OR**

Find $g \circ f$ and $f \circ g$ if $f: R 	o R$ and $g: R 	o R$ are given by functions $f(x) = x^2 + 2$ and $g(x) = \dfrac{x}{x-1}, x 
e 1$ respectively. Show that $g \circ f 
e f \circ g$.

Accepted Answer

>

> [!TLDR]
> Verify the three defining properties of an equivalence relation — reflexivity, symmetry, transitivity — directly from the divisibility condition.
>
> **Main question:** $R = \{(a,b): (a-b)	ext{ is divisible by }5\}$ on $\mathbb{Z}$.
>
> **Reflexive:** For any $a\in\mathbb{Z}$, $a-a=0$, and $0$ is divisible by $5$ (since $0 = 5	imes0$). So $(a,a)\in R$ for all $a$. Reflexive. ✓
>
> **Symmetric:** Suppose $(a,b)\in R$, i.e. $a-b = 5k$ for some integer $k$. Then $b-a = -(a-b) = -5k = 5(-k)$, which is also divisible by $5$. So $(b,a)\in R$. Symmetric. ✓
>
> **Transitive:** Suppose $(a,b)\in R$ and $(b,c)\in R$, i.e. $a-b=5m$ and $b-c=5n$ for integers $m,n$. Adding:
> $(a-b)+(b-c) = a-c = 5m+5n = 5(m+n)$
> which is divisible by $5$. So $(a,c)\in R$. Transitive. ✓
>
> Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation on $\mathbb{Z}$.
>
> **OR (alternative question):** $f(x)=x^2+2$, $g(x)=\dfrac{x}{x-1}$ ($x
e1$). Find $g\circ f$ and $f\circ g$, and show they differ.
>
> $g\circ f(x) = g(f(x)) = g(x^2+2) = \dfrac{x^2+2}{(x^2+2)-1} = \dfrac{x^2+2}{x^2+1}$
> …

Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

The three checks

A worked template

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