Basics of Relations & Functions

A relation is defined mathematically as a subset of a Cartesian product, and a function is a special kind of relation. These concepts are important because many real-world situations such as matching students to roll numbers, prices to products, and inputs to outputs can be represented through relations and functions.

A relation from set A to set B is any subset of the Cartesian product \[A \times B\].

So, if \[R \subseteq A \times B\], then R is a relation from A to B.

A function from set A to set B is a relation in which every element of A has exactly one image in B.

Condition for a Function

• every element of the domain must be used

• no element of the domain can have more than one image

• different elements may have the same image

An ordered pair is a pair of objects whose components occur in a special order. It is written by listing the two components in the specified order, separating them by a comma and enclosing the pair in parentheses. In the ordered pair (a, b), a is called the first component and b the second component.

• Two ordered pairs are equal only if their corresponding components are equal, so (a,b) = (c,d) if and only if a = c and b = d.

• In general, (a,b) ≠ (b,a)(a,b).

If A and B are two non-empty sets, then their Cartesian product is written as \[A \times B\].

• It is the set of all ordered pairs \[(a, b)\] such that \[a \in A\] and \[b \in B\].

• \[A \times B = \{(a, b) : a \in A, b \in B\}\].

• Domain: The set of all first components of the ordered pairs in a relation R is called the domain of the relation R.

• Codomain: If R is a relation from A to B, then the set B is called the co–domain of the relation R.

• Range: The set of all second components of all ordered pairs in a relation R is called the range of the relation.

Find x and y when (x + 3, 2) = (4, y − 3)

Solution: Using the definition of equality of two ordered pairs, we have

  (x + 3, 2) = (4, y − 3)
  ⇒ x + 3 = 4 and 2 = y − 3
  ⇒ x = 1 and y = 5

The Cartesian product A × A has 9 elements, among which are found (−1, 0) and (0, 1). Find the set A and the remaining elements of A × A.

Solution. Let n(A) = m.

Given n(A × A) = 9 ⇒ n(A) · n(A) = 9
⇒ m² = 9 ⇒ m = 3 (∵ m > 0)

Given (−1, 0) ∈ A × A ⇒ −1 ∈ A and 0 ∈ A.
Also, (0, 1) ∈ A × A ⇒ 0 ∈ A and 1 ∈ A.

Thus, −1, 0, 1 ∈ A but n(A) = 3.
Therefore, A = {−1, 0, 1}.

The remaining elements of A × A are
(−1, −1), (−1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1).

Let \[A = \{1, 2, 3, 4, 5\}\] and \[B = \{1, 4, 5\}\]. Let R be a relation such that \[(x, y) \in R\] implies \[x < y\]. List the elements of R and find its domain and range.

Solution:

\[R = \{(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)\}\]

• Domain = \[\{1, 2, 3, 4\}\] (set of all first elements)

• Range = \[\{4, 5\}\] (set of all second elements)

• Codomain = \[B = \{1, 4, 5\}\]

Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find

1. A × (B ∩ C)
2. (A × B) ∩ (A × C)

Solution:

(i) By the definition of the intersection of two sets,
  B ∩ C = {4}.

Therefore,
  A × (B ∩ C) = {(1, 4), (2, 4), (3, 4)}.

(ii) Now A × B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
and A × C = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}.

Therefore, (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}.

Given \[A = \{1, 2, 3, 4, 5\}\], find the ordered pairs satisfying \[x + y = 5\] where \[x, y \in A\]. Is this relation a function?

Solution:

The relation \[R = \{(1, 4), (2, 3), (3, 2), (4, 1)\}\]

Checking for function:

• Input 1 has exactly one output: 4 ✓

• Input 2 has exactly one output: 3 ✓

• Input 3 has exactly one output: 2 ✓

• Input 4 has exactly one output: 1 ✓

• Input 5 has no output (not used)

Since inputs 1, 2, 3, and 4 each have exactly one output and no input has multiple outputs, this is a function from the subset \[\{1, 2, 3, 4\}\] to \[A\].