Revision: Relations and Functions >> Relations and Functions Maths Commerce (English Medium) Class 12 CBSE

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

• 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.

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 \[b \in B\].

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

A relation R on set A is said to be an equivalence relation if
(i) it is reflexive,
(ii) it is symmetric and
(iii) it is transitive.

Relation R on set A satisfying all the above three properties is an equivalence relation.

A relation R on a set A is said to be symmetric if

(a, b) ∈ R

⇒  (b, a)  ∈ R for all a b ∈ A

i.e.  aRb ⇒ bRa for all a , b ∈ A

A relation R on a set A is said to be transitive if

(a, b) ∈ R and (b, c) ∈ R

⇒ (a, c) ∈ R for all a, b , c ∈ R

i.e. aRb and bRc

⇒ aRc for all a, b, c ∈ R

A relation R on a (non-empty) set A is called an equivalence relation iff it is :

(i) reflexive, (ii) symmetric and (iii) transitive, i.e. iff

1. a R a for all a ∈ A
2. a R b implies b R a for all a, b ∈ A and
3. a R b, b R c implies a R c for all a, b, c ∈ A

If R is an equivalence relation on a set A, then the equivalence class of an element \[a \in A\] is the set of all elements of A related to a.

This means an equivalence class collects all elements that are considered “equivalent” under the given relation.

Let f: A → B and g: B → C be any two functions. Then, the composition of f and g, denoted by gof, is defined as a function gof: A → C given by

gof(x) = g[f(x)], ∀ x ∈ A

• Domain (gof) = Domain (f)
• g∘f(x) = g(f(x)) → first apply f, then g

A Function is called a self-inverse function if its inverse is the exact same as the original function.

• Condition: \[(f \circ f)(x) = I_x = x\].

• Examples: \[f(x) = \frac{5}{x}\] and \[g(x) = 7 - x\].

A function \[f: X \to Y\] is defined to be invertible if there exists a function \[g: Y \to X\] such that:

Where \[I_X\] and \[I_Y\] are identity functions on sets X and Y. The function g is the inverse of f, denoted as \[f^{-1}\].

An inverse function takes
us back where we started 

f : R → R, given by f(x) = [x]

It is seen that f(1.2) = [1.2] = 1 and f(1.9) = [1.9] = 1.

∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.

∴ f is not one-one.

Now, consider 0.7 ∈ R.

It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.

∴ f is not onto.

Hence, the greatest integer function is neither one-one nor onto.

To prove:

(f + g)oh = foh + goh

Consider:

((f + g)oh)(x)

= (f + g)(h(x))

= f(h(x)) + g(h(x))

= (foh)(x) + (goh)(x)

= {(foh) + (goh)}(x)

∴ ((f + g)oh)(x) = {(foh) + (goh)}(x)   ∀ x ∈ R

Hence, (f + g)oh = foh + goh.

To prove:

(f · g)oh = (foh)·(goh)

Consider:

((f · g)oh)(x)

= (f · g)(h(x))

= f(h(x))·g(h(x))

= (foh)(x)·(goh)(x)

= {(foh)·(goh)}(x)

∴ ((f · g)oh)(x) = {(foh)·(goh)}(x)   ∀ x ∈ R

Hence, (f · g) oh = (foh)·(goh).

• Empty relation and universal relation are called trivial relations.

• Identity relation always contains all self-pairs and is always reflexive.

• To check reflexivity, see whether every (a,a) is present in the relation.

• To check symmetry, see whether every (a,b) has the reverse pair (b,a).

• To check transitivity, see whether (a,b) and (b,c) together give (a,c).

• A relation is an equivalence relation only when all three properties—reflexive, symmetric, and transitive—are satisfied together.

• An equivalence relation must be reflexive, symmetric, and transitive.

• Equality is a standard example of an equivalence relation.

• Relations like “greater than” or “is mother of” are not equivalence relations because they do not satisfy all three properties.

• Equivalence classes are sets of mutually related elements.

• Equivalence relations and partitions of sets are closely connected.

• In \[g \circ f\], first apply f, then apply g.

• \[(g \circ f)(x) = g(f(x))\].

• Composition is defined only when the output of the first function is acceptable as input to the second function.

• In general, \[g \circ f \neq f \circ g\].

• Composition is associative whenever defined.

• Identity function leaves a function unchanged under composition.