# Relations

Let A B and be two non-empty sets. Then, a relation R from A to B is a subset of A $\times$ B.

Thus, R is a relation from A to B ⇒ R $\subseteq$ A $\times$ B. If R is a relation from a non-empty set A to a non-empty set B and if (a, b) $\in$ R, then we write aRb which is read as ‘a is related to b by the relation R.’ If (a, b) $\notin$ R , then we write aRb and it is read as ‘a is not related to b by the relation R’.

e.g., If R is a relation between two sets A = {1, 2, 3} and B = {1, 4, 9} defined as “square root of”.

Here, 1R1, 2R4, 3R9.

So, R = {(1, 1), (2, 4), (3, 9)}

**Domain and Range of Relations**

Let Rbe a relation from A to B. The domain of Ris the set of all those elements a $\in$ A such that (a, b) $\in$ R for some b $\in$ B.

So, Domain of R = {a $\in$ A : (a, b) $\in$ R, ∀ b $\in$ B}

and range of R is the set of all those elements b $\in$ B such that (a, b) $\in$ R for some a $\in$ A.

So, Range of R = {b $\in$ B : (a, b) $\in$ R, ∀ a $\in$ A} .

Here, B is called the codomain of R.

e.g., Let A = {1, 2, 3} and B = {3, 5, 6}

Let aRb ⇒ a < b.

Then, R = {(1, 5), (2, 5), (3, 5), (1, 6), (2, 6), (3, 6)}

So, Domain of R = {1, 2, 3}, range of R = {5,6} and codomain of R = {3, 5, 6}

**Note:**

- Let A and B be two non-empty finite sets having p and q elements respectively.
- Total number of relations form A to B = $pq^2$.

**Sample Problem 1**

(a) {0, 1, 2, 3, 4, 5}, {5, 6, 7, 8, 9, 10}

(b) {1, 2, 3, 4, 5}, {5, 6, 7, 8, 9, 10}

(c) {0, 1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}

(d) None of the above

**Sample Problem 2**

*R*= {(x, y): x $\in$ I, $x^2 + y^2 \leq 4$} is a relation in I, then domain of R is

(a) {0, 1, 2}

(b) {-2, -1, 0}

(c) {-2, -1, 0, 1, 2}

(d) None of these

*R*= {(x, y): x $\in$ I, $x^2 + y^2 \leq 4$} = {(0, 0), (0, -1), (0, 1), (0, -2), . . ., (-2, 0)}

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