Let S be the set of all triangles in a plane and let R be the relation on S defined by (Δ₁, Δ₂) ∈ R⇔ triangle Δ₁ is congruent to triangle Δ₂.
Reflexivity : For each triangle Δ ∈ S, we have Δ = Δ ⇒ (Δ, Δ) ∈ R for all Δ ∈ S.
⇒ R is reflexive on S.
Symmetry : Let Δ₁ Δ₂ ∈ S such that (Δ₁, Δ₂) ∈ R. Then (Δ₁ Δ₂) ∈ R ⇒ (Δ₁, Δ₂) ∈ R
∴ R is symmetric on S
Transitivity : Let Δ₁, Δ₂, Δ₃, ∈ S such that (Δ₁, Δ₂) ∈ R and (Δ₂, Δ₃) ∈ R
Then (Δ₁, Δ₂) ∈ Rand (Δ₂, Δ₃) ∈ R ⇒ (Δ₁, Δ₃) ∈ R.
∴ R is transitive on S.
Since R is reflexive, symmetric and transitive, R is an equivalence relation.

Clearly, A is the set of all first components in ordered pairs A x B and B is the set of all second components in ordered pairs A x B.
∴ A = {a, b} and B = {1, 2, 3}.

Set of concentric circles in a plane is an infinite set since number of concentric circles in a plane is infinite

Given f : R ⇾ R, defined by f(x) = cos x
We know f(0) = cos 0 = 1
and f(2π) = cos 2π = 1
∴ f(0) = f(2π) ⇒ 0 ≠ (2π)
∴ f is not one-one.
Since the values of cos x tie between -1 and 1, the range of f(x) is not equal to its co-domains.
∴ f is not onto.
Hence, f is neither one-one nor onto.

Let f and g denote the identity function and the modulus function.
Then f : R ⟶ R is defined as f(x) = x and g : R ⟶ R is defined as g(x) = |x|
g(x) = 0 ⇒ |x| = 0 ⇒ x =0
∴ The quotient of f by g is \({f\over g}:R-\{0\}⟶R\) and is defined as
 \(\left(f\over g \right)(x )={f(x)\over g(x)}={x\over |x|}=\begin{cases} {x\over x}=1\ if\ xx>0 \\{x\over -x}=-1\ if\ x<0 \end{cases}\)