17.12 - Miscellaneous Examples

In the previous section, we saw identity element and inverse element. In this section, we will see some miscellaneous examples.

Solved example 17.41
If R₁ and R₂ are equivalence relations in a set A, show that R₁∩R₂ is also an equivalence relation.
Solution:
Let us assume that, the set A = {a₁, a₂, a₃, a₄, . . . , a_n}
Part (i): Proving that, the intersection is reflexive.
1. R₁ is an equivalence relation in set A.
• So all reflexive pairs like (a₁,a₁), (a₂,a₂), (a₃,a₃), . . .(a_n,a_n) will be present in the set R₁.
2. R₂ is also an equivalence relation in set A.
• So all reflexive pairs like (a₁,a₁), (a₂,a₂), (a₃,a₃), . . .(a_n,a_n) will be present in the set R₂ also.
3. The same reflexive pairs are present in both R₁ and R₂.
• So those reflexive pairs will be present in R₁∩R₂ also.
Therefore, R₁∩R₂ is reflexive.

Part (ii): Proving  that, the intersection is symmetric.
1. Suppose that, a random pair (a₃,a₇) is present in the intersection.
   ♦ Then this (a₃,a₇) will be present in the set R₁.
   ♦ This (a₃,a₇) will be present in the set R₂ also.
2. (a₃,a₇) is present in R₁.
• But R₁ is an equivalence relation. So the symmetric pair (a₇,a₃) will be present in R₁.
3. (a₃,a₇) is present in R₂.
• But R₂ is an equivalence relation. So the symmetric pair (a₇,a₃) will be present in R₂.
4. From (2) and (3), we see that:
(a₇,a₃) is present in both R₁ and R₂.
5. From (1) and (4) we see that:
The symmetric pairs (a₃,a₇) and (a₇,a₃) are present in the intersection.
6. In this way, all symmetric pairs will be present in the intersection.
So the intersection is a symmetric relation.

Part (iii): Proving that, the intersection is transitive.
1. Suppose that, two random pairs (a₃,a₇) and (a₇,a₂) are present in the intersection.
   ♦ Then these two pairs will be present in R₁.
   ♦ These two pairs will be present in R₂ also.
2. (a₃,a₇) and (a₇,a₂) are present in R₁.
• But R₁ is a symmetric relation. So the transitive pair (a₃,a₂) will be present in R₁.   
3. (a₃,a₇) and (a₇,a₂) are present in R₂.
• But R₂ is a symmetric relation. So the transitive pair (a₃,a₂) will be present in R₂.
4. From (2) and (3), we see that:
• (a₃,a₂) will be present in both R₁ and R₂.
• So it will be present in the intersection also.
5. From (1) and (4),we see that:
• The intersection contains (a₃,a₇), (a₇,a₂) and (a₃,a₂)
Therefore, the intersection is transitive.     
◼ From parts (i), (ii) and (iii), we see that, the intersection is reflexive, symmetric and transitive. So the intersection is an equivalence relation.

Solved example 17.42
Let R be the relation on set A of ordered pairs of positive integers defined by (x,y)R(u,v) if and only if xv = yu. Show that R is an equivalence relation.
Solution:
1. Set A is a set of ordered pairs.
• Each of those ordered pairs will contain +ve integers.
For example: (3,5), (1,2), (7,11) etc.,

2. The given relation is on A.
• That means, we need to consider the set A × A
• Each element in A × A will be a pair of ordered pairs.
For example: [(11,3),(5,4)], [(7,5),(8,2)], [(14,8),(9,3)] etc.,
• In general, we can write:
Each element in A × A will be a pair of ordered pairs in the form [(x,y),(u,v)]

3. The set R will contain those elements from A × A, which satisfy the condition: xv = yu.
• We need to prove that R is an equivalence relation.

4. First we check whether R is reflexive.
(i) If R is reflexive, then a possible random element in R is: [(a₃,a₇),(a₃,a₇)].
(ii) Let us check whether this element satisfies the condition xv = yu.
• Here, x = u = a₃ and y = v = a₇
• We can write:
xv = a₃ a₇  and  yu = a₇ a₃ = a₃ a₇
(iii) We see that xv = yu.
• So all reflexive elements are eligible to be included in R.
Therefore, R is a reflexive relation.   

5. Next we check whether R is symmetric.
(i) Suppose that a random element [(a₃,a₇),(a₂,a₉)] is present in R.
• It's symmetric element is [(a₂,a₉),(a₃,a₇)]. Is this symmetric element present in R? Let us check.
(ii) [(a₃,a₇),(a₂,a₉)] is present in R.
• That means, this element satisfies the condition: xv = yu
• That means, a₃ a₉ = a₇ a₂.
(iii) If the symmetric element is to be present in R, it must also satisfy the condition xv = yu.
• That means, a₂ a₇ must be equal to a₉ a₃.
• From (ii), we see that, they are indeed equal.
(iv) So we can write:
• If [(a₃,a₇),(a₂,a₉)] is present in R, then the symmetric element [(a₂,a₉),(a₃,a₇)] will also be present in R.
• So all symmetric elements are eligible to be included in R.
Therefore, R is a symmetric relation.

6. Finally, we check whether R is transitive.
(i) Suppose that two random elements [(a₃,a₇),(a₂,a₉)] and [(a₂,a₉),(a₈,a₁₁)] are present in R.
• Then the transitive element is [(a₃,a₇),(a₈,a₁₁)]. Is this transitive element present in R? Let us check.
(ii) [(a₃,a₇),(a₂,a₉)] is present in R.
• That means, this element satisfies the condition: xv = yu
• That means, a₃ a₉ = a₇ a₂.
(iii) Similarly, [(a₂,a₉),(a₈,a₁₁)] is present in R.
• That means, this element satisfies the condition: xv = yu
• That means, a₂ a₁₁ = a₉ a₈.
(iv) If the transitive element written in (i) is to be present in R, then it should also satisfy the condition xv = yu.
• That means, a₃ a₁₁ must be equal to a₇ a₈
• That means, $\frac{a_3}{a_7}~\text{must be equal to}~\frac{a_8}{a_{11}}$
(v) From (ii) we get: $\frac{a_3}{a_7}~=~\frac{a_2}{a_9}$
(vi) From (iii) we get: $\frac{a_2}{a_9}~=~\frac{a_8}{a_{11}}$
(vii) Combining the results in (v) and (vi), we get:
$\frac{a_2}{a_9}~=~\frac{a_8}{a_{11}}~=~\frac{a_3}{a_7}$
• So the condition mentioned in (iv) is satisfied.
(vii) That means, all transitive elements will be present in R.
Therefore, R is a transitive relation.

◼ Based on the above 6 steps, we see that, R is reflexive, symmetric and transitive. So R is an equivalence relation.

Solved example 17.43
Let X = {1,2,3,4,5,6,7,8,9}. Let R₁ be a relation in X given by R₁ = {(x,y): x-y is divisible by 3} and R₂ be another relation on X given by R₂ ={(x,y): {x,y}⊂{1,4,7} or {x,y}⊂{2,5,8} or {x,y}⊂{3,6,9}}. Show that R₁ = R₂.
Solution:
1. We can write set R₁ easily. Ordered pairs like (1,4), (4,1), (5,8), are some of the elements of R₁.
• But in this problem, we do not have to write the elements of R₁. We just need to know the nature of the elements. We see that, for all those elements, (x-y) will be divisible by 3.
2. Next we consider R₂
(i) Any ordered pair (x,y) for which, {x,y} is a subset of {1,4,7}, will be an element of R₂.
(ii) Any ordered pair (x,y) for which, {x,y} is a subset of {2,5,8}, will be an element of R₂.
(iii) Any ordered pair (x,y) for which, {x,y} is a subset of {3,6,9}, will be an element of R₂.
3. Consider the sets {1,4,7}.
• We can take any two elements from this set. The difference between those two elements will be divisible by 3.
• Similar is the case with the other two sets {2,5,8} and {3,6,9}
• Union of the three sets will give X.
• So R₁ will be a subset of R₂
4. While writing the ordered pairs of R₂, we see that, difference between the members of each ordered pair is divisible by 3.
• So R₂ will be a subset of R₁.
5. Now we can compare the results.
   ♦ In (3), we see that: R₁ ⊂ R₂.
   ♦ In (4), we see that: R₂ ⊂ R₁.
• So we can write: R₁ = R₂

Solved example 17.44
Let f: X→Y be a function. Define a relation R in X given by R = {(a,b): f(a) = f(b)}. Examine whether R is an equivalence relation or not.
Solution:
1. Let X = {x₁, x₂, x₃, . . . } and Y = {y₁, y₂, y₃, . . .}
2. Then a possible example set for f is {(x₁, y₅), (x₂, y₈), (x₃, y₁₁), . . .}
• This means:
    ♦ When x₁ is the input, the function f gives y₅ as the output.
    ♦ When x₂ is the input, the function f gives y₈ as the output.
    ♦ When x₃ is the input, the function f gives y₁₁ as the output.
so on . . .
3. Now we can write the set R.
• Set R will contain ordered pairs of the form (a,b).
• Consider any one ordered pair (a,b). That ordered pair is eligible to be in R because, f(a) = f(b).
• So both a and b will be from set X. This is because, all inputs of f are taken from X.
4. So R is a relation on X.
• That means, the elements of R are taken from the set X×X.
• So a random ordered pair taken from R will be (x₅,x₉)
• This ordered pair is eligible to be in R because, f(x₅) = f(x₉)
5. Now we check whether R is reflexive.
• A possible random element in X×X is (x₅,x₅)
• This element will satisfy the condition f(a) = f(b).
This is because, f(x₅) = f(x₅)
• So all reflexive pairs will be present in R.
Therefore, R is reflexive.
6. Next we check whether R is symmetric.
• If a random pair (x₅,x₉) is present in R, then the symmetric pair (x₉,x₅) will also be present in R.
This is because:
f(x₅) = f(x₉) ⇒ f(x₉) = f(x₅)
• So all symmetric pairs will be present in R.
Therefore, R is symmetric.
7. Finally we check whether R is transitive.
• Consider two random pairs from R: (x₅,x₉) and (x₉,x₂)
• Will the transitive pair (x₅,x₂) be present in R?
• Since both (x₅,x₉) and (x₉,x₂) are present in R, we can write:
f(x₅) = f(x₉) = f(x₂)
• So it is clear that, the transitive pair (x₅,x₂) will be present in R
Therefore, R is transitive.
◼ Since R is reflexive, symmetric and, transitive, it is an equivalence relation.

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Solved example 17.45
Determine which of the following binary operations on the set N are associative and which are commutative.
$(a)~a * b = 1 ~ \forall ~ a,b \in N~~~~~(b)~a * b = \frac{a+b}{2} ~ \forall ~ a,b \in N$
Solution:
Part (i):
1. Checking whether commutative or not.
• For an operation to be commutative, the condition which should be satisfied is:
(a∗b) = (b∗a)
• For our present case, a and b should be natural numbers.
• It is given that, if we perform the operation '*' between any two natural numbers a and b, then the result will be 1.
• So the same operation between b and a will also give 1.
• That means, (a∗b) = (b∗a)
• So the given ∗ is a commutative binary operation.
2. Checking whether associative or not.
• For an operation to be associative, the condition which should be satisfied is:
(a∗b)∗c = a∗(b∗c)
• For our present case, a, b and c should be natural numbers.
• First we calculate (a∗b)∗c:
(a ∗ b) = 1
So (a∗b)∗c = (1∗c) = 1
• Next we calculate a∗(b∗c):
(b ∗ c) = 1
So a∗(b∗c) = (a∗1) = 1
• We see that: (a∗b)∗c = a∗(b∗c)
So the given ∗ is an associative binary operation.
◼ Based on the above 2 steps, we can write:
The given ∗ is both commutative and associative.

Part (ii):
1. Checking whether commutative or not.
• For an operation to be commutative, the condition which should be satisfied is:
(a∗b) = (b∗a)
• For our present case, a and b should be natural numbers.
• For any two natural numbers, $\frac{a+b}{2}$ will be equal to $\frac{b+a}{2}$ .
• That means, (a∗b) = (b∗a)
• So the given ∗ is a commutative binary operation.
2. Checking whether associative or not.
• For an operation to be associative, the condition which should be satisfied is:
(a∗b)∗c = a∗(b∗c)
• For our present case, a, b and c should be natural numbers.
• First we calculate (a∗b)∗c:
$(a*b)~=~\frac{a+b}{2}$
$\text{So}~(a*b)*c~=~\left(\frac{a+b}{2}\right) * c~=~\frac{\frac{a+b}{2}~+~c}{2}~=~\frac{a+b+2c}{4}$
• Next we calculate a∗(b∗c):
$(b*c)~=~\frac{b+c}{2}$
$\text{So}~a*(b*c)~=~a * \left(\frac{b+c}{2}\right)~=~\frac{a~+~\frac{b+c}{2}}{2}~=~\frac{2a+b+c}{4}$
• We see that: (a∗b)∗c ≠ a∗(b∗c)
So the given ∗ is not an associative binary operation.
◼ Based on the above 2 steps, we can write:
The given ∗ is commutative but not associative.

Solved example 17.46
Find the number of all one-one functions from set A = {1,2,3} to itself.
Solution:
1. Take any one element from the domain. It can be mapped to the codomain in 3 different ways.
2. Take a second element from the domain. It can be mapped to the codomain in 2 ways. This is because, the way chosen in step 1 should not be repeated. For example, (1,2) and (2,2) cannot be selected. Then it would not be a one-one function.
3. Take the third element from the domain. It can be mapped to the codomain in one way.
4. So the total number of ways = 3 × 2 × 1 = 3! = 6

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In the next section,we will see a few more solved examples.

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17.11 - Identity Element and Inverse Element

In the previous section, we saw commutative property and associative property. We saw some solved examples also. In this section, we will see identity element and inverse element for binary operations.

Identity element

This can be explained in 6 steps:
1. We know that, any number 'a' can be added to the number zero in any order. We can write (a+0) or (0+a). Both will give the same answer.
• For example: (8+0) = (0+8) = 8
2. But for subtraction, 'any order' is not possible. (a-0) need not be equal to (0-a).
• For example: (8-0) ≠ (0-8)     
3. We know that, any number 'a' can be multiplied to the number 1 in any order. We can write (a × 1) or (1 × a). Both will give the same answer.
• For example: (8 × 1) = (1 × 8) = 8
4. But for division, 'any order' is not possible. (a÷1) need not be equal to (1÷a).
• For example: (8÷1) ≠ (1÷8)
5. We can write a summary:
• For binary addition involving any number 'a' and zero, order is not important.    
• For binary subtraction involving any number 'a' and zero, order is important.    
    ♦ We must write:
        ✰ Subtract 0 from a. 
        ✰ or   
        ✰ Subtract a from 0.    

• For binary multiplication involving any number 'a' and 1, order is not important.    
• For binary division involving any number 'a' and one, order is important.    
    ♦ We must write:
        ✰ Divide 8 by 1. 
        ✰ or   
        ✰ Divide 1 by 8.

6. Based on the above steps, we can write:
Given a binary operation ∗: A × A → A,
an element e ∈ A, if it exists, is called identity for the operation ∗, if:
a ∗ e = a = e ∗ a, ∀ a ∈ A.    

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Let us see a solved example:
Solved example 17.38
Show that zero is the identity for addition on R and 1 is the identity for multiplication on R. But there is no identity element for the operations
–: R × R → R and ÷: R × R → R.
Solution:
Part (i)
1. If a number e has to qualify as the identity element, it must satisfy two conditions:
(i) a ∗ e = a, ∀ a ∈ R.
(ii) e ∗ a = a, ∀ a ∈ R.
2. We want to prove that zero is the identity for addition on R.
• We see that, zero satisfies both conditions because:
(i) a + 0 = a, ∀ a ∈ R.
(ii) 0 + a = a, ∀ a ∈ R.
• So zero is the identity for addition on R.
3. Next we want to prove that 1 is the identity for multiplication on R.
• We see that, 1 satisfies both conditions because:
(i) a × 1 = a, ∀ a ∈ R.
(ii) 1  ×  a = a, ∀ a ∈ R.
• So 1 is the identity for multiplication on R.

Part (ii)
1. We want to prove that, there is no identity element for subtraction on R.
• Let us check whether the element ‘x’ satisfies the two conditions:
(i) a - x = a, ∀ a ∈ R.
(ii) x - a = a, ∀ a ∈ R.
• No real number x can simultaneously satisfy the two conditions.
• So there is no identity element for subtraction on R.
2. Next we want to prove that, there is no identity element for division on R.
• Let us check whether the element ‘x’ satisfies the two conditions:
(i) a ÷ x = a, ∀ a ∈ R.
(ii) x ÷ a = a, ∀ a ∈ R.
• No real number x can simultaneously satisfy the two conditions.
• So there is no identity element for division on R.

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We saw that, zero is the identity for the addition operation on R. But instead of R, if the set is N, then we cannot consider zero as the identity. This is because, set N does not contain element zero. In fact, the addition operation on N does not have any identity.

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Inverse element

This can be explained in 6 steps:
1. We know that, any number 'a' can be added to the number (-a) in any order. We can write (a+ -a) or (-a+a). Both will give the same answer zero.
• For example: (8+ -8) = (-8+8) = 0
   ♦ Zero is the identity for addition.
2. But for subtraction, 'any order' is not possible. (a- -a) need not be equal to (-a-a).
• For example: (8- -8) ≠ (-8-8)
• Recall that, subtraction does not have identity element.
• Since there is no identity element, we can exclude subtraction from our present discussion on inverse element.       
3. We know that, any number 'a' can be multiplied by the number '1/a' in any order. We can write (a × 1/a) or (1/a × a). Both will give the same answer 1.
• For example: (8 × 1/8) = (1/8 × 8) = 1
   ♦ 1 is the identity for multiplication.
4. But for division, 'any order' is not possible. (a ÷ 1/a) need not be equal to (1/a ÷ a).
• For example: (8 ÷ 1/8) ≠ (1/8 ÷ 8)
• Recall that, division does not have identity element.
• Since there is no identity element, we can exclude division from our present discussion on inverse element.       
5. We can write a summary:
• For binary addition involving any two numbers 'a' and (-a), order is not important. The result will be the addition identity zero.    
• We exclude binary subtraction from our present discussion. This is because, there is no subtraction identity.

• For binary multiplication involving any two numbers 'a' and '1/a', order is not important. The result will be the multiplication identity 1.
• We exclude binary division from our present discussion. This is because, there is no division identity.

6. Based on the above steps, we can write:
• Given a binary operation ∗: A × A → A, with the identity element e in A.
• An element a is said to be invertible with respect to the operation ∗, if there exists an element b in a such that a∗b = e = b∗a.
• b is called the inverse of a and is denoted by a^-1.

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Let us see a solved example:
Solved example 17.39
Show that –a is the inverse of a for the addition operation ‘+’ on R and 1/a is the inverse of a ≠ 0 for the multiplication operation ‘×’ on R.
Solution:
1. If a number b has to qualify as the inverse element of a, it must satisfy two conditions:
(i) a ∗ b = e, ∀ a,b ∈ R.
(ii) b ∗ a = e, ∀ a,b ∈ R.
• In the above two conditions, we specify set R because, it is given that: the operation is on R.
2. We want to prove that '-a' is the inverse of 'a' for addition on R.
• We see that, '-a' satisfies both conditions because:
(i) a + -a = 0, ∀ a, -a ∈ R.
(ii) -a + a = 0, ∀ a, -a ∈ R.
(0 is the e for addition)
• So '-a' is the inverse of 'a' for addition on R.
3. Next we want to prove that '1/a' is the inverse of 'a' for multiplication on R.
• We see that, '1/a' satisfies both conditions because:
(i) a × 1/a = 1, ∀ a, 1/a ∈ R.
(ii) 1/a  ×  a = 1, ∀ a, 1/a ∈ R.
(1 is the e for multiplication)
• So '1/a' is the inverse of 'a' for multiplication on R.

Solved example 17.40
Show that '–a' is not the inverse of a ∈ N for the addition operation + on N and '1/a' is not the inverse of a ∈ N for multiplication operation × on N, for a ≠ 1.
Solution:
1. If a number b has to qualify as the inverse element of a, it must satisfy two conditions:
(i) a ∗ b = e, ∀ a,b ∈ N.
(ii) b ∗ a = e, ∀ a,b ∈ N.
• In the above two conditions, we specify set N because, it is given that: the operation is on N.
2. Consider addition.
• We see that, '-a' can be substituted in the place of b. The two equations will be satisfied.
• But '-a' is not an element of N. So we cannot use '-a'.
3. Consider multiplication.
• We see that, '1/a' can be substituted in the place of b. The two equations will be satisfied.
• But '1/a' is not an element of N. So we cannot use '1/a'

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The link below gives a few more solved examples

Exercise 17.4 

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17.10 - Commutative and Associative Properties

In the previous section, we saw the basics of binary operations. We saw some solved examples also. In this section, we will see a few more solved examples. We will also see some properties of binary operations.

Solved example 17.33
Show that the ∨: R × R → R given by (a,b) → max {a,b} and the ∧: R × R → R given by (a,b) → min {a,b} are binary operations.
Solution:
Part (i): ∨: R × R → R given by (a,b) → max {a,b}
1. R × R will contain ordered pairs of the form (a,b), where a and b are two real numbers.
2. (a,b) is the input. Output will be the largest among a and b.
• So the output will be a real number. It will be a unique number in the codomain R.
3. So we can write:
    ♦ The given operation,
    ♦ carries the element (a,b) of the domain R×R,
    ♦ to an unique element of the codomain R
• Therefore, the given operation is a binary operation.

Part (ii): ∨: R × R → R given by (a,b) → min {a,b}
1. R × R will contain ordered pairs of the form (a,b), where a and b are two real numbers.
2. (a,b) is the input. Output will be the smallest among a and b.
• So the output will be a real number. It will be a unique number in the codomain R.
3. So we can write:
    ♦ The given operation,
    ♦ carries the element (a,b) of the domain R×R,
    ♦ to an unique element of the codomain R
• Therefore, the given operation is a binary operation.

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• Let us write some examples related to the above solved example:
    ♦ ∨(5,9) = 9
    ♦ ∧(5,9) = 5
    ♦ ∨(5,-9) = 5
    ♦ ∧(5,-9) = -9
• The above examples help us to get a better understanding about the max and min functions.

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Now we will learn about operation table. We will learn it by analyzing an example. It can be written in 7 steps:
1. Consider the following function:
∨: A × A → A given by ∨(a,b) → max {a,b}
    ♦ Where A = {1,2,3}
2. At the top of the table, in the red cell, we write the function.
3. In the magenta cell, we write the symbol of the binary operation. In our present case, it is ‘∨’.
• This is shown in the fig.17.11(I) below:

Fig.17.11

4. Next we fill up the yellow row and yellow column.
• In the yellow column, we fill up the elements of A. They are filled from top to bottom.
    ♦ These elements are taken as ‘a’ of the function.
    ♦ In the table, these elements are denoted as ‘a_i’.
• In the yellow row also, we fill up the elements of A. They are filled from left to right.
    ♦ These elements are taken as ‘b’ of the function.
    ♦ In the table, these elements are denoted as ‘a_j’.
5. Now we fill up the green rows and columns.
    ♦ The row numbers are denoted by the letter i.  
    ♦ The column numbers are denoted by the letter j.
• Let us see an example: We want to fill the green cell with i = 1 and j = 2.
    ♦ When the row number i is 1, we see that, a_i = a₁ = 1.
    ♦ When the column number j is 2, we see that, a_j = a₂ = 2.
    ♦ When we have both a_i and a_j, we can find the largest among them. It is 2.
    ♦ So the cell should be filled with 2.
This is shown in fig.17.11(II) above. In this way, all green cells can be filled.
6. Note that, if set A has n elements, then the green cells will be arranged in n rows and n columns.
7. Now we will write about the general case. It can be written in 3 steps:
(i) Consider the binary operation *: A × A → A
where A = {a₁, a₂, a₃, . . . a_n}
(ii) Since there are n elements in A, the green cells will be arranged in n rows and n columns.
(iii) The green cell at the intersection of the i^th row and j^th column will have the value: (a_i * a_j)

• The converse can be written in 3 steps:
(i) Given any binary operation table with n rows and n columns.
(ii) Based on that operation table, we can define a binary operation as follows:
*: A × A → A given by: (a_i * a_j) = value in the i^th row and j^th column.
(iii) Set A will have n elements.

Commutative binary operation

This can be explained in 6 steps:
1. We know that, two numbers a and b can be added in any order. We can write (a+b) or (b+a). Both will give the same answer.
• For example: (5+7) = (7+5) = 12
2. But for subtraction, order is important. (a-b) need not be equal to (b-a).
• For example: (5-7) ≠ (7-5)     
3. We know that, two numbers a and b can be multiplied in any order. We can write (ab) or (ba). Both will give the same answer.
• For example: (5 × 7) = (7 × 5) = 35
4. But for division, order is important. (a÷b) need not be equal to (b÷a).
• For example: (5÷7) ≠ (7÷5)
5. We can write a summary:
• ‘Addition of 5 and 7’ is meaningful.     
• ‘Subtraction of 5 and 7’ is meaningless.
    ♦ We must write:
        ✰ Subtract 7 from 5. 
        ✰ or   
        ✰ Subtract 5 from 7.    

• ‘Multiplication of 5 and 7’ is meaningful.     
• ‘Division of 5 and 7’ is meaningless.
    ♦ We must write:
        ✰ Divide 7 by 5. 
        ✰ or   
        ✰ Divide 5 by 7.

6.Based on the above steps, we can write:
A binary operation ∗ on the set X is called commutative, if a ∗ b = b ∗ a, for every a, b ∈ X

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Let us see some solved examples:
Solved example 17.34
Show that +: R × R → R and ×: R × R → R are commutative binary operations, but –: R × R → R and ÷: R × R  → R are not commutative.
Solution:
Part (i):
1. Let (a,b) ∈ R × R
We can write: (a+b) = (b+a) for all a, b ∈ R.
So ‘+’ is a commutative binary operation.
2. Let (a,b) ∈ R × R
We can write: (a × b) = (b × a) for all a, b ∈ R.
So ‘×’ is a commutative binary operation.
Part (ii):
1. Consider (5,7) ∈ R × R
We have: (5 - 7) ≠ (7 – 5).
So ‘-’ is not a commutative binary operation.
2. Consider (5,7) ∈ R × R
We have: (5÷7) ≠ (7÷5).
So ‘÷’ is not a commutative binary operation.

Solved example 17.35
Show that ∗ : R × R → R defined by a ∗ b = a + 2b is not commutative.
Solution:
1. Consider (5,7) ∈ R × R
2. We are going to calculate (a∗b). We get:
(a∗b) = (5∗7) = [5 + (2×7)] = [5 + 14] = 19
3. Next, we are going to calculate (b∗a). We get:
(b∗a) = (7∗5) = [7 + (2×5)] = [7 + 10] = 17
4. From (2) and (3), we see that: (a∗b) ≠ (b∗a).
So the given ‘∗’ is not a commutative binary operation.

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Associative binary operation

This can be explained in 6 steps:
1. We know that, three numbers a, b and c can be added in any convenient groups.
• We can write (a+b) +c or a + (b+c). Both will give the same result. Rearranging the parentheses will not change the result.
• For example: (5+7) + 2 = 5 + (7+2) = 14
2. But for subtraction, we cannot rearrange the parentheses.
(a-b) - c need not be equal to a - (b-c).
• For example: (5-7) - 2 ≠ 5 - (7-2)     
3. We know that, three numbers a, b and c can be multiplied in any convenient groups.
• We can write (ab)c or a(bc). Both will give the same result. Rearranging the parentheses will not change the result.
• For example: (5 × 7) × 2 = 5 × (7 × 2) = 70
4. But for division, we cannot rearrange the parentheses.
(a÷b) ÷ c need not be equal to a ÷ (b÷c).
• For example: (5÷7)÷2 ≠ 5÷(7÷2)    
5. We can write a summary:
• For addition and multiplication, rearranging the parentheses will not change the result.
• For subtraction and division, rearranging the parentheses will change the result.
6. Based on the above steps, we can write:
A binary operation ∗ on the set X is called associative,
if (a ∗ b) ∗ c = a ∗ (b ∗ c), for all a, b, c ∈ X

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Let us see some solved examples:
Solved example 17.36
Show that +: R × R → R and ×: R × R → R are associative binary operations, but – : R × R → R and ÷ : R × R → R are not associative.
Solution:
Part (i):
1. Let a, b, c ∈ R
We can write: (a+b) + c = a + (b+c) for all a, b, c ∈ R.
So ‘+’ is an associative binary operation.
2. Let (a,b) ∈ R × R
We can write: (ab)c = a(bc) for all a, b, c ∈ R.
So ‘×’ is an associative binary operation.
Part (ii):
1. Consider 5,7,2 ∈ R
We have: (5 - 7) - 2 ≠ 5 - (7 – 2).
So ‘-’ is not an associative binary operation.
2. Consider 5,7,2 ∈ R
We have: (5÷7)÷2 ≠ 5÷(7÷2).
So ‘÷’ is not an associative binary operation.

Solved example 17.37
Show that ∗ : R × R → R defined by a ∗ b = a + 2b is not associative.
Solution:
1. Let a= 5, b = 7 and c = 3.
2. We are going to calculate (a∗b)∗c.We get:
(a∗b) = (5∗7) = [5 + (2×7)] = [5 + 14] = 19
So (a∗b)∗c = 19∗3 = [19 + (2×3)] = [19 + 6] = 25
3. Next, we are going to calculate a∗(b∗c). We get:
(b∗c) = (7∗3) = [7 + (2×3)] = [7 + 6] = 13
So a∗(b∗c) = (5∗13) = 5 + (2×13)] = [5 + 26] = 31
4. From (2) and (3), we see that: (a∗b)∗c ≠ a∗(b∗c).
So the given ‘∗’ is not an associative binary operation.

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Importance of associative property of binary operation

This can be written in 3 steps:
1. Suppose that, we are given the following expression:
a₁ ∗ a₂ ∗ a₃ ∗ . . . ∗ a_n.
2. We see that parentheses are not present.
• But no doubts will arise because, associative property is valid for ‘∗’.
• Since no parentheses are present, we can straight away evaluate the expression by substituting the appropriate symbol in the place of ‘∗’
3. Due to the validity of the associative property, all algebraic expressions will contain parentheses where ever they are required.
4. Parentheses are compulsory in the following two situations:
(i) When subtraction or division occur.
(ii) When more than one type of operation occur.

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In the next section, we will see identity of a binary operation.

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17.9 - Binary Operations

In the previous section, we completed a discussion on composite functions and invertible functions. In this section, we will see binary operations.

Some basics can be written in 3 steps:
1. We are already familiar with the four fundamental operations:
addition, subtraction, multiplication and division.
2. Let us write the three main features of these operations:
(i) Given any two numbers a and b.
• We can find four new numbers:
   ♦ a+b
   ♦ a-b
   ♦ ab
   ♦ a/b where b ≠ 0.
(ii) Consider any of the four operations. That operation must be carried out between two numbers only.
(iii) If there is a third number, we must follow the procedure given below:
• If three numbers a, b and c are to be added, then we must first add a and b. To the sum of a and b, we can add c.
   ♦ That means: a+b+c = (a+b) + c.
• If three numbers a, b and c are to be multiplied, then we must first multiply a and b. The product (ab) can be multiplied by c.
   ♦ That means: abc = (ab)c.
3. The word ‘binary’ means two.
So addition, subtraction, multiplication and division are examples of binary operation.

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Now we can write a general definition which is applicable to all binary operations. It can be written in 7 steps:

1. Let us denote any binary operation by one symbol ‘*’
• That is., the symbol ‘*’ can be:
'+' or '-' or '×' or '÷' or any other binary operation.

2. * can be considered as a function.
• This can be explained using an example.
   ♦ Consider the function f(x) = x² + 3. This function tells us to square the input x and then add 3.
   ♦ In a similar way, the function * will tell us what to do when two numbers a and b are given.
• For the function f, the input is a single value x. So we write f(x)
• But for the function *, there will be two input values a and b. So we write *(a,b)

3. For the function f, we specify the domain. Input values for f are taken from the domain.
• In a similar way, we must specify the domain for function * also. This can be done in 4 steps:
(i) Let set A contain all the numbers which can be subjected to the binary operation *.
(ii) But for the function *, we cannot input individual numbers from A. We can input only pairs.
(iii) Recall that the set A × A will contain all the possible pairs that can be formed from the numbers in A.
(iv) So A × A can be taken as the domain.

4. The output of function * is not a pair. It is a single number. So set A can be taken as the codomain.

5. So we have the domain and codomain. We can write four of the many possible ways in which the function * can be defined:
(i) +: A × A →A is defined as +(a,b) = a+b   
(ii) -: A × A →A is defined as -(a,b) = a-b   
(iii) ×: A × A →A is defined as ×(a,b) = a × b   
(iv) ÷: A × A →A is defined as ÷(a,b) = a÷b

6. So it is clear that, * can be described as a function.
    ♦ It has input and output.
    ♦ It has domain and codomain.
• However, from now on wards, we will call '*' as a binary operation. We will not call '*' as a function.

7. It is important to note that, only one output must be present for each input.
• That means, in the Venn diagram, only one arrow must diverge from each element of the domain.
• So for every input (a,b), there will be a unique output.

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Let us see a solved example:
Solved example 17.29
Show that addition, subtraction and multiplication are binary operations on R, but division is not a binary operation on R. Further, show that division is a binary operation on the set R_* of nonzero real numbers.
Solution:
Part (i):
• Binary operation on R means:
Domain is R × R and codomain is R
• Let us check the four binary operations:
1. Is the following operation possible ?
+: R × R →R is defined as +(a,b) = a+b   
Answer:
R × R will contain ordered pairs of the form (a,b), where a and b are two real numbers. It is possible to add any two real numbers. The sum will also be a real number. The sum will be a unique number in the codomain R. So this operation is possible.  

2. Is the following operation possible ?
-: R × R →R is defined as -(a,b) = a-b   
Answer:
R × R will contain ordered pairs of the form (a,b), where a and b are two real numbers. It is possible to find the difference between any two real numbers. The difference will also be a real number. The difference will be a unique number in the codomain R. So this operation is possible.

3. Is the following operation possible ?
×: R × R →R is defined as  ×(a,b) = a × b   
Answer:
R × R will contain ordered pairs of the form (a,b), where a and b are two real numbers. It is possible to find the product of any two real numbers. The product will also be a real number. The product will be a unique number in the codomain R. So this operation is possible. 

4. Is the following operation possible ?
÷: R × R →R is defined as  ÷(a,b) = a ÷ b   
Answer:
R × R will contain ordered pairs of the form (a,b), where a and b are two real numbers. This operation is not defined when b = 0. So this operation is not possible when the domain is R × R.

Part (ii):
• Is the following operation possible ?
÷: R_* × R_* → R_* is defined as  ÷(a,b) = a ÷ b   
Answer:
R_* × R_* will contain ordered pairs of the form (a,b), where a and b are two real numbers. But R_* is the set of non-zero real numbers. This set will not contain zero. So there will not be any ordered pairs in which b is zero. That means, when this binary operation is carried out, there will not be any division by zero. The quotient will be a unique number in the codomain R_*. So this operation is possible.

Solved example 17.30
Show that subtraction and division are not binary operations on N.
Solution:
• Binary operation on N means:
Domain is N × N and codomain is N.
• Let us check the binary operations of subtraction and division:
1. Is the following operation possible ?
-: N × N → N is defined as -(a,b) = a-b   
Answer:
N × N will contain ordered pairs of the form (a,b), where a and b are two natural numbers. It is possible to find the difference between any two natural numbers. But the difference need not be a natural number. For example, if a = 5 and b = 9, then (a-b) will be -4. This -4 is not a natural number. It will not be present in the codomain N. So this operation is not possible.

2. Is the following operation possible ?
÷: N × N → N is defined as  ÷(a,b) = a ÷ b
Answer:
N × N will contain ordered pairs of the form (a,b), where a and b are two natural numbers. It is possible to find the quotient when any natural number is divided by any natural number. But the quotient need not be a natural number. For example, if a = 5 and b = 9, then 5/9 is not a natural number. 5/9 is not present in the codomain N. So this operation is not possible.

Solved example 17.31
Show that *: R × R →R is defined as *(a,b) = a+4b² is a binary operation.  
Solution:
1. R × R will contain ordered pairs of the form (a,b), where a and b are two real numbers.
2. First we find 4b². Then we add it to a.
• The sum will be a real number. The sum will be a unique number in the codomain R.
3. So for any pair (a,b), the given operation is valid. That means, it is a binary operation.

Solved example 17.32
Let P be the set of all subsets of a given set X.
Show that:
(i) ∪: P × P → P given by (A, B) → A ∪ B
(ii) ∩ : P × P → P given by (A, B) → A ∩ B
are binary operations on the set P.
Solution:
• Note that, in the discussion so far in this section, we denoted the input as (a,b). Here a and b are numbers.
• But for the present problem, input is given as (A,B). This is because, A and B are sets. We are inputting sets.

Part (i):
1. (A,B) is an ordered pair. This pair is taken from the set P × P.
• When we are given the input (A,B), we can find the union of A and B.
• This A∪B will be a set.
• A∪B will be present in P. This is because, P is the set of all subsets.
• Also, A∪B will be unique. We will get only one set when we calculate the union.
2. So we can write:
    ♦ The given operation,
    ♦ carries the element (A,B) of the domain P×P,
    ♦ to an unique element A∪B of the codomain P
• Therefore, the given operation is a binary operation.

Part (ii):
1. (A,B) is an ordered pair. This pair is taken from the set P × P.
• When we are given the input (A,B), we can find the intersection of A and B.
• This A∩B will be a set.
• A∩B will be present in P. This is because, P is the set of all subsets.
• Also, A∩B will be unique. We will get only one set when we calculate the union.
2. So we can write:
    ♦ The given operation,
    ♦ carries the element (A,B) of the domain P×P,
    ♦ to an unique element A∩B of the codomain P
• Therefore, the given operation is a binary operation.

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In the next section, we will see a few more solved examples. We will also see some properties of binary operations.

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17.8 - Properties of Invertible Functions

In the previous section, we saw two solved examples related to invertible functions. We saw composite of composite functions also. In this section, we will see two more solved examples. We will also see a new property of invertible functions.

Solved example 17.27
Consider two functions:

f: {1,2,3}→{a,b,c} defined as:
f(1) = a, f(2) = b, f(3) = c

g: {a,b,c}→{apple, ball, cat} defined as:
g(a) = apple, g(b) = ball, g(c) = cat.

I. Show that, f, g and (g∘f) are invertible.
II. Find out f^-1, g^-1 and (g∘f)^-1.
III. Show that (g∘f)^-1 = f^-1∘g^-1

Solution:
Part (I): Proving that f, g and (g∘f) are invertible.
1. f is a function. Domain is {1,2,3} and codomain is {a,b,c}.
• We know that f is a set. The elements of this set are ordered pairs of the form (x,y).
• Based on the given data, we can write all the ordered pairs in f:
f = {(1,a), (2,b), (3,c)}
• We see that:
   ♦ Each element in the domain has a unique image. So f is one-one.
   ♦ All elements in the codomain are images. So f is onto.
• Since f is both one-one and onto, it is invertible.

2. g is a function. Domain is {a,b,c} and codomain is {apple, ball, cat}.
• We know that g is a set. The elements of this set are ordered pairs of the form (x,y).
• Based on the given data, we can write all the ordered pairs in g:
g = {(a,apple), (b,ball), (c,cat)}
• We see that:
   ♦ Each element in the domain has a unique image. So g is one-one.
   ♦ All elements in the codomain are images. So g is onto.
• Since g is both one-one and onto, it is invertible.

3. (g∘f)(x) = g(f(x))
• For (g∘f),
   ♦ domain is the domain of f which is {1,2,3}
   ♦ codomain is the codomain of g which is {apple, ball, cat}
• So we get:
(g∘f)(1) = g(f(1)) = g(a) = apple
(g∘f)(2) = g(f(2)) = g(b) = ball
(g∘f)(3) = g(f(3)) = g(c) = cat

• We know that (g∘f) is a set. The elements of this set are ordered pairs of the form (x,y).
• Based on the above results, we can write all the ordered pairs in (g∘f):
(g∘f) = {(1,apple), (2,ball), (3,cat)}
• We see that:
   ♦ Each element in the domain has a unique image. So (g∘f) is one-one.
   ♦ All elements in the codomain are images. So (g∘f) is onto.
• Since (g∘f) is both one-one and onto, it is invertible.

Part (II): Finding f^-1, g^-1 and (g∘f)^-1.
1. Finding f^-1:
(i) In part (I), we saw that:
f = {(1,a), (2,b), (3,c)}
• So we can write the reverse of f. We will call it p.
• Since f is one-one and onto, the reverse can be easily written:
p = {(a,1), (b,2), (c,3)}
• The domain of p is {a,b,c}. And codomain is {1,2,3}

(ii) Now we have both f and p. We can calculate (p∘f)
(p∘f) means, output of f is used as the input of p.
• The two functions are:
f: {1,2,3}→{a,b,c}
p: {a,b,c}→{1,2,3}
• We know that, (p∘f) directly connects the domain of f to the codomain of p. So (p∘f) is from {1,2,3} to {1,2,3}.
• When the input is 1, we get:
(p∘f)(1) = p(f(1)) = p(a) = 1
• When the input is 2, we get:
(p∘f)(2) = p(f(2)) = p(b) = 2
• When the input is 3, we get:
(p∘f)(3) = p(f(3)) = p(c) = 3

• So whatever input we give, the output will be the same. That means, (p∘f) is an identity function.
• The inputs are taken from {1,2,3}. So we can write:
(p∘f) = I_{1,2,3}.

(iii) Similarly, we can calculate (f∘p)
(f∘p) means, output of p is used as the input of f.
• The two functions are:
f: {1,2,3}→{a,b,c}
p: {a,b,c}→{1,2,3}
• We know that, (f∘p) directly connects the domain of p to the codomain of f. So (f∘p) is from {a,b,c} to {a,b,c}.
• When the input is a, we get:
(f∘p)(a) = f(p(a)) = f(1) = a
• When the input is 2, we get:
(f∘p)(b) = f(p(b)) = f(2) = b
• When the input is 3, we get:
(f∘p)(c) = f(p(c)) = f(3) = c

• So whatever input we give, the output will be the same. That means, (f∘p) is an identity function.
• The inputs are taken from {a,b,c}. So we can write:
(f∘p) = I_{a,b,c}.

(iv). Let us write a summary:
• We are given a function f: {1,2,3}→{a,b,c}
• We wrote the reverse function p: {a,b,c}→{1,2,3}
• From (ii), we got: (p∘f) = I_{1,2,3}
• From (iii), we got: (f∘p) = I_{a,b,c}
• So f is invertible.
• Also, the inverse of f = f^-1 = p
• That means:
The inverse of f: {1,2,3}→{a,b,c} is:
p: {a,b,c}→{1,2,3}

2. Finding g^-1:
(i) In part (I), we saw that:
g = {(a,apple), (b,ball), (c,cat)}
• So we can write the reverse of g. We will call it q.
• Since g is one-one and onto, the reverse can be easily written:
q = {(apple,a), (ball,b), (cat,c)}
• The domain of q is {apple, ball, cat}. And codomain is {a,b,c}

(ii) Now we have both g and q. We can calculate (q∘g)
(q∘g) means, output of g is used as the input of q.
• The two functions are:
g: {a,b,c}→{apple, ball, cat}
q: {apple, ball, cat}→{a,b,c}
• We know that, (q∘g) directly connects the domain of g to the codomain of q. So (q∘g) is from {a,b,c} to {a,b,c}.
• When the input is a, we get:
(q∘g)(a) = q(g(a)) = q(apple) = a
• When the input is b, we get:
(q∘g)(b) = q(g(b)) = q(ball) = b
• When the input is c, we get:
(q∘g)(c) = q(g(c)) = q(cat) = c

• So whatever input we give, the output will be the same. That means, (q∘g) is an identity function.
• The inputs are taken from {a,b,c}. So we can write:
(q∘g) = I_{a,b,c}.

(iii) Similarly, we can calculate (g∘q)
(g∘q) means, output of q is used as the input of g.
• The two functions are:
g: {a,b,c}→{apple, ball, cat}
q: {apple, ball, cat}→{a,b,c}
• We know that, (g∘q) directly connects the domain of q to the codomain of g. So (g∘q) is from {apple, ball, cat} to {apple, ball, cat}.
• When the input is apple, we get:
(g∘q)(apple) = g(q(apple)) = g(a) = apple
• When the input is 2, we get:
(g∘q)(ball) = g(q(ball)) = g(b) = ball
• When the input is 3, we get:
(g∘q)(cat) = g(q(cat)) = g(c) = cat

• So whatever input we give, the output will be the same. That means, (g∘q) is an identity function.
• The inputs are taken from {apple, ball, cat}. So we can write:
(g∘q) = I_{{apple, ball, cat}}.

(iv). Let us write a summary:
• We are given a function g: {a,b,c}→{apple, ball, cat}
• We wrote the reverse function q: {apple, ball, cat}→{a,b,c}
• From (ii), we got: (q∘g) = I_{a,b,c}
• From (iii), we got: (g∘q) = I_{{apple, ball, cat}}
• So g is invertible.
• Also, the inverse of g = g^-1 = q
• That means:
The inverse of g: {a,b,c}→{apple, ball, cat} is:
q: {apple, ball, cat}→{a,b,c}

3. Finding (g∘f)^-1:
(i) In part (I), we saw that:
(g∘f) = {(1,apple), (2,ball), (3,cat)}
• So we can write the reverse of (g∘f). We will call it r.
• Since (g∘f) is one-one and onto, the reverse can be easily written:
r = {(apple,1), (ball,2), (cat,3)}
• The domain of r is {apple, ball, cat}. And codomain is {1,2,3}

(ii) Now we have both (g∘f) and r. We can calculate (r∘(g∘f))
(r∘(g∘f)) means, output of (g∘f) is used as the input of r.
• The two functions are:
(g∘f): {1,2,3}→{apple, ball, cat}
r: {apple, ball, cat}→{1,2,3}
• We know that, (r∘(g∘f)) directly connects the domain of (g∘f) to the codomain of r. So (r∘(g∘f)) is from {1,2,3} to {1,2,3}.
• When the input is 1, we get:
(r∘(g∘f))(1) = r((g∘f)(1)) = r(apple) = 1
• When the input is 2, we get:
(r∘(g∘f))(2) = r((g∘f)(2)) = r(ball) = 2
• When the input is 3, we get:
(r∘(g∘f))(3) = r((g∘f)(3)) = r(cat) = 3

• So whatever input we give, the output will be the same. That means, (r∘(g∘f)) is an identity function.
• The inputs are taken from {1,2,3}. So we can write:
(r∘(g∘f)) = I_{1,2,3}.

(iii) Similarly, we can calculate ((g∘f)∘r)
((g∘f)∘r) means, output of r is used as the input of (g∘f).
• The two functions are:
(g∘f): {1,2,3}→{apple, ball, cat}
r: {apple, ball, cat}→{1,2,3}
• We know that, ((g∘f)∘r) directly connects the domain of r to the codomain of (g∘f). So ((g∘f)∘r) is from {apple, ball, cat} to {apple, ball, cat}.
• When the input is apple, we get:
((g∘f)∘r)(apple) = (g∘f)(r(apple)) = (g∘f)(1) = apple
• When the input is 2, we get:
((g∘f)∘r)(ball) = (g∘f)(r(ball)) = (g∘f)(2) = ball
• When the input is 3, we get:
((g∘f)∘r)(cat) = (g∘f)(r(cat)) = (g∘f)(3) = cat

• So whatever input we give, the output will be the same. That means, ((g∘f)∘r) is an identity function.
• The inputs are taken from {apple, ball, cat}. So we can write:
((g∘f)∘r) = I_{{apple, ball, cat}}.

(iv). Let us write a summary:
• We are given a function (g∘f): {1,2,3}→{apple, ball, cat}
• We wrote the reverse function r: {apple, ball, cat}→{1,2,3}
• From (ii), we got: (r∘(g∘f)) = I_{1,2,3}
• From (iii), we got: ((g∘f)∘r) = I_{{apple, ball, cat}}
• So (g∘f) is invertible.
• Also, the inverse of (g∘f) = (g∘f)^-1 = r
• That means:
The inverse of (g∘f): {1,2,3}→{apple, ball, cat} is:
r: {apple, ball, cat}→{1,2,3}

Part III: Showing that, (g∘f)^-1 = f^-1∘g^-1.
1. From part (II)(1), we have:
f^-1 = p
Where p: {a,b,c}→{1,2,3}
• In set form, we can write:
f^-1 = {(a,1), (b,2), (c,3)}


2. From part (II)(2), we have:
g^-1 = q
Where q: {apple, ball, cat}→{a,b,c}
• In set form, we can write:
q^-1 = {(apple,a), (ball,b), (cat,c)}

3. From part (II)(3), we have:
(g∘f)^-1 = r
Where r: {apple, ball, cat}→{1,2,3}
• In set form, we can write:
(g∘f)^-1 = {(apple,1), (ball,2), (cat,3)}

4. Now we can calculate f^-1∘g^-1.
(f^-1∘g^-1)(x) = f^-1(g^-1(x))
• For (f^-1∘g^-1),
   ♦ domain is the domain of g^-1 which is {apple, ball, cat}
   ♦ codomain is the codomain of f^-1 which is {1,2,3}
• So we get:
(f^-1∘g^-1)(apple) = f^-1(g^-1(apple)) = f^-1(a) = 1
(f^-1∘g^-1)(ball) = f^-1(g^-1(ball)) = f^-1(b) = 2
(f^-1∘g^-1)(cat) = f^-1(g^-1(cat)) = f^-1(c) = 3

5. We can write the result in (4), in the form of a set:
(f^-1∘g^-1) = {(apple,1), (ball,2), (cat,3)}

6. Comparing the results in (3) and (5), we get:
(g∘f)^-1 = (f^-1∘g^-1)

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The result obtained in the above solved example 17.27 can be used in general. We can write:
If f: X→Y and g: Y→Z are two invertible functions, then (g∘f) is also invertible. The inverse of (g∘f) is (f^-1∘g^-1)

The proof can be written in 7 steps:
1. First we will write some basic details:
(i) Given that, f: X→Y is an invertible function.
• So we can write:
    ♦ f(x) = y
    ♦ f^-1(y) = x
(ii) Given that, g: Y→Z is an invertible function.
• So we can write:
    ♦ g(y) = z
    ♦ g^-1(z) = y
(iii) Note that, domain of g is same as codomain of f.

2. Now we recall an important property of inverse functions.
• If the inverse of f is f^-1, then two conditions will be satisfied:
(i) (f^-1∘f) = I_X
(ii) (f∘f^-1) = I_Y
Where
    ♦ X is the domain of f
    ♦ Y is the domain of f^-1 

• Similarly, if the inverse of g is g^-1, then two conditions will be satisfied:
(i) (g^-1∘g) = I_Y
(ii) (g∘g^-1) = I_Z
Where
    ♦ Y is the domain of g
    ♦ Z is the domain of g^-1

3. Next step is to determine the domain and codomain of the composite functions.
(i) Domain of (g∘f):
• (g∘f) connects the domain of f with the codomain of g.
    ♦ So the domain of (g∘f) is the domain of f, which is X.
    ♦ Also the codomain of (g∘f) is the codomain of g, which is Y. 
(ii) Domain of (f^-1∘g^-1):
• (f^-1∘g^-1) connects the domain of g^-1 with the codomain of f^-1.
    ♦ So the domain of (f^-1∘g^-1) is the domain of g^-1, which is Z.
    ♦ Also the codomain of (f^-1∘g^-1) is the codomain of f^-1, which is X. 

4. Now we can use the property mentioned in step (2).
• That is:
If the inverse of (g∘f) is (f^-1∘g^-1), then two conditions will be satisfied:
(i) (f^-1∘g^-1)∘(g∘f) = I_X
(ii) (g∘f)∘(f^-1∘g^-1) = I_Z
Where
    ♦ X is the domain of (g∘f)
    ♦ Z is the domain of (f^-1∘g^-1)

5. First we show that (f^-1∘g^-1)∘(g∘f) = I_X.
$\begin{array}{ll}{}    &{(f^{-1} \circ g^{-1})(g \circ f)}    & {~=~}    &{\Bigl((f^{-1} \circ g^{-1}) \circ g \Bigr) \circ f ~\color {magenta}{\text{- - - (A)}}}    &{} \\
{}    &{}    & {~=~}    &{\Bigl(f^{-1} \circ (g^{-1} \circ g) \Bigr) \circ f ~\color {magenta}{\text{- - - (B)}}}    &{} \\
{}    &{}    & {~=~}    &{(f^{-1} \circ I_{Y}) \circ f ~\color {magenta}{\text{- - - (C)}}}    &{} \\
{}    &{}    & {~=~}    &{I_X ~\color {magenta}{\text{- - - (D)}}}    &{} \\
\end{array}               
$

◼ Remarks:
• Line marked as A:
Here we use the formula: h∘(g∘f) = (h∘g)∘f
• Line marked as B:
Here also, we use the formula: h∘(g∘f) = (h∘g)∘f
• Line marked as C:
Here we use the fact that: (g^-1∘g) = I_Y
• Line marked as D:
We must simplify (f^-1∘I_Y)∘f. This can be done in two steps:
(i) First we will determine (f^-1∘I_Y)
• Here, the output of I_Y is used as the input for f^-1.
• Also this function connects the domain of I_Y with the codomain of f^-1.
    ♦ Domain of I_Y is Y
    ♦ Codomain of f^-1 is X
So we get: (f^-1∘I_Y) = f^-1(I_Y) = f^-1(y) = x.
(ii) Now we can determine (f^-1∘I_Y)∘f
• Here, the output of f is used as the input for (f^-1∘I_Y).
• Also this function connects the domain of f with the codomain of (f^-1∘I_Y).
    ♦ Domain of f is X
    ♦ Codomain of (f^-1∘I_Y) is also X
So we get: (f^-1∘I_Y)∘f = I_X.

6. In a similar way, we can show that:
(g∘f)∘(f^-1∘g^-1) = I_Z.

7. So both conditions mentioned in (4) are satisfied.
• Therefore it is proved that:
If f: X→Y and g: Y→Z are two invertible functions, then (g∘f) is also invertible. The inverse of (g∘f) is (f^-1∘g^-1)  

Solved example 17.28
Let S = {1,2,3}. Determine whether the functions f: S→S defined as below have inverses. Find f^-1 if it exists.
(a) f = {(1,1), (2,2), (3,3)}
(b) f = {(1,2), (2,1), (3,1)}
(c) f = {(1,3), (3,2), (2,1)}
Solution:
Part (a): f = {(1,1), (2,2), (3,3)}
• Here f is both one-one and onto. So f^-1 exists.
• We can write: f^-1 = {(1,1), (2,2), (3,3)}

Part (b): f = {(1,2), (2,1), (3,1)}
• Here f(2) = f(3) = 1.
So f is not one-one and onto. Therefore, f^-1 does not exist.

Part (c): f = {(1,3), (3,2), (2,1)}
• Here f is both one-one and onto. So f^-1 exists.
• We can write: f^-1 = {(3,1), (2,3), (1,2)}

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Link to a few more solved examples is given below:

Exercise 17.3

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In the next section, we will see binary operations.

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17.7 - Composite of Composite Functions

In the previous section, we saw invertible functions. We saw some solved examples also. In this section, we will see a few more solved examples. Based on those solved examples, we will see composite of composite functions.

Solved example 17.25

Let f: N→R be a function defined as f(x)=4x²+12x+15. Show that f: N→S, where, S is the range of f, is invertible. Find the inverse of f.  
Solution:
1. Based on the given data, we can write 5 points:
(i) f is a set. It contains ordered pairs of the form (x,y).
(ii) The domain of f is N. So x can be natural numbers only.
(iii) For every x, the corresponding y can be calculated using the equation y = 4x²+12x+15
(iv) Since y is calculated using this equation, it will also be a natural number.
(v) The y values thus calculated, forms the set S. Set S is the codomain. In this problem, it is specifically given that, S is the range. So codomain is the range. Recall that, range is a subset of codomain. Range will contain only the outputs.  

2. Now we have the details about f. So we can write the reverse of f. We will call it g.
g: S→N
• The details about g can be written in 5 steps:
(i) g is a set. It contains ordered pairs of the form (y,x)
(ii) The domain of g is set S. The set S is already formed when function f is defined.
(iii) Each element y in S was derived from an element x in N.
• So when we work in reverse from S to N, we can write:
Every element y in S will have a corresponding element x in N.
(iv) y was derived using the equation: y = 4x²+12x+15
We can bring x to the left side as follows:

$$\begin{array}{ll}{}    &{y}    & {~=~}    &{4x^2 + 12x + 15}    &{} \\
{\implies}    &{y}    & {~=~}    &{4x^2 + 12x + 9 + 6}    &{} \\
{\implies}    &{y}    & {~=~}    &{(2x + 3)^2 + 6}    &{} \\
{\implies}    &{(2x+3)^2}    & {~=~}    &{y-6}    &{} \\
{\implies}    &{2x+3}    & {~=~}    &{\sqrt{y-6}}    &{} \\
{\implies}    &{x}    & {~=~}    &{\frac{\sqrt{y-6}~-~3}{2}~\color{magenta}{\text{- - - (I)}}}    &{} \\
\end{array}               
$$

---

◼ Remarks:
• Line marked as I:
Here we must note two points (a) and (b):
(a) Input x must be natural numbers. So the smallest input possible is ‘1’.
• We have y = 4x²+12x+15.
So the smallest possible output corresponding to the smallest possible input will be:
y = 4(1)²+12(1)+15 = 31.
• This value of y is greater than ‘6’. So (y-6) will never become -ve. Therefore we can apply the square root $\sqrt{y-6}$.

(b) When we apply the square root, we usually write: $\pm \sqrt{y-6}$
• But in the present case, the -ve sign is not applicable because, x is a natural number. It cannot be -ve.

---

• So when working in reverse, the x corresponding to y can be obtained as: $x = \frac{\sqrt{y-6}~-~3}{2}$
(v) Based on this, we can define g as follows:
g: S→N is defined as: $x~=~g(y)~=~ \frac{\sqrt{y-6}~-~3}{2}$
• The inputs for g are taken from S. The outputs will be present in N.

3. Now we have both f and g. We can calculate (g∘f)
• (g∘f) means, output of f is used as the input of g.
• The two functions are: f: N→S and g: S→N
• We know that, (g∘f) directly connects the domain of f to the codomain of g. So (g∘f) is from N to N.
• We denoted the inputs from N as x. So we are calculating (g∘f)(x).
• We get:
$$\begin{array}{ll}{}    &{(g \circ f)(x)}    & {~=~}    &{g(f(x))}    &{} \\
{}    &{}    & {~=~}    &{g(4x^2 + 12x + 15)}    &{} \\
{}    &{}    & {~=~}    &{\frac{\sqrt{(4x^2 + 12x + 15) -6}~-~3}{2}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\sqrt{(4x^2 + 12x + 9)}~-~3}{2}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\sqrt{(2x + 3)^2}~-~3}{2}}    &{} \\
{}    &{}    & {~=~}    &{\frac{(2x + 3)~-~3}{2}}    &{} \\
{}    &{}    & {~=~}    &{\frac{2x}{2}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$

• So whatever input x we give, the output will be the same x. That means, (g∘f) is an identity function.
• The inputs are taken from N. So we can write:
(g∘f) = I_N.

4. Similarly, we can calculate (f∘g)
• (f∘g) means, output of g is used as the input of f.
• The two functions are: g: S→N and f: N→S
• We know that, (f∘g) directly connects the domain of g to the codomain of f. So (f∘g) is from S to S.
• We denoted the inputs from S as y. So we are calculating (f∘g)(y).
• We get:
$$\begin{array}{ll}{}    &{(f \circ g)(y)}    & {~=~}    &{f(g(y))}    &{} \\
{}    &{}    & {~=~}    &{f \left(\frac{\sqrt{y-6} - 3}{2} \right)}    &{} \\
{}    &{}    & {~=~}    &{4 \left[\frac{\sqrt{y-6} - 3}{2} \right]^2 ~+~12 \left[\frac{\sqrt{y-6} - 3}{2} \right] ~+~15 }    &{} \\
{}    &{}    & {~=~}    &{(\sqrt{y-6} - 3)^2 ~+~6(\sqrt{y-6} - 3) ~+~15 }    &{} \\
{}    &{}    & {~=~}    &{(\sqrt{y-6})^2 ~-~ 6 \sqrt{y-6} ~+~ 9 ~+~ 6\sqrt{y-6} ~-~ 18 ~+~15 }    &{} \\
{}    &{}    & {~=~}    &{y-6 ~-~ 6 \sqrt{y-6} ~+~ 9 ~+~ 6\sqrt{y-6} ~-~ 18 ~+~15 }    &{} \\
{}    &{}    & {~=~}    &{y-6 ~+~ 9  ~-~ 18 ~+~15 }    &{} \\
{}    &{}    & {~=~}    &{y}    &{} \\
\end{array}               
$$
• So whatever input y we give, the output will be the same y. That means, (f∘g) is an identity function.
• The inputs for (f∘g) are taken from Y. So we can write:
(f∘g) = I_Y.

5. Let us write a summary:
• We are given a function f: N → S
• We wrote the reverse function g: S → N
• From (3), we got: (g∘f) = I_X
• From (4), we got: (f∘g) = I_Y
• So f is invertible.
• Also, the inverse of f = f^-1 = g
• That means:
The inverse of $y = f(x) = 4x^2 + 12x + 15$ is:
$x = g(y) = \frac{\sqrt{y-6}~-~3}{2}$

Solved example 17.26
Consider f: N→N, g: N→N and h: N→R defined as:
f(x) = 2x,
g(y) = 3y+4 and
h(z) = sin z
for all x, y and z in N.
Show that h∘(g∘f) = (h∘g)∘f.
Solution:
1. (g∘f) = g(f(x)) = g(2x) = 3(2x) + 4 = 6x+4
• (g∘f) connects the domain of f with the codomain of g. So (g∘f) is from N to N.
• The inputs taken from the domain of f are denoted as x.
• So we can write:
(g∘f): N→N is defined as:
(g∘f)(x) = 6x + 4 for all x∈N

2. Now we can find h∘(g∘f)
h∘(g∘f) = h((g∘f)(x)) = h(6x+4) = sin(6x+4)
• h∘(g∘f) connects the domain of (g∘f) with the codomain of h. So h∘(g∘f) is from N to R.
• The inputs taken from the domain of (g∘f) are denoted as x.
• So we can write:
h∘(g∘f): N→R is defined as:
(h∘(g∘f))(x) = sin(6x + 4) for all x∈N

3. (h∘g) = h(g(y)) = h(3y+4) = sin(3y+4)
• (h∘g) connects the domain of g with the codomain of h. So (h∘g) is from N to R.
• The inputs taken from the domain of g are denoted as y.
• So we can write:
(h∘g): N→R is defined as:
(h∘g)(y) = sin(3y+4) for all y∈N

4. Now we can find (h∘g)∘f
(h∘g)∘f = (h∘g)(f(x)) = (h∘g)(2x) = sin(3(2x) + 4) = sin(6x+4)
• (h∘g)∘f connects the domain of f with the codomain of (h∘g). So (h∘g)∘f is from N to R.
• The inputs taken from the domain of f are denoted as x.
• So we can write:
(h∘g)∘f: N→R is defined as: (h∘g)∘f(x) = sin(6x + 4) for all x∈N

5. From (2) and (4), we see that:
(h∘(g∘f))(x) = ((h∘g)∘f)(x) for all x∈N.

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The result obtained in the above solved example 17.26 can be used in general. We can write:
If f: X→Y, g: Y→Z and h: Z→S are functions, then
h∘(g∘f) = (h∘g)∘f

The proof can be written in 5 steps:
1. We are given:
    ♦ f: X→Y
    ♦ g: Y→Z
    ♦ h: Z→S
So we can write:
    ♦ f(x) = y
    ♦ g(y) = z
    ♦ h(z) = s

2. We have to prove that: h∘(g∘f) = (h∘g)∘f

3. First we consider the LHS. We will determine (g∘f) in the LHS.
• (g∘f) connects the domain of f with the codomain of g.
• That means, (g∘f) is from X to Z.
• The inputs will be from X. The outputs should be in Z.
• So we get:
(g∘f)(x) = g(f(x)) = g(y) = z

4. Now we can determine h∘(g∘f)
• h∘(g∘f) connects the domain of (g∘f) with the codomain of h.
•  That means, h∘(g∘f) is from X to S.
• The inputs will be from X. The outputs should be in S.
• So we get:
h∘(g∘f)(x) = h((g∘f)(x)) = h(z) = s

5. Next we consider the RHS. We will determine (h∘g) in the RHS.
• (h∘g) connects the domain of g with the codomain of h.
• That means, (h∘g) is from Y to S.
• The inputs will be from Y. The outputs should be in S.
• So we get:
(h∘g)(y) = h(g(y)) = h(z) = s

6. Now we can determine (h∘g)∘f
• (h∘g)∘f connects the domain of f with the codomain of (h∘g).
•  That means, (h∘g)∘f is from X to S.
• The inputs will be from X. The outputs should be in S.
• So we get:
((h∘g)∘f)(x) = (h∘g)(f(x)) = (h∘g)(y) = s

7. Let us compare the results:
• From (4) we get:
h∘(g∘f)(x) = h((g∘f)(x)) = h(z) = s
• From (6) we get:
((h∘g)∘f)(x) = (h∘g)(f(x)) = (h∘g)(y) = s
• Results are the same. So we can write:
h∘(g∘f)(x) = ((h∘g)∘f)(x) for all x∈X.

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In the next section, we will see two more solved examples.

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17.6 - Invertible Function

In the previous section, we saw properties of composite functions. In this section, we will see invertible functions.

First we will see a solved example.

Solved example 17.22
Let f: {1,2,3}  →  {a,b,c}  be  one-one  and  onto  function given by f(1) = a, f(2) = b and f(3) = c. Show that there exists a function g: {a, b, c} → {1, 2, 3}such that g∘f = I_X and f∘g = I_Y, where, X =  {1, 2, 3} and Y = {a, b, c}.
Solution:
1. We have two sets:
• X = {1,2,3}
• Y = {a,b,c}
2. We have a function which is one-one and onto:
• f: X→Y defined by:
    ♦ f(1) = a
    ♦ f(2) = b
    ♦ f(3) = c
3. We can think about a new function in the reverse order. This function is also one-one and onto:
• g: Y→X defined by:
    ♦ g(a) = 1
    ♦ g(b) = 2
    ♦ g(c) = 3
4. Now we can calculate (g∘f)
• Recall that, (g∘f) means, output of f is used as the input of g.
    ♦ (g∘f)(1) = g(f(1)) = g(a) = 1
    ♦ (g∘f)(2) = g(f(2)) = g(b) = 2
    ♦ (g∘f)(3) = g(f(3)) = g(c) = 3
• We see that:
Whatever input we give for the function (g∘f), the output will be the same. So (g∘f) is an identity function.
• But we must specify the domain for the identity function. We see that, the input values for (g∘f) are taken from the set X. So the domain is X. That means, it is an identity function on set X.
• We can write: (g∘f) = I_X.
5. Similarly, we can calculate (f∘g)
• Recall that, (f∘g) means, output of g is used as the input of f.
    ♦ (f∘g)(a) = f(g(a)) = f(1) = a
    ♦ (f∘g)(b) = f(g(b)) = f(2) = b
    ♦ (f∘g)(c) = f(g(c)) = f(3) = c
• We see that:
Whatever input we give for the function (f∘g), the output will be the same. So (f∘g) is an identity function.
• But we must specify the domain for the identity function. We see that, the input values for (f∘g) are taken from the set Y. So the domain is Y. That means, it is an identity function on set Y.
• We can write: (f∘g) = I_Y.

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The result obtained from the above solved example, can be applied in general. It can be written in 2 steps:
1. We have a function f: X→Y
• The function f, is one-one and onto.
2. Then there exists another function g: Y→X such that:
    ♦ (g∘f) = I_X
    ♦ (f∘g) = I_Y

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The converse is also true. It can be written in 2 steps:
1. There exists two functions: f: X→Y and g: Y→X
• The two functions together satisfy two conditions:
    ♦ (g∘f) = I_X
    ♦ (f∘g) = I_Y
2. Then f is both one-one and onto.

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Now we can write about invertible function. It can be written in 4 steps:
1. We have a function f: X→Y
2. There exists another function g: Y→X such that:
    ♦ (g∘f) = I_X
    ♦ (f∘g) = I_Y
3. Then f is an invertible function.
4. The function g is called inverse of f.
Inverse of f is denoted as: f^-1.
5. Note the subscripts X and Y in step(2)
    ♦ X is the domain of the original function f.
    ♦ Y is the domain of the inverse function g.

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Checking whether a function is invertible or not

This can be written in 4 steps:
1. Consider the condition:
    ♦ (g∘f) = I_X
    ♦ (f∘g) = I_Y
2. This condition can be satisfied only if f is one-one and onto.
3. So we can write:
• If f is invertible, then it will be one-one and onto.
• Conversely, if f is one-one and onto, then it will be invertible.
4. This information can be used to prove a function to be invertible.
• All we need to do is: prove that, it is one-one and onto.
• This is especially helpful in situations where we do not need to find f^-1.

Solved example 17.23
Let f: N → Y be a function defined as f(x) = 4x + 3, where, Y = {y ∈ N : y = 4x + 3 for some x ∈ N}.
Show that f is invertible. Find the inverse.
Solution:
1. Based on the given data, we can write 5 points:
(i) f is a set. It contains ordered pairs of the form (x,y).
(ii) The domain of f is N. So x can be natural numbers only.
(iii) For every x, the corresponding y can be calculated using the equation y = 4x + 3
(iv) Since y is calculated using this equation, it will also be a natural number.
(v) The y values thus calculated, forms the set Y. Set Y is the codomain.
• The codomain Y will not contain natural numbers like 1, 2, 8, 9 etc., This is because, these numbers do not satisfy the equation y = 4x + 3.
• Since the codomain in this case contain only the outputs, we can say that, codomain is same as the range.

2.Now we have the details about f. So we can write the reverse of f. We will call it g.
g: Y→N
• The details about g can be written in 5 steps:
(i) g is a set. It contains ordered pairs of the form (y,x)
(ii) The domain of g is set Y. The set Y is already formed when function f is defined.
(iii) Each element y in Y was derived from an element x in N.
• So when we work in reverse from Y to N, we can write:
Every element y in Y will have a corresponding element x in N.
(iv) y was derived using the equation: y = 4x + 3
• So when working in reverse, the x corresponding to y can be obtained as: $x = \frac{y-3}{4}$
(v) Based on this, we can define g as follows:
g: Y→N is defined as: $x~=~g(y)~=~ \frac{y-3}{4}$
• The inputs for g are taken from Y. The outputs will be present in N.

3. Now we have both f and g. We can calculate (g∘f)
• (g∘f) means, output of f is used as the input of g.
• The two functions are: f: N→Y and g: Y→N
• We know that, (g∘f) directly connects the domain of f to the codomain of g. So (g∘f) is from N to N.
• We denoted the inputs from N as x. So we are calculating (g∘f)(x).
• We get:
$$\begin{array}{ll}{}    &{(g \circ f)(x)}    & {~=~}    &{g(f(x))}    &{} \\
{}    &{}    & {~=~}    &{g(4x+3)}    &{} \\
{}    &{}    & {~=~}    &{\frac{[(4x+3)-3]}{4}}    &{} \\
{}    &{}    & {~=~}    &{\frac{4x}{4}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$

• So whatever input x we give, the output will be the same x. That means, (g∘f) is an identity function.
• The inputs are taken from N. So we can write:
(g∘f) = I_N.

4. Similarly, we can calculate (f∘g)
• (f∘g) means, output of g is used as the input of f.
• The two functions are: g: Y→N and f: N→Y
• We know that, (f∘g) directly connects the domain of g to the codomain of f. So (f∘g) is from Y to Y.
• We denoted the inputs from Y as y. So we are calculating (f∘g)(y).
• We get:
$$\begin{array}{ll}{}    &{(f \circ g)(y)}    & {~=~}    &{f(g(y))}    &{} \\
{}    &{}    & {~=~}    &{f \left(\frac{y-3}{4} \right)}    &{} \\
{}    &{}    & {~=~}    &{f \left(4 \times \frac{y-3}{4} + 3 \right)}    &{} \\
{}    &{}    & {~=~}    &{f \left(y-3 + 3\right)}    &{} \\
{}    &{}    & {~=~}    &{y}    &{} \\
\end{array}               
$$
• So whatever input y we give, the output will be the same y. That means, (f∘g) is an identity function.
• The inputs for (f∘g) are taken from Y. So we can write:
(f∘g) = I_Y.

5. Let us write a summary:
• We are given a function f: N → Y
• We wrote the reverse function g: Y → N
• From (3), we got: (g∘f) = I_X
• From (4), we got: (f∘g) = I_Y
• So f is invertible.
• Also, the inverse of f = f^-1 = g
• That means:
The inverse of $f(x) = 4x + 3$ is:
$g(y) = \frac{y-3}{4}$

Solved example 17.24
Let Y = {n² : n∈N} ⊂ N. Consider f: N→Y as f(n) = n². Show that f is invertible. Find the inverse of f.
Solution:
1. Based on the given data, we can write 5 points:
(i) f is a set. It contains ordered pairs of the form (x,y).
(ii) The domain of f is N. So x can be natural numbers only.
(iii) For every x, the corresponding y can be calculated using the equation y = x²
(iv) Since y is calculated using this equation, it will also be a natural number.
(v) The y values thus calculated, forms the set Y. Set Y is the codomain.
• The codomain Y will not contain natural numbers like 2, 5, 7, 8 etc., This is because, these numbers do not satisfy the equation y = x², where x is a natural number.
• Since the codomain in this case contain only the outputs, we can say that, codomain is same as the range.

2. Now we have the details about f. So we can write the reverse of f. We will call it g.
g: Y→N
• The details about g can be written in 5 steps:
(i) g is a set. It contains ordered pairs of the form (y,x)
(ii) The domain of g is set Y. The set Y is already formed when function f is defined.
(iii) Each element y in Y was derived from an element x in N.
• So when we work in reverse from Y to N, we can write:
Every element y in Y will have a corresponding element x in N.
(iv) y was derived using the equation: y = x²
• So when working in reverse, the x corresponding to y can be obtained as: $x = \sqrt{y}$
(v) Based on this, we can define g as follows:
g: Y→N is defined as: $x~=~g(y)~=~ \sqrt{y}$
• The inputs for g are taken from Y. The outputs will be present in N.

3. Now we have both f and g. We can calculate (g∘f)
• (g∘f) means, output of f is used as the input of g.
• The two functions are: f: N→Y and g: Y→N
• We know that, (g∘f) directly connects the domain of f to the codomain of g. So (g∘f) is from N to N.
• We denoted the inputs from N as x. So we are calculating (g∘f)(x).
• We get:
$$\begin{array}{ll}{}    &{(g \circ f)(x)}    & {~=~}    &{g(f(x))}    &{} \\
{}    &{}    & {~=~}    &{g(x^2)}    &{} \\
{}    &{}    & {~=~}    &{\sqrt{x^2}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$

• So whatever input x we give, the output will be the same x. That means, (g∘f) is an identity function.
• The inputs are taken from N. So we can write:
(g∘f) = I_N.

4. Similarly, we can calculate (f∘g)
• (f∘g) means, output of g is used as the input of f.
• The two functions are: g: Y→N and f: N→Y
• We know that, (f∘g) directly connects the domain of g to the codomain of f. So (f∘g) is from Y to Y.
• We denoted the inputs from Y as y. So we are calculating (f∘g)(y).
• We get:
$$\begin{array}{ll}{}    &{(f \circ g)(y)}    & {~=~}    &{f(g(y))}    &{} \\
{}    &{}    & {~=~}    &{f \left(\sqrt{y}\right)}    &{} \\
{}    &{}    & {~=~}    &{y}    &{} \\
\end{array}               
$$
• So whatever input y we give, the output will be the same y. That means, (f∘g) is an identity function.
• The inputs for (f∘g) are taken from Y. So we can write:
(f∘g) = I_Y.

5. Let us write a summary:
• We are given a function f: N → Y
• We wrote the reverse function g: Y → N
• From (3), we got: (g∘f) = I_X
• From (4), we got: (f∘g) = I_Y
• So f is invertible.
• Also, the inverse of f = f^-1 = g
• That means:
The inverse of $f(x) = x^2$ is:
$g(y) = \sqrt{y}$

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In the next section, we will see a few more solved examples.

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