WBCHSE Class 12 Maths Solutions For Functions

Relations and Functions – Chapter 2 Functions

Function Let A and B be two nonempty sets. Then, a rule f which associates to each element x ∈ A, a unique element, denoted by f(x) of B, is called a function from A to B and we write, f: A → B.

f(x) is called the image of x, while x is called the pre-image of f(x).

Domain, Codomain and Range of a Function

Let f: A → B. Then, A is called the domain of f and B is called the codomain of f.

And, f(A) = {f(x): x ∈ A} is called the range of f.

Example 1 Let A = {1,2,3,4} and B = {1,4,9,16,25}.

Consider the rule f: A → B: f(x) = x² ∀ x ∈ A.

Then, each element in A has its unique image in B.

So, f is a function from A to B.

f(1) = 1² = 1, f(2) = 2² = 4, f(3) = 3² = 9, f(4) = 4² = 16.

Dom(f) = {1,2,3,4} = A, codomain (f) = {1,4,9,16,25} = B

and range (f) = {1,4,9,16}.

Clearly, 25 ∈ B does not have its pre-image in A.

Read and Learn More  Class 12 Math Solutions

Example 2 Let N be the set of all natural numbers.

Let f: N → N : f(x) = 2x ∀ x ∈ N.

Then, every element in N has its unique image in N.

So, f is a function from N to N.

Clearly, f(1) = 2, f(2) = 4, f(3) = 6 …, and so on.

Dom(f) = N, codomain (f) = N,

range (f) = {2,4,6,8,10..}.

Various Types of Functions

Many-One Function A function f: A → B is said to be many-one if two or more than two elements in A have the same image in B.

Example Let A = {-1,1,2,3} and B = {1,4,9}.

Let f: A → B: f(x) = x² ∀ x ∈ A.

Then, each element in A has a unique image under f in B.

∴ f is a function from A to B such that

f(-1) = (-1)² = 1; f(1) = 1² = 1; f(2) = 2² = 4 and f(3) = 3² = 9.

Two different elements, namely -1 and 1, have the same image 1 ∈ B.

Hence, f is many-one.

One-One or Injective Function

A function f: A → B is said to be one-one if distinct elements in A have distinct images in B.

f is one-one when f(x₁) = f(x₂) ⇒ x₁ = x₂.

Example Let N be the set of all natural numbers.

Let f: N → N : f(x) = 2x ∀ x ∈ N.

Then, f(x₁) = f(x₂) ⇒ 2x₁ = 2x₂

⇒ x₁ = x₂

∴ f is one-one

Onto or Surjective Function

A function f: A → B is said to be onto if every element in B has at least one pre-image in A.

Thus, if f is onto, then for each y ∈ B ∃ at least one element x ∈ A such that y = f(x).

Also, f is onto ⇔ range (f) = B.

Example Let N be the set of all natural numbers and let E be the set of all even natural numbers.

Let f: N → E: f(x) = 2x ∀ x ∈ N.

Then, y = 2x ⇒ x = \(\frac{1}{2}\)y.

Thus, for y ∈ E there exists \(\frac{1}{2}\)y ∈ N such that

\(f\left(\frac{1}{2} y\right)=\left(2 \times \frac{1}{2} y\right)=y \text {. }\)

∴ f is onto.

WBBSE Class 12 Functions Solutions

Into Function, A function f: A → B is said to be into if there exists even a single element in B having no pre-image in A.

Clearly, f is into ⇔ range (f) ⊂ B.

Example Let A = {2,3,5,7} and B = {0,1,3,5,7}.

Let f: A → B : f(x) = (x-2). Then,

f(2) = (2-2) = 0, f(3) = (3-2) = 1, f(5) = (5-2) = 3 and f(7) = (7-2) = 5.

Thus, every element in A has a unique image in B.

Now, ∃ 7 ∈ B having no pre-image in A.

∴ f is into

Note that range (f) = {0,1,3,5} ⊂ B.

Bijective Function

A one-on-one function is said to be bijective or a one-to-one correspondence.

Constant Function A function f: A → B is called a constant function if every element of A has the same image in B.

Example Let A = {1,2,3} and B = {5,7,9}.

Let f: A → B: f(x) = 5 for all x ∈ A.

Every element in A has the same image.

So, f is a constant function.

Remark The range of a constant function is a singleton set.

Identity Function The function I_A: A → A : I_A(x) = x ∀ x ∈ A is called an identity function on A.

Domain (I_A) = A and range (I_A) = A.

Equal Functions Two functions f and g having the same domain D are said to be equal if f(x) = g(x) ∀ x ∈ D.

Solved Examples

Example 1 Let f: N → N: f(x) = 2x for all x ∈ N. Show that f is one-one and into.

Solution

Given

Let f: N → N: f(x) = 2x for all x ∈ N.

We have

f(x₁) = f(x₂) ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂.

∴ f is one-one.

Let y = 2x. Then, x = \(\frac{y}{2}\).

If we put y = 3, then x = \(\frac{3}{2}\) ∉ N.

Thus, 3 ∈ N has no pre-image in N.

∴ f is into.

Hence, f is one-one and into.

Example 2 Show that the function f: R → R: f(x) = x² is neither one-one nor onto.

Solution

Given

f: R → R: f(x) = x²

We have f(-1) = (-1)² = 1 and f(1) = 1² = 1.

Thus, two different elements in R have the same image.

∴ f is not one-one.

If we consider -1 in the codomain R, then it is clear that there is no element in R whose image is -1.

∴ f is not onto.

Hence, f is neither one-one nor onto.

Example 3 Show that the function f: R → R: f(x) = |x| is neither one-one nor onto.

Solution

Given

f: R → R: f(x) = |x|

We have f(-1) = |-1| = 1 and f(1) = |1| = 1.

Thus, two different elements in R have the same image.

∴ f is not one-one.

If we consider -1 in the codomain R, then it is clear that there is no real number x whose modulus is -1.

Thus, -1 ∈ R has no pre-image in R.

∴ f is not onto.

Hence, f is neither one-one nor onto.

Understanding Functions in Class 12 Maths

Example 4 For any real number x, we define [x] = greatest integer function f: R → R: f(x) = [x] is neither one-one nor onto.

Solution

Given

f: R → R: f(x) = [x]

Clearly, [1.2] = 1 and [1.3] = 1.

Thus, two different real numbers have the same image.

∴ f(1.2) = 1 and f(1.3) = 1.

Thus, two different real numbers have the same image.

∴ f is not one-one.

Clearly, there is no real number x such that f(x) = [x] = 1.1.

So, f is not onto.

Hence, f is neither one-one nor onto.

Example 5 Let R₀ be the set of all nonzero real numbers. Show that f: R₀ → R₁₀ : f(x) = \(\frac{1}{x}\) is a one-one onto function.

Solution

Given

Let R₀ be the set of all nonzero real numbers.

We have

f(x₁) = f(x₂) ⇒ \(\frac{1}{x_1}=\frac{1}{x_2}\) ⇒ x₁ = x₂.

∴ f is one-one.

Again, y = \(\frac{1}{x}\) ⇒ x = \(\frac{1}{y}\).

Now, if y is a nonzero real number, then x = (1/y) is a nonzero real number such that f(1/y) = y.

Thus, each y in R₀ has its pre-image in R₀.

So, f is onto

Hence, f is one-one onto.

Example 6 Show that the function f: R → R; f(x) = x³ is one-one and onto.

Solution

Given

f: R → R; f(x) = x³

We have

f(x₁) = f(x₂) ⇒ x₁³ = x₂³

⇒ (x₁³ – x₂³) = 0

⇒ (x₁ – x₂)(x₁² + x₁x₂ + x₂²) = 0

⇒ \(\left(x_1-x_2\right)\left[\left(x_1+\frac{x_2}{2}\right)^2+\frac{3}{4} x_2^2\right]=0\)

⇒ (x₁ – x₂) = 0 [∵ \(\left(x_1+\frac{x_2}{2}\right)^2+\frac{3}{4} x_2^2 \neq 0\)]

⇒ x₁ = x₂

∴ f is one-one.

Let y ∈ R and let y = x³. Then, x = y^1/3 ∈ R.

Thus, for each y in the codomain R there exists y^1/3 in R such that f(y^1/3) = (y^1/3)³ = y.

∴ f is onto.

Hence, f is one-one onto.

Example 7 Show that the function f: R → R: f(x) = 3 – 4x is one-one onto and hence bijective.

Solution

Given

f: R → R: f(x) = 3 – 4x

We have

f(x₁) = f(x₂) ⇒ 3 – 4x₁ = 3 – 4x₂

⇒ -4x₁ = -4x₂

⇒ x₁ = x₂.

∴ f is one-one.

Now, let y = 3 – 4x. Then, \(x=\frac{(3-y)}{4}\)

Thus, for each y ∈ R (codomain of f), there exists \(x=\frac{(3-y)}{4}\) ∈ R

such that f(x) = \(f\left(\frac{3-y}{4}\right)=\left\{3-4 \cdot \frac{(3-y)}{4}\right\}\) = 3 – (3-y) = y.

This shows that every element in codomain of f has its pre-image in dom(f).

∴ f is onto.

Hence, the given function is bijective.

Example 8 Show that the function f: N → N, defined by

\(f(x)=\left\{\begin{array}{l}
x+1, \text { if } x \text { is odd } \\
x-1, \text { if } x \text { is even }
\end{array}\right.\) is one-one and onto.

Solution

Suppose f(x₁) = f(x₂).

Case 1 When x₁ is odd and x₂ is even

In this case, f(x₁) = f(x₂) ⇒ x₁ + 1 = x₂ + 1

⇒ x₁ = x₂.

This is a contradiction, since the difference between an odd integer and an even integer can never be 2.

Thus, in this case, f(x₁) ≠ f(x₂).

Similarly, when x₁ is even and x₂ is odd, then f(x₁) ≠ f(x₂).

Case 2 When x₁ and x₂ are both odd

In this case, f(x₁) = f(x₂) ⇒ x₁ + 1 = x₂ + 1

⇒ x₁ = x₂.

∴ f is one-one.

Case 3 When x₁ and x₂ are both even

In this case, f(x₁) = f(x₂) ⇒ x₁ – 1 = x₂ – 1

⇒ x₁ = x₂.

∴ f is one-one.

In order to show that f is onto, let y ∈ N (the codomain).

Case 1 When y is odd

In this case, (y+1) is even.

∴ f(y+1) = (y+1) – 1 = y.

Case 2 When y is even

In this case, (y-1) is odd.

∴ f(y-1) = y – 1 + 1 = y.

Thus, each y ∈ N (codomain of f) has its pre-image in dom(f).

∴ f is onto.

Hence, f is one-one onto.

Step-by-Step Solutions to Function Problems

Example 9 Show that f: N → N, defined by

\(f(n)=\left\{\begin{array}{l}
\frac{n+1}{2}, \text { if } n \text { is odd } \\
\frac{n}{2}, \text { if } n \text { is even }
\end{array}\right.\) is a many-one onto function.

Solution

We have

\(f(1)=\frac{(1+1)}{2}=\frac{2}{2}=1 \text { and } f(2)=\frac{2}{2}=1 \text {. }\)

Thus, f(1) = f(2) while 1 ≠ 2.

∴ f is many-one.

In order to show that f is onto, consider an arbitrary element n ∈ N.

If n is odd then (2n – 1) is odd, and

\(f(2 n-1)=\frac{(2 n-1+1)}{2}=\frac{2 n}{2}=n \text {. }\)

If n is even, then 2n is even and

\(f(2 n)=\frac{2 n}{2}=n.\)

Thus, for each n ∈ N (whether even or odd) there exists its pre-image in N.

∴ f is onto.

Hence, f is many-one onto.

Example 10 Show that the signum function f: R → R, defined by

\(f(x)=\left\{\begin{array}{r}
1, \text { if } x>0 \\
0, \text { if } x=0 \\
-1, \text { if } x<0
\end{array}\right.\) is neither one-one nor onto.

Solution

Clearly, f(2) = 1 and f(3) = 1.

Thus, f(2) = f(3) while 2 ≠ 3.

∴ f is not one-one.

Range(f) = {1,0,-1} ⊂ R.

So, f is into.

Hence, f is neither one-one nor onto.

Example 11 Let A = R – {R} and B = R – {1}. Let f: A → B: f(x) = \(\frac{x-2}{x-3}\) for all values of x ∈ A. Show that f is one-one and onto.

Solution

Given

Let A = R – {R} and B = R – {1}. Let f: A → B: f(x) = \(\frac{x-2}{x-3}\) for all values of x ∈ A.

f is one-one, since

f(x₁) = f(x₂) ⇒ \(\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}\)

⇒ (x₁-2)(x₂-3) = (x₁-3)(x₂-2)

⇒ x₁x₂ – 3x₁ – 2x₂ + 6 = x₁x₂ – 2x₁ – 3x₂ + 6

⇒ x₁ = x₂.

Let y ⊂ B such that y = \(\frac{x-2}{x-3}\).

Then, (x-3)y = (x-2) ⇒ x = \(\frac{(3 y-2)}{(y-1)}\).

Clearly, x is defined when y ≠ 1.

Also, x = 3 will give us 1 = 0, which is false.

∴ x ≠ 3.

And, \(f(x)=\frac{\left(\frac{3 y-2}{y-1}-2\right)}{\left(\frac{3 y-2}{y-1}-3\right)}=y \text {. }\)

Thus, for each y ∈ B, there exists x ∈ A such that f(x) = y.

∴ f is onto.

Hence, f is one-one onto.

Example 12 Let A and B be two nonempty sets. Show that the function f: (A x B) → (B x A): f(a,b) = (b,a) is a bijective function.

Solution

Given:

Let A and B be two nonempty sets.

f is one-one, since

f(a₁,b₁) = f(a₂,b₂) ⇒ (b₁,a₁) = (b₂,a₂)

⇒ a₁ = a₂ and b₁ = b₂

⇒ (a₁,b₁) = (a₂,b₂).

In order to show that f is onto, let (b,a) be an arbitrary element of (B x A).

Then, (b,a) ⊂ (B x A)

⇒ b ∈ B and a ∈ A

⇒ (a,b) ∈ (A x B).

Thus, for each (b,a) ∈ (B x A), there exists (a,b) ∈ A x B such that f(a,b) = (b,a).

∴ f is onto.

Thus, f is one-one onto and hence bijective.

Example 13 Find the domain and range of the real function f(x) = \(\sqrt{9-x^2}\).

Solution

Given

f(x) = \(\sqrt{9-x^2}\).

It is clear that f(x) = \sqrt{9-x^2} is not defined when (9 – x²) < 0, i.e., when x² > 9, i.e., when x > 3 or x < -3.

∴ dom(f) – {x ∈ R: -3 ≤ x ≤ 3}.

Also, y = \(\sqrt{9-x^2}\) ⇒ y² = (9-x²)

⇒ x = \(\sqrt{9-y^2}\).

Clearly, x is not defined when (9-y²) < 0.

But, (9-y²) < 0 ⇒ y² > 9

⇒ y > 3 or y < -3.

∴ range(f) = {y ∈ R: -3 ≤ y ≤ 3}.

Types of Functions Explained

Example 14 Find the domain and range of the real function, defined by f(x) = \(\frac{1}{\left(1-x^2\right)}\).

Solution

Given

f(x) = \(\frac{1}{\left(1-x^2\right)}\).

Clearly, \(\frac{1}{\left(1-x^2\right)}\) is not defined when 1 – x² = 0, i.e., when x = ±1.

∴ dom(f) = R – {-1,1}.

Also, \(\frac{1}{\left(1-x^2\right)} \Rightarrow\left(1-x^2\right)=\frac{1}{y}\)

⇒ \(x=\sqrt{1-\frac{1}{y}} .\)

Clearly, x is not defined when \(\left(1-\frac{1}{y}\right)\) < 0 or 1 < \frac{1}{y} or y < 1.

∴ range(f) = R – {y ∈ R : y ≥ 1}.

Example 15 Consider a function f : X → Y and define a relation R in X be R = {(a,b): f(a) = f(b)}. Show that R is an equivalence relation.

Solution

Given:

A function f : X → Y

Here, R satisfies the following properties:

(1) Reflexity

Let a ∈ X. Then,

f(a) = f(a) ⇒ (a,a) ∈ R.

∴ R is reflexive.

(2) Symmetry

Let (a,b) ∈ R. Then,

(a,b) ∈ R ⇒ f(a) = f(b) ⇒ f(b) = f(a) ⇒ (b,a) ∈ R.

∴ R is symmetric.

(3) Transitivity

Let (a,b) ∈ R and (b,c) ∈ R. Then,

(a,b) ∈ R, (b,c) ∈ R

⇒ f(a) = f(b) and f(b) = f(c)

⇒ f(a) = f(c)

⇒ (a,c) ∈ R.

∴ R is transitive.

Hence, R is an equivalence relation.

Composition of Functions

Let f: A → B and g: B → C be two given functions. Then, the composition of f and g, denoted by g o f is the function, defined by (g o f): A → C : (g o f)(x) = g{f(x)} ∀ x ∈ A.

Clearly dom(g o f) = dom (f).

Also, g o f is defined only when range (f) ⊆ dom(g).

Remark (f o g) is defined only when range (g) ⊆ dom (f).

And, dom(f o g) = dom(g).

Solved Examples

Example 1 Let f:{1,3,4} → g:{1,2,5} → {1,3} be defined as f = {(1,2), (3,5), (4,1)} and g = {(1,3),(2,3),(5,1)}. Find (g o f) and (f o g).

Solution

Given

Let f:{1,3,4} → g:{1,2,5} → {1,3} be defined as f = {(1,2), (3,5), (4,1)} and g = {(1,3),(2,3),(5,1)}.

Here range (f) = {1,2,5} and dom(g) = {1,2,5}.

Clearly, range(f) ⊆ dom(g).

∴ (g o f) is defined and dom(g o f) = dom(f) = {1,3,4}.

Now, (g o f)(1) = g{f(1)} = g(2) = 3;

(g o f)(3) = g{f(3)} = g(5) = 1;

(g o f)(4) = g{f(4)} = g(1) = 3.

Hence, (g o f) = {(1,3), (3,1), (4,3)}.

Again, range(g) = {1,3} and dom(f) = {1,3,4}.

Clearly, range(g) ⊆ dom(f).

∴ (f o g) is defined and dom(f o g) = dom(g) = {1,2,5}.

Now, (f o g)(1) = f{g(1)} = f(3) = 5;

(f o g)(2) = f{g(2)} = f(3) = 5;

(f o g)(5) = f{g(5)} = f(1) = 2.

Hence, (f o g) = {(1,5), (2,5), (5,2)}.

Example 2 Let R be the set of all real numbers. Let f: R → R: f(x) = cos x and let g: R → R: g(x) = 3x². Show that (g o f) ≠ (f o g).

Solution

Given:

Let R be the set of all real numbers. Let f: R → R: f(x) = cos x and let g: R → R: g(x) = 3x²

Let x be an arbitrary real numbers. Then,

(g o f)(x) = g{f(x)} = g(cos x) = 3(cos x)² = 3cos²x.

(f o g)(x) = f{g(x)} = f(3x²) = cos(3x²).

Taking x = 0, we have

(g o f)(0) = 3cos²0 = (3 x 1) = 3.

(f o g)(0) = cos(3 x 0) = cos 0 = 1.

∴ (g o f)(0) ≠ (f o g)(0).

Hence, g o f ≠ f o g.

Example 3 Let R be the set of all real numbers. Let f: R → R: f(x) = sin x and g: R → R: g(x) = x². Prove that g o f ≠ f o g.

Solution

Given

Let R be the set of all real numbers. Let f: R → R: f(x) = sin x and g: R → R: g(x) = x².

Let x be an arbitrary real number. Then,

(g o f)(x) = g{f(x)} = g(sin x) = (sin x)².

(f o g)(x) = f{g(x)} = f(x²) = sinx².

Clearly, (sin x)² ≠ sinx².

Hence, g o f ≠ f o g.

Example 4 Let f: R → R: f(x) = 8x³ and g: R → R: g(x) = x^1/3. Find (g o f) and (f o g) and show that g o f ≠ f o g.

Solution

Given

Let f: R → R: f(x) = 8x³ and g: R → R: g(x) = x^1/3.

Let x ∈ R. Then, we have

(g o f)(x) = g{f(x)} = g(8x³) = (8x³)^1/3 = 2x.

(f o g)(x) = f{g(x)} = f(x^1/3) = 8(x^1/3)³ = 8x.

∴ g o f ≠ f o g.

Example 5 Let f: R → R: f(x) = (x² – 3x + 2), find (f o f)(x).

Solution

Given

Let f: R → R: f(x) = (x² – 3x + 2)

We have

(f o f)(x) = f{f(x)} = f(x² 3x + 2) = f(y), where y = (x² – 3x + 2)

= (y² – 3y + 2)

= (x² – 3x + 2)² – 3(x² – 3x + 2) + 2

= (x⁴ – 6x³ + 10x² – 3x).

Example 6 If f: R → R: f(x) = (3 – x³)^1/3, show that (f o f)(x) = x.

Solution

Given

f: R → R: f(x) = (3 – x³)^1/3

We have

(f o f)(x) = f{f(x)} = f(3 – x³)^1/3

= f(y), where y = (3 – x³)^1/3

= (3 – y³)^1/3 = [3 – (3 – x³)]^1/3 [∵ y³ = (3 – x³)]

= (x³)^1/3 = x.

Hence, (f o f)(x) = x.

Common Questions on Functions and Their Solutions

Example 7 Let f: A → B, and let I_A and I_B be identity functions on A and B respectively. Prove that (f o I_A) = f and (I_B o f) = f.

Solution

Given

Let f: A → B, and let I_A and I_B be identity functions on A and B respectively.

Let x ∈ A and let f(x) = y. Then,

(f o I_A)(x) = f{I_A(x)}

= f(x) [∵ I_A(x) = x]

∴ (f o I_A) = f.

And, (I_B o f)(x) = I_B{f(x)}

= I_B(y) [∵ f(x) = y]

= y [∵ I_B(y) = y]

= f(x) [∵ y = f(x)].

∴ (IB o f) = f.

Hence, (f o I_A) = f and (I_B o f) = f.

Example 8 (Associativity) Let f: A → B, g: B → C and h: C → D. Then, prove that (h o g) o f = h o (g o f).

Solution

Given

Let f: A → B, g: B → C and h: C → D.

Let x ∈ A. Then,

{(h o g) o f}(x) = (h o g){f(x)}

= h[g{f(x)}]

= h[(g o f)(x)]

= {h o (g o f)}(x).

∴ (h o g) o f = h o (g o f).

Example 9 Let f: Z → Z: f(n) = 3n and let g: Z → Z, defined by

g(n)= \(\begin{cases}\frac{n}{3}, & \text { if } n \text { is a multiple of } 3 \\ 0, & \text { if } n \text { is nota multiple of } 3\end{cases}\)

Show that g o f = I_z and f o g ≠ I_z.

Solution

Let n be an arbitrary element of Z. Then,

(g o f)(n) = g{f(n)}

= g(3n) = \(\frac{3 n}{3}\) = n

= I_z(n).

∴ (g o f) = I_z.

Also, we have

(f o g)(1) = f{g(1)}

= f(0) [∵ g(1) = 0]

= (3 x 0) = 0 [∵ f(n) = 3n].

I_z(1) = 1 [∵ Iz(x) = x ∀ x ∈ Z].

∴ f o g ≠ I_z.

Example 10 Let \(A=R-\left\{\frac{3}{5}\right\} \text { and } B=R-\left\{\frac{7}{5}\right\}\). Let f: A → B: f(x) = \(\frac{7 x+4}{5 x-3}\) and g: B → A: g(y) = \(\frac{3 y+4}{5 y-7}\).

Show that (g o f) = I_A and (f o g) = I_B.

Solution

Let x ∈ A. Then,

(g o f)(x) = g[f(x)]

= \(8\left(\frac{7 x+4}{5 x-3}\right)\)

= g(y), where y = \(\frac{7 x+4}{5 x-3}\) …(1)

= \(\frac{3 y+4}{5 y-7}=\frac{3\left(\frac{7 x+4}{5 x-3}\right)+4}{5\left(\frac{7 x+4}{5 x-3}\right)-7}\) [using (1)]

= \(\frac{(21 x+12+20 x-12)}{(5 x-3)} \times \frac{(5 x-3)}{(35 x+20-35 x+21)}\)

= \(\frac{41 x}{41}\) = x = I_A(x).

∴ (g o f) = I_A.

Again, let y ∈ B. Then,

(f o g)(y) = f[g(y)]

= \(f\left(\frac{3 y+4}{5 y-7}\right)\)

= f(z), where z = \(\frac{3 y+4}{5 y-7}\) …(2)

= \(\frac{7 z+4}{5 z-3}=\frac{7\left(\frac{3 y+4}{5 y-7}\right)+4}{5\left(\frac{3 y+4}{5 y-7}\right)-3}\)

= \(\frac{(21 y+28+20 y-28)}{(5 y-7)} \times \frac{(5 y-7)}{(15 y+20-15 y+21)}\)

= \(\frac{41 y}{41}\) = y = I_B(y).

∴ (f o g) = I_B.

Hence, (g o f) = I_A and (f o g) = I_B.

Example 11 Let f: A → B and g: B → A such that (g o f) = I_A. Show that is one-one and g is onto.

Solution

Given

Let f: A → B and g: B → A such that (g o f) = I_A.

We have

f(x₁) = f(x₂) ⇒ g{f(x₁)} = g{f(x₂)}

⇒ (g o f)(x₁) = (g o f)(x₂)

⇒ I_A(x₁) = I_A(x₂)

⇒ x₁ = x₂.

∴ f is one-one.

In order to show that g is onto, let a ∈ A and let f(a) = b ∈ B.

Then, g(b) = g[f(a)] = (g o f)(a)

= I_A(a) [∵ g o f = I_A].

Thus, for each a ∈ A, there exists b ∈ B such that g(b) = a.

∴ g is onto.

Invertible Function

Let f: A → B. If there exists a function g: B → A such that g o f = I_A and f o g = I_B, then f is called an invertible function and g is called the inverse of f. We write, f^-1 = g.

Remark Clearly, f^-1 o f I_A and f o f^-1 = I_B.

Example Let f: R → R: f(x) = 2x + 3.

Let y = f(x). Then,

y = f(x) ⇒ y = 2x + 3

⇒ x = \(\frac{1}{2}\)(y – 3)

⇒ f^-1(y) = \(\frac{1}{2}\)(y – 3) [∵ y = f(x) ⇒ x = f^-1(y)]

Thus, we define:

f^-1: R → R: f^-1(y) = \(\frac{1}{2}\)(y-3).

Theorem 1 If f: A → B is one-one onto, then prove that f is an invertible function.

Proof

Let y ∈ B. Then, f being one-one onto, there exists a unique x ∈ A such that f(x) = y.

We define g: B → A: g(y) = x. Then,

(g o f)(x) = g[f(x)]

= g(y) [∵ f(x) = y]

= x [∵ g(y) =x]

= I_A(x).

∴ (g o f) = I_A

(f o g)(y) = f[g(y)]

= f(x)  [∵ g(y) = x]

= y    [∵ f(x) = y]

= I_B

∴ (f o g) = I_B.

Hence, f is invertible and f^-1 = g.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Applications of Functions in Real Life

Theorem 2 If f: A → B is an invertible function, then prove that f is one-one onto.

Proof

Let f: A → B be an invertible function. Then, there exists a function g: B → A such that

g o f = I_A and f o g = I_B.

Now, f(x₁) = f(x₂)

⇒ g{f(x₁)} = g{f(x₂)}

⇒ (g o f)(x₁) = (g o f)(x₂)

⇒ I_A(x₁) = I_A(x₂) [∵ g o f = I_A]

⇒ x₁ = x₂

∴ f is one-one.

Let y ∈ B. Then, g(y) ∈ A. Let g(y) = x. Then,

g(y) = x

⇒ f{g(y)} = f(x)

⇒ (f o g)(y) = f(x)

⇒ IB(y) = f(x) [∵ f o g = IB]

⇒ y = f(x).

Thus, for each y ∈ B there exists x ∈ A such that y = f(x).

∴ f is onto

Hence, f is one-one onto.

Remark f is invertible ⇔ f is one-one onto.

Solved Examples

Example 1 Let f: R → R: f(x) = 4x + 3 for all x ∈ R. Show that f is invertible and find f^-1.

Solution

Given

Let f: R → R: f(x) = 4x + 3 for all x ∈ R.

We have

f(x₁) = f(x₂) ⇒ 4x₁ + 3 = 4x₂ + 3

⇒ 4x₁ = 4x₂ ⇒ x₁ = x₂.

∴ f is one-one.

Again, y = 4x + 3 ⇒ x = \(\frac{(y-3)}{4}\)

Now, if y ∈ R (codomain of f), then there exists x = \(\frac{(y-3)}{4}\) ∈ R such that f(x) = \(f\left(\frac{y-3}{4}\right)=\left\{4 \cdot \frac{(y-3)}{4}+3\right\}=y \text {. }\)

∴ f is onto.

Thus, f is one-one onto and therefore invertible.

Now, y = f(x) ⇒ y = 4x + 3

⇒ x = \(\frac{(y-3)}{4}\)

⇒ \(f^{-1}(y)=\frac{(y-3)}{4}\) [∵ f(x) = y ⇒ x = f^-1(y)].

Thus, we define:

\(f^{-1}: R \rightarrow R: f^{-1}(y)=\frac{(y-3)}{4}\) for all y ∈ R.

Example 2 Let R₊ be the set of all positive real numbers. Let f: R₊ → [4, ∞[:f(x) = x² + 4. Show that f is invertible and find f^-1.

Solution

Given

Let R₊ be the set of all positive real numbers. Let f: R₊ → [4, ∞[:f(x) = x² + 4.

We have

f(x₁) = f(x₂) ⇒ x₁² + 4 = x₂² + 4

⇒ x₁² = x₂²

⇒ x₁² – x₂² = 0

⇒ (x₁ – x₂)(x₁ + x₂) = 0

⇒ x₁ – x₂ = 0 [∵ (x₁ + x₂) ≠ 0]

⇒ x₁ = x₂.

∴ f is one-one

Now, y = x² + 4 ⇒ x = \(\sqrt{y-4}\) in R⁺ such that

f(x) = f(\(\sqrt{y-4}\))=(\(\sqrt{y-4})^2+4=(y-4)+4=y \text {. }\)

∴ f is onto.

Thus, f is one-one onto and therefore invertible.

Now, y = f(x) ⇒ y = x² + 4

⇒ x = \(\sqrt{y-4}\)

⇒ \(f^{-1}(y)=\sqrt{y-4}\)

∴ \(f^{-1}:\left[4, \infty\left[\rightarrow R_{+}: f^{-1}(y)=\sqrt{y-4} .\right.\right.\)

Example 3 Let R⁺ be the set of all positive real numbers. Let f: R⁺ → R⁺ : f(x) = e^x for all x ∈ R⁺. Show that f is invertible and hence find f^-1.

Solution

Given

Let R⁺ be the set of all positive real numbers. Let f: R⁺ → R⁺ : f(x) = e^x for all x ∈ R⁺.

f is one-one, since

f(x₁) = f(x₂) ⇒ e^x₁ = e^x₂ ⇒ x₁ = x₂.

Now, for each y ∈ R⁺, there exists a positive real number, namely log y such that

f(log y) = e^logy = y.

∴ f is onto.

Thus, f is one-one onto and hence invertible.

We define:

f^-1: R⁺ → R⁺: f^-1(y) = log y for all y ∈ R⁺.

Example 4 Let A = \(\left\{x: x \in R, \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}\right\} \text { and } B=\{y: y \in R,-1 \leq y \leq 1\} \text {. }\)

Show that the function f: A → B: f(x) = sin x is invertible and hence find f^-1.

Solution

Here, A = \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) and B = [-1,1].

Also, f: A → B: f(x) = sin x.

f is one-one, since

f(x₁) = f(x₂) ⇒ sin x₁ = sin x₂

⇒ x₁ = x₂ {∵ x₁, x₂ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)}.

∴ f is one-one.

Also, range (f) = [-1,1] = B. So, f is onto.

Thus, f is one-one onto and hence invertible.

Now, y = f(x) ⇒ y = sin x

⇒ x = sin^-1y

⇒ f^-1(y) = sin^-1y.

Thus, we define:

f^-1: [-1,1] → \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) : f^-1(y) = sin^-1y.

Example 5 Let f: N → Y: f(x) = x², where y = range(f). Show that f is invertible and find f^-1.

Solution

Given

Let f: N → Y: f(x) = x², where y = range(f).

We have

f(x₁) = f(x₂) ⇒ x₁² = x₂²

⇒ x₁² – x₂² = 0

⇒ (x₁ – x₂)(x₁ + x₂) = 0

⇒ x₁ – x₂ = 0 [∵ x₁ + x₂ ≠ 0]

⇒ x₁ = x₂.

∴ f is one-one.

Since range(f) = Y, so f is onto.

Thus, f is one-one onto and therefore invertible.

Let y ∈ Y. Then, there exists x ∈ N such that f(x) = y.

Now, y = f(x) ⇒ y = x²

⇒ x = √y

⇒ f^-1(y) = √y. [∵ f(x) = y ⇒ x = f^-1(y)]

Thus, we define

f^-1: Y → N: f^-1(y) = √y.

Example 6 Let f: [1,1] → Y: f(x) = \(\frac{x}{(x+2)}\), x ≠ -2 and Y = range(f). Show that f is invertible and find f^-1.

Solution

Given

Let f: [1,1] → Y: f(x) = \(\frac{x}{(x+2)}\), x ≠ -2 and Y = range(f).

We have

f(x₁) = f(x₂) ⇒ \(\frac{x_1}{x_1+2}=\frac{x_2}{x_2+2}\)

⇒ x₁x₂ + 2x₁ = x₁x₂ + 2x₂

⇒ 2(x₁ – x₂) = 0

⇒ x₁ – x₂ = 0

⇒ x₁ = x₂.

∴ f is one-one.

Since range(f) = Y, so f is onto.

Thus, f is one-one onto and therefore invertible.

Let y ∈ Y. Then, there exists x ∈ [-1,1] such that f(x) = y.

Now, y = f(x) ⇒ y = \(\frac{x}{(x+2)}\)

⇒ x = \(\frac{2 y}{(1-y)}\)

⇒ f^-1(y) = \(\frac{2 y}{(1-y)}\)

Thus, we define:

\(f^{-1}:[-1,1] \rightarrow Y: f^{-1}(y)=\frac{2 y}{(1-y)}, y \neq 1 .\)

Example 7 Let f: N → Y: f(x) = 4x² + 12x + 15 and Y = range(f). Show that f is invertible and find f^-1.

Solution

Given

f: N → Y: f(x) = 4x² + 12x + 15 and Y = range(f).

f is one-one, since

f(x₁) = f(x₂) ⇒ 4x₁² + 12x₁ + 15 = 4x₂² + 12x² + 15

⇒ 4(x₁² – x₂²) + 12(x₁ – x₂) = 0

⇒ (X₁² – x₂²) + 3(x₁ – x₂) = 0

⇒ (x₁ – x₂)(x₁ + x₂ + 3) = 0

⇒ x₁ – x₂ = 0 [∵ x₁ + x₂ + 3 ≠ 0]

⇒ x₁ = x₂.

Also, range(f) = Y. So, f is onto.

Thus, f is one-one onto and therefore invertible.

Let y ∈ Y. Then, f being onto, there exists x such that y = f(x).

Now, y = f(x) ⇒ y = 4x² + 12x + 15

⇒ y = (2x+3)² + 6

⇒ (2x+3) = \(\sqrt{y-6}\)

⇒ x = \(\frac{1}{2}(\sqrt{y-6}-3)\)

⇒ f^-1(y) = \(\frac{1}{2}(\sqrt{y-6}-3) .\)

Thus, we define:

\(f^{-1}: Y \rightarrow N: f^{-1}(y)=\frac{1}{2}(\sqrt{y-6}-3)\)

f is invertible and  \(f^{-1}: Y \rightarrow N: f^{-1}(y)=\frac{1}{2}(\sqrt{y-6}-3)\)

Example 8 Let f: R → R: f(x) = 10x + 7. Find the function g: R → R such that g o f = f o g = I_R.

Solution

Given

Let f: R → R: f(x) = 10x + 7.

Clearly, g = f^-1 …(1)

Now, f(x₁) = f(x₂) ⇒ 10x₁ + 7 = 10x₂ + 7

⇒ 10x₁ = 10x₂

⇒ x₁ = x₂.

∴ f is one-one.

Now, y = f(x) ⇒ y = 10x + 7

⇒ x = \(\frac{(y-7)}{10}\)

Clearly, for each y ∈ R (codomain of f) there exists x ∈ R such that

f(x) = \(f\left(\frac{y-7}{10}\right)=\left\{10 \cdot\left(\frac{y-7}{10}\right)+7\right\}=y\)

∴ f is onto.

Thus, f is one-one onto and therefore, f^-1 exists.

We define: f^-1: R → R: f^-1(y) = \(\frac{y-7}{10}\).

Hence, g: R → R: g(y) = \(\frac{y-7}{10}\) [using (1)].

Example 9 Let f: W → W : f(n)= \(\begin{cases}(n-1), & \text { when } n \text { is odd } \\ (n+1), & \text { when } n \text { is even. }\end{cases}\)

Show that f is invertible. Find f^-1.

Solution

Let f(n₁) = f(n₂).

Case 1 When n₁ is odd and n₂ is even

In this case, f(n₁) = f(n₂) ⇒ n₁ – 1 = n₂ + 1

⇒ n₁ – n₂ = 2.

If n₁ is odd and n₂ is even, then (n₁ – n₂) ≠ 2.

Thus, we arrive at a contradiction.

So, in this case, f(n₁) ≠ f(n₂).

Similarly, when n₁ is even and n₂ is odd, then f(n₁)≠f(n₂).

Case 2 When n₁ and n₂ are both odd

In this case, f(n₁) = f(n₂) ⇒ n₁ – 1 = n₂ – 1

⇒ n₁ = n₂.

Case 3 When n₁ and n₂ are both even

In this case, f(n₁) = f(n₂) ⇒ n₁ + 1 = n₂ + 1

⇒ n₁ = n₂.

Thus, from all the cases, we get f(n₁) = f(n₂) ⇒ n₁ = n₂.

∴ f is one-one.

Now, we show that f is onto.

Let n ∈ W.

Case 1 When n is odd

In this case, (n-1) is even

and f(n-1) = (n-1) + 1 = n …(1)

Case 2 When n is even

In this case, (n+1) is odd

and f(n+1) = (n+1) – 1 = n …(2)

Thus, each n ∈ W has its pre-image in W.

∴ f is onto.

Thus, f is one-one onto and hence invertible.

Clearly, we have

\(f^{-1}(n)=\left\{\begin{array}{l}
(n-1), \text { when } n \text { is odd } \\
(n+1), \text { when } n \text { is even }
\end{array}\right.\) [using (1) and (2)].

Example 10 Let A = {1,2,3} and let f: A → A, defined by f = {(1,2), (2,3), (3,1)}. Find f^-1, if it exists.

Solution

Given

Let A = {1,2,3} and let f: A → A, defined by f = {(1,2), (2,3), (3,1)}.

We have f(1) = 2, f(2) = 3 and f(3) = 1.

Dom(f) = {1,2,3} = A and range (f) = {1,2,3} = A.

Clearly, different elements in A have different images.

∴ f is one-one.

Range(f) = A ⇒ f is onto.

Thus, f is one-one onto and therefore invertible.

Now, f(1) = 2, f(2) = 3 and f(3) = 1

⇒ f^-1(2) = 1, f^-1(3) = 2 and f^-1(1) = 3.

Hence, f^-1 = {(2,1), (3,2), (1,3)}.

Some Results On Invertible Functions

Theorem 1 Prove that an invertible function has a unique inverse.

Proof

Let f: A → B, which is one-one onto and therefore, invertible.

If possible, let it have two inverses, say g and h.

Then, (f o g) = I_B and (f o h) = I_B

⇒ (f o g)(y) = (f o h)(y) [each = I_B(y)]

⇒ f{g(y)} = f{h(y)} for all y ∈ B

⇒ g(y) = h(y) for all y ∈ B [∵ f is one-one]

∴ g = h.

Hence, f has a unique inverse.

Theorem 2 Let f be an invertible function. Then, prove that (f^-1)^-1 = f.

Proof

Let f: A → B, which is invertible.

In order to prove that (f^-1)^-1 = f, it is sufficient to show that f^-1 o f = I_A and f o f^-1 = I_B.

Clearly, f: A → B is one-one onto.

∴ f^-1: B → A is one-one onto.

Let x ∈ A and let f(x) = y. Then, f^-1(y) = x.

∴ (f^-1 o f)(x) = f^-1{f(x)}

= f^-1(y) [∵ f(x) = y]

= x

= I_A(x).

∴ f^-1 of = I_A.

Again, let y ∈ B.

Then, f being onto, there exists x ∈ A such that f(x) = y and therefore, f^-1(y) = x.

∴ (f o f^-1)(y) = f{f^-1(y)}

= f(x) [∵ f^-1(y) = x]

= y

= I_B(y).

∴ f o f^-1 = I_B.

Thus, f^-1 o f = I_A and f o f^-1 = I_B.

Hence, (f^-1)^-1 = f.

Theorem 3 Let f: A → B and g: B → C be one-one onto function. Prove that (g o f): A → C which is one-one onto and (g o f)^-1 = f^-1 o g^-1.

Proof

Let f: A → B be one-one onto and g: B → C be one-one onto.

We first show that g o f is one-one onto.

(g o f) is one-one since

(g o f)(x₁) = (g o f)(x₂)

⇒ g{f(x₁)} = g{f(x₂)}

⇒ f(x₁) = f(x₂) [∵ g is one-one]

⇒ x₁ = x₂   [∵ f is one-one].

Let z ∈ C. Then, g being onto, there exists y ∈ B such that g(y) = z.

Now, f being onto, there exists x ∈ A such that f(x) = y.

∴ z = g(y)

= g{f(x)} [∵ y = f(x)]

= (g o f)(x).

Thus, for each z ∈ C, there exists x ∈ A such that (g o f)(x) = z.

∴ (g o f) is onto.

Thus, (g o f) is one-one onto.

Now, f(x) = y ⇒ f^-1(y) = x.

And, g(y) = z ⇒ g^-1(z) = y.

Also, (g o f)(x) = z ⇒ (g o f)^-1(z) = x.

∴ (f^-1 o g^-1)(z) = f^-1{g^-1(z)}

= f-1(y) [∵ g^-1(z) = y]

= x [∵ f^-1(y) = x]

= (g o f)^-1(z).

Hence, (g o f)^-1 = (f^-1 o g^-1).

Chapter 3 Binary Operations

Closure Property An operation * on a nonempty set S is said to satisfy the closure property, if

a ∈ S, b ∈ S ⇒ a * b ∈ S for all a, b ∈ S.

Also, in this case, we say that S is closed for *.

An operation * on a nonempty set S, satisfying the closure property is known as a binary operation.

Example 1 (1) Addition on the set N of all natural numbers is a binary operation, since a ∈ N, b ∈ N ⇒ a + b ∈ N for all, a, b ∈ N.

(2) Multiplication on N is a binary operation, since a ∈ N, b ∈ N ⇒ a x b ∈ N for all a, b ∈ N.

Similarly, addition as well as multiplication is a binary operation on each one of the sets Z, Q, R anc C of integers, rationals, reals and complex numbers respectively.

Example 2 Let S be a nonempty set and P(S) be its power set. Then, the union operation on P(S) is a binary operation, since

A ∈ P(S), B ∈ P(S) ⇒ A ∪ B ∈ P(S) for all A, B ∈ P(S).

Similarly, intersection on P(S) is a binary operation, as

A ∈ P(S), B ∈ P(S) ⇒ A ∩ B ∈ P(S) for all A, B ∈ P(S).

Example 3 Subtraction on N is not a binary operation, since

3 ∈ N, 5 ∈ N but (3 – 5) = -2 ∉ N.

Subtraction on the set Z of all integers is a binary operation, since

a ∈ Z, b ∈ Z ⇒ a – b ∈ Z for all a, b ∈ Z.

Example 4 Addition on the set S of all irrationals is not a binary operation, since

2 + √3 ∈ S, 2 – √3 ∈ S but (2 + √3) + (2 – √3) = 4 ∉ S.

Multiplication on the set S of all irrationals is not a binary operation, since

√2 ∈ S, -√2 ∈ S but (√2)(-√2) = -2 ∉ S.

Example 5 Let N be the set of all natural numbers. Then, the exponential operation (a,b) → a^b is a binary operation on N, since a ∈ N, b ∈ N ⇒ a^b ∈ N for all a, b ∈ N.

Let Z be the set of all integers. The exponential operation (a,b) → a^b is not a binary operation on Z, since 2 ∈ Z, -3 ∈ Z but 2-3 = \(\frac{1}{2^3}\)=\(\frac{1}{8}\) ∉ Z.

Properties of a binary operation

(1) Associative law A binary operation * on a nonempty set S is said to be associative, if

(a*b)*c = a*(b*c) for all a, b, c ∈ S.

(2) Commutative law A binary operation * on a nonempty set S is said to be commutative, if

a*b = b*a for all a, b ∈ S.

(3) Distributive law Let * and ○ be two binary operations on a nonempty set s. We say that * is distributive over ○, if

a*(b ○ c) = (a*b)○(a*c) for all a, b, c ∈ S.

Example 1 Let R be the set of all real numbers. Then,

(1) addition on R satisfies the closure property, the associative law and the commutative law,

(2) multiplication on R satisfies the closure property, the associative law and the commutative law,

(3) multiplication distributes addition on R, since

a.(b+c) = a.b + a.c for all a, b, c ∈ R.

Examples of Composite and Inverse Functions

Example 2 Let Z be the set of all integers. Then, subtraction on Z is clearly a binary operation. But, it is neither commutative nor associative, as (3-5) ≠ (5-3) and (3-4) – 5 ≠ 3 – (4-5).

Identity Element Let * be a binary operation on a nonempty set S. An element e ∈ S, if it exists such that

a * e = e * a = a for all a ∈ S, is called an identity element of S, with respect to *.

Example 1 (1) For addition on R, zero is the identity element in R, since a + 0 = 0 + a = a for a ∈ R.

(2) For multiplication on R, 1 is the identity element in R, since a x 1 = 1 x a = a for all a ∈ R.

Example 2 Let P(S) be the power set of a nonempty set S. Then, Φ is the identity element for union on P(S) as

A ∪ Φ = Φ ∪ A = A for all A ∈ P(S).

Also, S is the identity element for intersection on P(S), since

A ∩ S = S ∩ A = A for all A ∈ P(S).

Inverse Of An Element Let * be a binary operation on a nonempty set S and let e be the identity element.

Let a ∈ S. We say that a is invertible if there exists an element b ∈ S such that a * b = b * a = e.

Also, in this case, b is called the inverse of a, and we write a^-1 = b.

Example 1 Consider addition on Z.

Clearly, the additive identity is 0, since

a + 0 = 0 + a = a for all a ∈ Z.

Also, corresponding to each a ∈ Z, there exists -a ∈ Z such that a + (-a) = (-a) + a = 0.

Thus, the additive inverse of a is -a.

Example 2 Consider multiplication on Z.

Clearly, 1 is the multiplicative identity on Z, since

a x 1 = 1 x a = a for all a ∈ Z.

Since 1 x 1 = 1, so the multiplicative inverse of 1 is 1.

Since (-1) x (-1) = 1, so the multiplicative inverse of (-1) is (-1).

No integer other than 1 and -1 has its multiplicative inverse in Z.

Solved Examples

Example 1 Let Z be the set of all integers. Then, addition on Z satisfies the following properties:

(1) Closure Property

We know that the sum of two integers is always an integer,

i.e., a ∈ Z, b ∈ Z ⇒ a + b ∈ Z for all a, b ∈ Z.

(2) Associative Law

For all a, b, c ∈ Z, we have

(a+b)+c = a+(b+c).

(3) Commutative Law

For all a, b ∈ Z, we have

a + b = b + a.

(4) Existence of Additive Identity

Clearly, 0 ∈ Z is the additive identity, since

0 + a = a + 0 = a for all a ∈ Z.

(5) Existence of Additive Inverse

For each a ∈ Z, there exists -a ∈ Z such that a + (-a) = (-a) + a = 0.

So, -a is the additive inverse of a.

Example 2 Let R₀ be the set of all nonzero real numbers. Then, multiplication on R₀ satisfies the following properties:

(1) Closure Property

We know that the product of two nonzero real numbers is a nonzero real numbers is a nonzero real number.

Thus, a ∈ R₀, b ∈ R₀ ⇒ ab ∈ R₀ for all a, b ∈ R₀.

(2) Associative Law

For all a, b, c ∈ R₀, we have

(ab)c = a(bc).

(3) Commutative Law

For all a, b ∈ R₀, we have ab = ba.

(4) Existence of Multiplicative Identity

Clearly, 1 ∈ R₀ is the multiplicative identity, since

1 x a = a x 1 for all a ∈ R₀.

(5) Existence of Multiplicative Inverse

For each a ∈ R₀ there exists \(\frac{1}{a}\) ∈ R₀ such that

\(a \times \frac{1}{a}=\frac{1}{a} \times a=1 .\)

Thus, the multiplicative inverse of a is \frac{1}{a}.

Example 3 Show that ther operation * on Z, defined by a * b = a + b + 1 for all a, b ∈ Z

satisfies (1) the closure property, (2) the associative law and (3) the commutative law

(4) Find the identity element in Z.

(5) What is the inverse of an element a ∈ Z?

Solution

(1) Closure Property

Let a ∈ Z, b ∈ Z. Then,

a * b = a + b + 1.

Now, a ∈ Z, b ∈ Z ⇒ a + b ∈ Z

⇒ a + b + 1 ∈ Z.

∴ * on Z satisfies the closure property.

(2) Associative Law

For all a, b, c ∈ Z, we have

(a*b)*c = (a + b + 1)*c

= (a + b + 1) + c + 1

= a + b + c + 2.

a*(b*c) = a*(b + c + 1)

= a + (b + c + 1) + 1

= a + b + c + 2.

∴ (a*b)*c = a*(b*c).

(3) Commutative law

For all a, b ∈ Z, we have

a*b = a + b + 1

= b + a + 1 [∵ a+b = b+a]

= b*a

(4) Existence of Identity Element

Let e be the identity element in Z.

Then, a*e = a ⇒ a + e + 1 = a

⇒ e = -1.

Thus, -1 ∈ Z is the identity element for *.

(5) Existence of Inverse

Let a ∈ Z and let its inverse be b. Then,

a * b = -1 ⇒ a + b + 1 = -1

⇒ b = -(2+a).

Clearly, 2 ∈ Z, a ∈ Z ⇒ -(2+a) ∈ Z.

Thus, each a ∈ Z has -(2+a) ∈ Z as its inverse.

Example 4 Show that the operation * on Q – {1}, defined by

a * b = a + b – ab for all a, b ∈ Q – {1} Satisfies (1) the closure property, (2) the associative law, (3) the commulative law.

(4) What is the identity element?

(5) For each a ∈ Q – {1}, find the inverse of a.

Solution

(1) Closure Property

Let a ∈ Q – {1} and b ∈ Q – {1}.

We know that Q is closed for addition, subtraction and multiplication.

∴ a + b – ab ∈ Q.

But, a*b = 1 ⇒ a + b – ab = 1

⇒ a(1-b) = (1-b)

⇒ a = 1,

Which is a contradiction since 1 ∉ Q – {1}.

∴ a * b ≠ 1.

Thus, a ∈ Q – {1}, b ∈ Q – {1} ⇒ a*b ∈ Q – {1}.

∴ * is a binary operation on Q – {1}.

(2) Associative Law

Let a, b, c ∈ Q – {1}. Then,

(a*b)*c = (a + b – ab)*c

= (a + b – ab) + c – a(b + c – bc)

= (a + b + c) – (ab + bc + ac) + abc.

And, a*(b*c) = a*(b + c – bc)

= a + (b + c – bc) – a(b + c – bc)

= (a + b + c) – (ab + bc + ac) + abc.

∴ (a*b)*c = a*(b*c).

Hence, * is associative.

(3) Commutative Law

Let a, b ∈ Q – {1}. Then,

a*b = a + b – ab

= b + a – ba [∵ + and . are commutative on Q – {1}]

= b*a.

Hence, * is commutative.

(4) Existence of Identity Element

Let e be the identity element.

Then, for all a ∈ Q – {1}, we have

a*e = a ⇒ a + e – ae = a

⇒ e(1-a) = 0 ⇒ e = 0 ∈ Q – {1}.

Now, a*0 = a + 0 – a x 0 = a.

And, 0*a = 0 + a – 0 x a = a.

Thus, 0 is the identity element in Q – {1}.

(5) Existence of Inverse

Let a ∈ Q – {1} and let a^-1 = b. Then,

a*b = 0 ⇒ a + b – ab = 0

⇒ a = ab – b = (a-1)b

⇒ \(b=\frac{a}{(a-1)} \in Q-\{1\}\)

∴ \(a^{-1}=\frac{a}{(a-1)} \in Q-\{1\}\)

Thus, each a ∈ Q-{1} has its inverse in Q-{1}.

Real-Life Applications of Function Concepts

Example 5 On the set N of all natural numbers, define the operation * on N be m*n = gcd(m,n) for all m, n ∈ N. Show that * is commutative as well as associative.

Solution

Given:

On the set N of all natural numbers, define the operation * on N be m*n = gcd(m,n) for all m, n ∈ N.

(1) Commutativity

For all m, n ∈ N, we have

gcd(m,n) = gcd(n,m).

∴ m*n = n*m ∀ m, n ∈ N.

Hence, * is commutative on N.

(2) Associativity

LEt m, n, p ∈ N. Then,

(m*n)*p = [gcd(m,n)]*p

= gcd[gcd{(m,n), p}]

= gcd[{m, gcd(n,p)}]

[∵ gcd of three numbers = gcd{(gcd of any two, third)}]

= gcd(m,n*p) = m*(n*p).

Hence, * is associative on N.

Example 6 consider the set A = {-1,1} with multiplication operation. We may prepare its composition table as shown below.

\(\begin{array}{r|rr}
\times & 1 & -1 \\
\hline 1 & 1 & -1 \\
-1 & -1 & 1
\end{array}\)

Multiplication on A satisfies the following properties:

(1) Closure Property

Since all the entries of the composition table are in A, so A is closed for multiplication.

(2) Associative Law

Since multiplication of integers is associative, in particular, multiplication on A is associative.

(3) Commutative Law

Since, every row of the table coincides with the corresponding column,

i.e., 1st row coincides with 1st column, 2nd row coincides with 2nd column.

So, multiplication is commutative on A.

(4) Existence of Identity

Clearly, 1 is the identity element in A, since

1 x 1 = 1 and (-1) x 1 = 1 x (-1) = -1.

(5) Existence of Inverse

It is clear from the table that

1 x 1 = 1 ⇒ inverse of (-1) is (-1).

New Operations

Example 7 Let A = {1,2,3,4,5}. Define an operation ∧ by a ∧ b = min{a,b}. Then, we may prepare its composition table as given below.

\(\begin{array}{l|lllll}
\wedge & 1 & 2 & 3 & 4 & 5 \\
\hline 1 & 1 & 1 & 1 & 1 & 1 \\
2 & 1 & 2 & 2 & 2 & 2 \\
3 & 1 & 2 & 3 & 3 & 3 \\
4 & 1 & 2 & 3 & 4 & 4 \\
5 & 1 & 2 & 3 & 4 & 5
\end{array}\)

Closure Property

Since all the entires of the composition table are from given set A, so A is closed for the operation ∧.

Commutative Law

In the given table every row coincides with corresponding column,

i.e., 1st row coincides with 1st column, 2nd row coincides with 2nd column, and so on.

∴ ∧ on A is commutative.

Example 8 Let A = {1,2,3,4,5}. Define an operation ∨ by a ∨ b = max{a,b}. Prepare its composition table. Show that A is closed for the given operation and that the given operation is commutative.

Solution

Given

Let A = {1,2,3,4,5}. Define an operation ∨ by a ∨ b = max{a,b}.

We prepare the table as given below.

\(\begin{array}{c|ccccc}
\vee & 1 & 2 & 3 & 4 & 5 \\
\hline 1 & 1 & 2 & 3 & 4 & 5 \\
2 & 2 & 2 & 3 & 4 & 5 \\
3 & 3 & 3 & 3 & 4 & 5 \\
4 & 4 & 4 & 4 & 4 & 5 \\
5 & 5 & 5 & 5 & 5 & 5
\end{array}\)

Closure Property

Since all the entries of the composition table are from the given set, so closure property is satisfied.

Commutative Law

Clearly, every row coincides with the corresponding column. So, commutative law is satisfied.