Prove that the relation R in the set of integers Z given by R = {(a, b) : (a − b) is divisible by number 2} is an equivalence relation.

**OR**

Find g∘f and f∘g if f : R→R and g : R→R are given by f(x) = cos x and g(x) = 3x². Show that g∘f ≠ f∘g.

Question

Prove that the relation R in the set of integers Z given by R = {(a, b) : (a − b) is divisible by number 2} is an equivalence relation.

**OR**

Find g∘f and f∘g if f : R→R and g : R→R are given by f(x) = cos x and g(x) = 3x². Show that g∘f ≠ f∘g.

Accepted Answer

>

> [!TLDR]
> Check the three defining properties — reflexivity, symmetry, transitivity — directly from the divisibility condition; all three hold, so $R$ is an equivalence relation.
>
> Given $R = \{(a,b): a,b\in\mathbb{Z},\ (a-b)	ext{ is divisible by }2\}$.
>
> **Concept:** A relation $R$ on a set is an **equivalence relation** if it is reflexive, symmetric, and transitive.
>
> **Step 1 — Reflexivity:** For any $a\in\mathbb{Z}$, $a-a = 0$, and $0$ is divisible by $2$. So $(a,a)\in R$ for every $a$. Hence $R$ is reflexive.
>
> **Step 2 — Symmetry:** Suppose $(a,b)\in R$, i.e. $a-b = 2k$ for some integer $k$. Then $b-a = -(a-b) = -2k = 2(-k)$, which is also divisible by $2$. So $(b,a)\in R$. Hence $R$ is symmetric.
>
> **Step 3 — Transitivity:** Suppose $(a,b)\in R$ and $(b,c)\in R$, i.e. $a-b=2k$ and $b-c=2m$ for integers $k,m$. Adding:
> $$a-c = (a-b)+(b-c) = 2k+2m = 2(k+m)$$ …

Q.Prove that the relation R in the set of integers Z given by R = {(a, b) : (a − b) is divisible by number 2} is an equivalence relation. OR Find g∘f and f∘g if f : R→R and g : R→R are given by f(x) = cos x and g(x) = 3x². Show that g∘f ≠ f∘g.

Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

The three checks

A worked template

Q.Prove that the relation R in the set of integers Z given by R = {(a, b) : (a − b) is divisible by number 2} is an equivalence relation. OR Find g∘f and f∘g if f : R→R and g : R→R are given by f(x) = cos x and g(x) = 3x². Show that g∘f ≠ f∘g.

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