Equivalence Relations, Domain and Range

This lesson covers three standard Class 12 questions: proving a relation is an equivalence relation, finding the domain of a square root function, and finding the range of a function built from $\cos x$.

When is a relation an equivalence relation?

A relation $R$ on a set is an equivalence relation when it satisfies all three of the following.

• Reflexive: $(a,a) \in R$ for every $a$.
• Symmetric: if $(a,b) \in R$ then $(b,a) \in R$.
• Transitive: if $(a,b) \in R$ and $(b,c) \in R$ then $(a,c) \in R$.

The divisibility-by-4 relation

Let

$A = \{\, x \in \mathbb{Z} : 0 \le x \le 12 \,\} = \{0,1,2,\dots,12\}$

and define

$R = \{\,(a,b) : |a-b| \text{ is divisible by } 4\,\}.$

We show $R$ is an equivalence relation.

Reflexive

For any $a \in A$,

$|a-a| = 0,$

and $0$ is divisible by $4$, so $(a,a) \in R$. Hence $R$ is reflexive.

Symmetric

Suppose $(a,b) \in R$, so $|a-b|$ is divisible by $4$. Since

$|b-a| = |a-b|,$

$|b-a|$ is also divisible by $4$, so $(b,a) \in R$. Hence $R$ is symmetric.

Transitive

Suppose $(a,b) \in R$ and $(b,c) \in R$, so both $|a-b|$ and $|b-c|$ are divisible by $4$. Then their sum is divisible by $4$, and

$|a-c| = |(a-b) + (b-c)| \le |a-b| + |b-c|,$

with $a-c = (a-b) + (b-c)$ a sum of two multiples of $4$, so $|a-c|$ is divisible by $4$. Hence $(a,c) \in R$ and $R$ is transitive.

Since $R$ is reflexive, symmetric and transitive, $R$ is an equivalence relation.

Elements related to 1

We want every $x \in A$ with $(x,1) \in R$, that is $|x-1|$ divisible by $4$. The multiples of $4$ available are

$|x-1| \in \{0,4,8,12\}.$

Solving each:

$x-1 = 0 \Rightarrow x = 1, \quad x-1 = 4 \Rightarrow x = 5,$ $x-1 = 8 \Rightarrow x = 9, \quad x-1 = 12 \Rightarrow x = 13.$

Since $13 \notin A$, the set of all elements related to $1$ is

$\{1, 5, 9\}.$

Domain of $f(x) = \sqrt{25 - x^2}$

The expression under the square root must be non-negative:

$25 - x^2 \ge 0 \;\Rightarrow\; x^2 \le 25 \;\Rightarrow\; -5 \le x \le 5.$

So the domain is

$D = [-5, 5].$

Range of $f(x) = \dfrac{1}{2 - \cos x}$

Start from the bounds on cosine:

$-1 \le \cos x \le 1.$

Multiply through by $-1$ and reverse the inequalities:

$-1 \le -\cos x \le 1.$

Add $2$ to each part:

$1 \le 2 - \cos x \le 3.$

Take reciprocals, which again reverses the inequalities:

$\tfrac{1}{3} \le \frac{1}{2 - \cos x} \le 1.$

So

$\tfrac{1}{3} \le f(x) \le 1,$

and the range is

$\left[\tfrac{1}{3}, 1\right].$

Key takeaways

• A relation is an equivalence relation exactly when it is reflexive, symmetric and transitive.
• Under the divisibility-by-4 relation on $\{0,\dots,12\}$, the elements related to $1$ are $\{1,5,9\}$.
• For a square root function, the domain comes from keeping the inside non-negative: here $[-5,5]$.
• The range of $\dfrac{1}{2-\cos x}$ is $\left[\tfrac{1}{3}, 1\right]$, found by chaining inequalities from $-1 \le \cos x \le 1$.