Chapter 2.8 - Solved Examples on Functions

In the previous section, we saw algebra of real functions. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 2.34
Let R be the set of real numbers. Define the real function
f: R→R by f(x) = x + 10 and sketch the graph of this function.
Solution:
1. Given that f: R→R
So the input x values should be from R. The output f(x) values must also be present in R.
2. Let us input some convenient x values:
    ♦ When x = -10, f(x) = f(-10) = (-10 + 10) = 0
    ♦ When x = -9, f(x) = f(-9) = (-9 + 10) = 1
    ♦ When x = -7, f(x) = f(-7) = (-7 + 10) = 3
    ♦ When x = 0, f(x) = f(0) = (0 + 10) = 10
    ♦ When x = 2, f(x) = f(2) = (2 + 10) = 12
    ♦ When x = 3, f(x) = f(3) = (3 + 10) = 13
    ♦ When x = 4, f(x) = f(4) = (4 + 10) = 14
3. We can make a table using the above values:

Table.2.10

Such a table is convenient to plot the graph of the function.
4. The graph is shown in fig.2.20 below:

Fig.2.20

 

• We see that, the graph is a line.
• In our earlier analytical geometry classes, we have seen that:
Equation of a line takes the form y = mx + c, where m and c are constants.
• In our present case, the function is f(x) = x + 10. This is comparable to y = mx + c because:
    ♦ y is equivalent to f(x)
    ♦ m is equivalent to 1
    ♦ c is equivalent to 10
5. The function f defined by f(x) = mx + c, x ∈ R is called linear function.
• So our present case f(x) = x + 10 is a linear function.

Solved example 2.35
Let R be a relation from Q to Q defined by R = {(a, b) : a, b ∈ Q and a – b ∈ Z}. Show that
(i) (a, a) ∈ R for all a ∈ Q
(ii) (a, b) ∈ R implies that (b, a) ∈ R
(iii) (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R
Solution:
1. Q is the set of rational numbers. A relation R is defined from Q to Q.
• Details about this relation R can be written as follows:
   ♦ The set R contains all ordered pairs (a, b) such that,
   ♦ a is a rational number,
   ♦ b is also a rational number,
   ♦ a-b is an integer.
2. Recall that,
   ♦ rational numbers can have a decimal part
   ♦ integers do not have a decimal part
3. We can write infinite pairs of rational numbers in such a way that their differences are integers.
• Let us see some examples:
   ♦ For the pair (3.157, 11.157), difference = (3.157 – 11.157) = -8
   ♦ For the pair (-2.73, 10.27), difference = (-2.73 – 10.27) = 13
• All such pairs are eligible to be included in the set R
• Now we can write the answers of the given questions.
Part (i):
• The pairs are in the form (a, a) and a is a rational number.
   ♦ Here both the elements are the same. Their difference (a-a) will be zero.
• Zero is an integer. So all pairs in the form (a, a), where a is a rational number, are eligible to be included in R.
• We can write: (a, a) ∈ R for all a ∈ Q
Part (ii):
• If a and b are such that a - b is an integer, then b – a will also be an integer.
An example:
• Consider the pair (3.157, 11.157)
   ♦ The difference (3.157 – 11.157) = -8
   ♦ The difference (11.157 – 3.157) = 8
Another example:
• Consider the pair (-2.73, 10.27)
   ♦ The difference (-2.73 – 10.27) = -13
   ♦ The difference (10.27 –  - 2.73) = 13
• So (b, a) will be eligible to be included in R
• We can write: (a,b) ∈ R implies that (b, a) ∈ R
Part (iii):
• We have to prove that, (a - c) is an integer if
   ♦ (a – b) gives an integer
   ♦ (b – c) also gives an integer,
• Let us see an example:
   ♦ Consider the pair (-2.73, 10.27)
   ♦ The difference (a – b) = (-2.73 – 10.27) = -13
   ♦ Consider the pair (10.27, 6.27)
   ♦ The difference (b – c) = (10.27 – 6.27) = 4
   ♦ Now the difference (a – c) = (-2.73 – 6.27) = -9
• So (a,c) is eligible to be included in R.
• We can prove this algebraically also:
   ♦ (a - b) + (b - c) = a - b + b - c = (a - c)
   ♦ (-2.73 – 10.27) + (10.27 – 6.27) = (-2.73 – 6.27)
• We can write: (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R

Solved example 2.36
Let f = {(1,1), (2,3), (0, –1), (–1, –3)} be a linear function from Z into Z. Find f(x).
Solution:
1. Given that f is a linear function. So it will be in the form: f(x) = mx + c
All ordered pairs in the set f will satisfy this relation.
2. Consider the pair (1, 1). We get: 1 = m × 1 + c
⇒ 1 = m + c   
3. Consider the pair (2, 3). We get: 3 = m × 2 + c
⇒ 3 = 2m + c
4. Subtracting (2) from (3), we get:
3 - 1 = (2m + c) - (m + c)
⇒ 2 = m
5. Substituting for m in (2), we get: 1 = 2 + c
⇒ c = -1
6. Substituting for m and c in (1), we get: f(x) = 2x - 1
7. Note: We used the ordered pairs (1, 1) and (2, 3)
   ♦ It would be much more easier if we use (1, 1) and (0, -1)
   ♦ This is because, the term with m will become zero.

Solved example 2.37
Find the domain of the function $\mathbf\small{\rm{f(x)=\frac{x^2+3x+5}{x^2-5x+4}}}$
Solution:
1. Consider the denominator x² - 5x + 4
• If this denominator become zero, f(x) cannot be defined. So we must not input those x values which make x² - 5x + 4 zero.
2. So we need to find those x values which will make x² - 5x + 4 zero.
• For that, we equate it to zero and solve for x.
x² - 5x + 4 = 0
⇒ x² - 5x = -4
⇒ x² - 5x + (⁵⁄₂)² = -4 + (⁵⁄₂)²
⇒ (x - ⁵⁄₂)² = -4 + ²⁵⁄₄
⇒ (x - ⁵⁄₂)² = ⁹⁄₄
⇒ x- ⁵⁄₂ = ± ³⁄₂
⇒ x = (⁵⁄₂ + ³⁄₂) or (⁵⁄₂ - ³⁄₂)
⇒ x = ⁸⁄₂ or ²⁄₂
⇒ x = 4 or 1
3. So it is clear that, we must not use 4 and 1 as input x.
• We can write: domain = R - {4, 1}
• That means, domain is the set obtained by performing the 'difference' operation on the sets R and {4, 1}
   ♦ See difference of two sets.

Solved example 2.38
The function f is defined by:
$f(x) =
\begin{cases}
1-x,  & \text{if}\; x < 0 \\
1,  & \text{if}\; x = 0 \\
x+1, & \text{if} \; x > 0
\end{cases}$
Draw the graph of f(x).
Solution:
1. If we input an x value which is less than 0, that x value will be processed according to the function: f(x) = 1 - x
• Let us try some x values which are less than 0:
   ♦ When x = -5, f(x) = 1 - x = 1 - (-5) = 6
   ♦ When x = -3, f(x) = 1 - x = 1 - (-3) = 4
2. If we input an x value which is equal to 0, that x value will be processed according to the function: f(x) = 1
• Here there is no need to try different x values. Only one input x is possible and that is '0'.
• Also it is given that, when x = 0, f(x) = a constant, which is '1'.
• So there is no need to see how the x value is processed.
3. If we input an x value which is greater than 0, that x value will be processed according to the function: f(x) = x + 1
• Let us try some x values which are less than 0:
   ♦ When x = 2, f(x) = x + 1 = 2 + 1 = 3
   ♦ When x = 7, f(x) = x + 1 = 7 + 1 = 8
4. Based on the above steps, we can make a table:

Table 2.11

5. Using the table, we can draw the graph:

Fig.2.21

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Link to some more solved examples is given below:

Solved examples 2.39 to 2.49

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We have completed a discussion on relations and functions. In the next chapter, we will see trigonometric functions.

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Chapter 1.11 - Some Interesting Relations Between Three Sets

In the previous section, we saw some interesting relations between two sets. In this section, we will see such relations between three sets. We will see them in the form of solved examples.

Solved example 1.64
Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.
Solution:
• We will solve this problem in two parts (a) and (b).
   ♦ In part (a), we will prove that B ⊂ C
   ♦ In part (b), we will prove that C ⊂ B
Part (a):
1. Let x be an element of B
• Using symbols, we write this as: x ∈ B
2. Then x will be an element of A ∪ B also.
• Using symbols, we write this as: x ∈ B ⇒ x ∈ (A ∪ B)
3. Given that, A ∪ B = A ∪ C
• So x must be an element of A ∪ C also.
• Using symbols, we write this as:
x ∈ (A ∪ B) ⇒ x ∈ (A ∪ C)
4. If x is an element of A ∪ C, it can be an element of A or C
• Using symbols, we write this as:
x ∈ (A ∪ C) ⇒ x ∈ A or x ∈ C
5. Let us assume that, x is an element of A
• In (1), we have already said that x is an element of B
   ♦ So x is an element of both A and B
   ♦ In that case, x will be an element of A ∩ B
• Using symbols, we write this as: x ∈ (A ∩ B)
6. Given that, A ∩ B = A ∩ C
   ♦ So x must be an element of A ∩ C also.
• Using symbols, we write this as:
x ∈ (A ∩ B) ⇒ x ∈ (A ∩ C)
7. If x is an element of A ∩ C, it must be an element of both A and C
• Using symbols, we write this as:
x ∈ (A ∩ C) ⇒ x ∈ A and x ∈ C
• That means, if an element x which belongs to B, is present in A also, it will be present in C also
8. In step (5), instead of assuming A, let us assume that, x belongs to C
• Then we can write:
The element x, which belongs to B, is present in C also
9. So whatever be the assumption that we make in (5), the element x, which belongs to B, will be present in C
◼ Thus we get: B ⊂ C

Part (b):
1. Let y be an element of C
• Using symbols, we write this as: y ∈ C
2. Then y will be an element of A ∪ C also.
• Using symbols, we write this as: y ∈ C ⇒ y ∈ (A ∪ C)
3. Given that, A ∪ B = A ∪ C
• So x must be an element of A ∪ B also.
• Using symbols, we write this as:
y ∈ (A ∪ C) ⇒ y ∈ (A ∪ B)
4. If y is an element of A ∪ B, it can be an element of A or B
• Using symbols, we write this as:
y ∈ (A ∪ B) ⇒ y ∈ A or y ∈ B
5. Let us assume that, y is an element of A
• In (1), we have already said that y is an element of C
   ♦ So y is an element of both A and C
   ♦ In that case, y will be an element of A ∩ C
• Using symbols, we write this as: y ∈ (A ∩ C)
6. Given that, A ∩ B = A ∩ C
   ♦ So y must be an element of A ∩ B also.
• Using symbols, we write this as:
y ∈ (A ∩ C) ⇒ y ∈ (A ∩ B)
7. If y is an element of A ∩ B, it must be an element of both A and B
• Using symbols, we write this as:
y ∈ (A ∩ B) ⇒ y ∈ A and y ∈ B
• That means, if an element y which belongs to C, is present in A also, it will be present in B also
8. In step (5), instead of assuming A, let us assume that, y belongs to B
• Then we can write:
The element y, which belongs to C, is present in B also
9. So whatever be the assumption that we make in (5), the element y, which belongs to C, will be present in B
◼ Thus we get: C ⊂ B

◼ In part (a), we proved: B ⊂ C
◼ In part (b), we proved: C ⊂ B
◼ So we get: B = C

Solved example 1.65
Show that if A ⊂ B, then C – B ⊂ C – A
Solution:
1. Let x be an element of C - B
• Using symbols, we write this as: x ∈ (C - B)
2. C - B will not contain any element of B
• So we can write:
    ♦ x will be an element of C.
    ♦ But x will not be an element of B
• Using symbols, we write this as:
x ∈ (C - B) ⇒ x ∈ C and x ∉ B
3. Given that A is a subset of B
• So all elements of A will be present in B
• In (2), we saw that x is not present in B
    ♦ If x is not present in B, x will not be present in A either.
4. If x is not present in A,
   ♦ x will not be deleted from C when (C - A) is formed
• That means, x will be present in (C - A)
5. So we can write:
x, which is an element of C - B, is an element of C - A also
◼ Thus we get: C - B ⊂ C - A

Solved example 1.66
Show that A ∩ B = A ∩ C need not imply B = C.
Solution:
We can show this using an example. It can be written in 3 steps:
1. Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {3, 4, 7, 8}
2. We get: A ∩ B = {3, 4} and A ∩ C = {3, 4}
3. Here A ∩ B = A ∩ C but A ≠ C

Solved example 1.67
Let A and B be sets. If A ∩ X = B ∩ X = ɸ and A ∪ X = B ∪ X for some set X, show that A = B.
(Hints A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use Distributive law)
Solution:
• We will solve this problem in two parts (a) and (b)
Part (a):
1. We have: A = A ∩ (A ∪ X)
• (A ∪ X) can be replaced by (B ∪ X) because, it is given that they are equal.
• So we get: A =  A ∩ (B ∪ X)
2. The RHS can be expanded using the distributive law of intersection. We get:
A = (A ∩ B) ∪ (A ∩ X)
⇒ A = (A ∩ B) ∪ ɸ [∵ (A ∩ X) = ɸ]
⇒ A = (A ∩ B)

Part (b):
1. We have: B = B ∩ (B ∪ X)
• (B ∪ X) can be replaced by (A ∪ X) because, it is given that they are equal.
• So we get: B =  B ∩ (A ∪ X)
2. The RHS can be expanded using the distributive law of intersection. We get:
B = (B ∩ A) ∪ (B ∩ X)
⇒ B = (B ∩ A) ∪ ɸ [∵ (B ∩ X) = ɸ]
⇒ B = (A ∩ B)

• From part (a), we have A = (A ∩ B)
• From part (b), we have B = (A ∩ B)
◼ So we can write: A = B

Solved example 1.68
Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = ɸ.
Solution:
1. A ∩ B should be non-empty. That means, there must be at least one common element in A and B.
• So let A = {1, 2} and B = {2, 3}
2. B ∩ C should be non-empty. That means, there must be at least one common element in B and C.
3. A ∩ C should be non-empty. That means, there must be at least one common element in A and C.
• So let C = {1, 3}
4. So we get:
   ♦ A ∩ B = {2}
   ♦ B ∩ C = {3}
   ♦ A ∩ C = {1}
• They are all non-empty sets
5. Now, there is not even a single element which is common in all three sets.
◼ So we get: A ∩ B ∩ C = ɸ.

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In the next section, we will see the practical problems which involves three sets.

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Chapter 1.10 - Some Interesting Relations Between Two Sets

In the previous section, we saw some practical problems involving sets. In this section, we will see some interesting relations between sets. We will see them in the form of solved examples.

Solved example 1.56
Show that A ∪ B = A ∩ B implies A = B
Solution:
• We will solve this problem in two parts: Part (a) and Part (b).
    ♦ In part (a), we prove that A ⊂ B
    ♦ In part (b), we prove that B ⊂ A
Part (a):
1. Let a be an element of A.
    ♦ Using symbols, we write this as: a ∈ A
2. Since a is an element of A, the same a will be an element of A ∪ B also.
    ♦ Using symbols, we write this as: a ∈ A ∪ B
3. Given that A ∪ B = A ∩ B
    ♦ So a will be an element of A ∩ B also.
    ♦ Using symbols, we write this as: a ∈ A ∩ B
4. A ∩ B is the set which contains only the common elements of A and B.
    ♦ So, if a is an element of A ∩ B, the same a will be an element of B also.
    ♦ Using symbols, we write this as: a ∈ B
5. Comparing two statements:
    ♦ In (1), we wrote that, a is an element of A.
    ♦ In (4), we saw that, the same a is an element of B also.
• That means, every element in A is an element of B also.
    ♦ Using symbols, we write this as: A ⊂ B

Part (b):
1. Let b be an element of B.
    ♦ Using symbols, we write this as: b ∈ B
2. Since b is an element of B, the same b will be an element of A ∪ B also.
    ♦ Using symbols, we write this as: b ∈ A ∪ B
3. Given that A ∪ B = A ∩ B
    ♦ So b will be an element of A ∩ B also.
    ♦ Using symbols, we write this as: b ∈ A ∩ B
4. A ∩ B is the set which contains only the common elements of A and B.
    ♦ So, if b is an element of A ∩ B, the same b will be an element of A also.
    ♦ Using symbols, we write this as: b ∈ A
5. Comparing two statements:
    ♦ In (1), we wrote that, b is an element of B.
    ♦ In (4), we saw that, the same a is an element of A also.
• That means, every element in B is an element of A also.
    ♦ Using symbols, we write this as: B ⊂ A

◼ In part (a), we proved that, A ⊂ B
◼ In part (b), we proved that, B ⊂ A
◼ This is possible only if the two sets are the same. So we can write: A = B

Solved example 1.57
In each of the following, determine whether the statement is true or false. If it is
true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B , then x ∈ B
(ii) If A ⊂ B and B ∈ C , then A ∈ C
(iii) If A ⊂ B and B ⊂ C , then A ⊂ C
(iv) If A ⊄ B and B ⊄ C , then A ⊄ C
(v) If x ∈ A and A ⊄ B , then x ∈ B
(vi) If A ⊂ B and x ∉ B , then x ∉ A
Solution:
(i) If x ∈ A and A ∈ B , then x ∈ B
This is false. Let us see an example:
• A = {1, 2}, B = {3, {1, 2}, 4}
    ♦ Here A is an element of B.
    ♦ But '1' and '2' are not elements of B.

(ii) If A ⊂ B and B ∈ C , then A ∈ C
This is false. Let us see an example:
A = {1, 2}, B = {1, 2, 3}, C = {4, {1, 2, 3}, 5}
    ♦ Here B is an element of C.
    ♦ But {1, 2} is not an element of C.

(iii) If A ⊂ B and B ⊂ C , then A ⊂ C
This is true. Proof can be written in 3 steps:
1. Let a be an element of A.
    ♦ Using symbols, we write this as: a ∈ A
2. Given that A ⊂ B
    ♦ So all elements of A are present in B.
    ♦ So a ∈ B
3. Given that B ⊂ C
    ♦ So all elements of B are present in C.
    ♦ So a ∈ C
• Thus we get A ⊂ C
◼ The Venn diagram in fig.1.26(a) also proves this.
    ♦ The circle B encloses circle A. This is because A ⊂ B
    ♦ The circle C encloses circle B. This is because B ⊂ A
    ♦ So the circle C automatically encloses circle A, indicating A ⊂ C

Fig.1.26

(iv) If A ⊄ B and B ⊄ C, then A ⊄ C
This is false. Let us see an example:
• A = {1, 2}, B = {2, 3, 4}, C = {1, 2, 3, 5}
    ♦ Here A ⊄ B and B ⊄ C
    ♦ But A ⊂ C
◼ The Venn diagram in fig.1.26(b) also proves that this is false.
   ♦ The circle B does not enclose circle A because A ⊄ B
   ♦ The circle C does not enclose circle B because B ⊄ C
   ♦ But the circle C can enclose circle A because A ⊂ C

(v) If x ∈ A and A ⊄ B , then x ∈ B
This is false. It can be written in steps:
1. Given that x is an element of A.
2. Given that A is not a subset of B. So all elements of A are not present in B.
3. So x need not be present in B.
Let us see an example:
• A = {1, 2}, B = {2, 3, 4}
    ♦ Here A ⊄ B
    ♦ Also, '1' is not present in B.

(vi) If A ⊂ B and x ∉ B , then x ∉ A.
This is true. It can be proved in 2 steps:
1. Given that A is a subset of B. So all elements of A will be present in B.
2. So if x is not present in B, it cannot be present in A.

Solved example 1.58
Show that the following four conditions are equivalent :
(i) A ⊂ B  (ii) A – B = ɸ  (iii) A ∪ B = B  (iv) A ∩ B = A
Solution:
Part (a):
• In this part, we prove that: A ⊂ B  ⇒ A – B = ɸ
   ♦ It can be proved in 4 steps:
1. To find A - B, we first discard all elements of B.
2. Then we discard elements which are common in both A and B.
3. But when we discard the elements in B, we automatically discard the following two items:
   ♦ Common elements mentioned in (2).
   ♦ All elements of A.
4. So after the operation A - B, none of the elements of either A or B will remain.
• So we can write:
If A ⊂ B, the difference A - B will be a null set.
• In other words, A ⊂ B  ⇒ A – B = ɸ

Part (b):
• In this part, we prove that: A ⊂ B ⇒ A ∪ B = B
   ♦ It can be proved in 4 steps:
1. To find A ∪ B, we write all elements of A and B together, but repeating elements will be written only once.
2. Here, all elements of A will be repeating. So none of the elements of A need to be considered.
3. All we need to do is: Write the elements of B.
4. So we can write: A ⊂ B ⇒ A ∪ B = B

Part (c):
In this part, we prove that: A ⊂ B ⇒ A ∩ B = A
   ♦ It can be proved in 4 steps:
1. To find A ∩ B, we write all elements which are common to both A and B.
2. Here, all elements of A are the common elements. In fact, they are the only common elements.
3. All we need to do is: Write the elements of A.
4. So we can write: A ⊂ B ⇒ A ∩ B = A

◼ Thus we see that, all given conditions are equivalent.

Solved example 1.59
Assume that P(A) = P(B). Show that A = B
Solution:
• We will solve this problem in two parts: Part (a) and Part (b).
    ♦ In part (a), we prove that A ⊂ B
    ♦ In part (b), we prove that B ⊂ A
Part (a):
1. P(A) will contain all subsets of A.
• So P(A) will contain A.
• Using symbols, we write this as: A ∈ P(A)
2. If A is an element of P(A), the same A will be an element of P(B) also.
   ♦ Because, it is given that P(A) = P(B)
• Using symbols, we write this as:
A ∈ P(A) ⇒ A ∈ P(B) [∵ P(A) = P(B)]
3. If A is an element of P(B), A will be a subset of B.
• Using symbols, we write this as: A ∈ P(B) ⇒ A ⊂ B
◼ Thus we proved that A ⊂ B

Part (b):
1. P(B) will contain all subsets of B.
• So P(B) will contain B.
• Using symbols, we write this as: B ∈ P(B)
2. If B is an element of P(B), the same B will be an element of P(A) also.
   ♦ Because, it is given that P(A) = P(B)
• Using symbols, we write this as:
B ∈ P(B) ⇒ B ∈ P(A) [∵ P(A) = P(B)]
3. If B is an element of P(A), B will be a subset of A.
• Using symbols, we write this as: B ∈ P(A) ⇒ B ⊂ A
◼ Thus we proved that B ⊂ A

• So we get: A ⊂ B and B ⊂ A
   ♦ This is possible only if A and B are equal.
◼  That means, A = B

Solved example 1.60
Is it true that for any sets A and B, P(A) ∪ P(B) = P(A ∪ B)? Justify your answer.
Solution:
This is false. Let us see an example. It can be written in steps:
1. Let A = {1, 2} and B = {2, 3}
• Then the power sets are:
   ♦ P(A) = {{1}, {2}, {1, 2}, ɸ}
   ♦ P(B) = {{2}, {3}, {2, 3}, ɸ}
2. So P(A) ∪ P(B) will be: {{1}, {2}, {3}, {1, 2}, {2, 3}, ɸ}
3. (A ∪ B) will be: {1, 2, 3}
• Then the power set of (A ∪ B) will be:
P(A ∪ B) = {{1}, {2}, {3}, {1, 2, 3}, {1, 2}, {2, 3}, {1, 3}, ɸ}
4. We see that, result in (2) is different from result in (3).

Solved example 1.61
Show that for any sets A and B,
A - B = A ∩ B'
Solution:
1. Consider the LHS: A - B
   ♦ To find A - B, we discard all elements of B.
   ♦ Then we discard those elements of A which are present in B also.
• The result will be the left side crescent in fig.1.27(b) below:

Fig.1.27

(The dotted B circle is imaginary. It does not come in the result. It is shown only to indicate relative positions)
2. Consider the RHS A ∩ B'
• B' is shown in fig.1.28(a) below. It is the red shaded portion.

Fig.1.28

• We want the intersection of two regions:
   ♦ Region B'
   ♦ Region A
3. The resulting intersection is the region,
   ♦ Where there is shading from both regions B' and A.
• So the result will be as in fig.1.28(b)
4. We see that:
Fig.1.27(b) is same as fig.1.28(b)
◼ So we can write: A - B = A ∩ B'
◼ Similarly, we can prove that: B - A = B ∩ A'

Solved example 1.62
Show that for any sets A and B,
(a) A = (A ∩ B) ∪ (A – B)
(b) A ∪ (B – A) = (A ∪ B)
Solution:
Part (a):
1. We have to prove that A = (A ∩ B) ∪ (A – B)
• Consider (A - B) in the RHS
• From the results in the previous solved example 1.61, we can write (A ∩ B') in the place of (A - B)
2. So the RHS becomes:
(A ∩ B) ∪ (A ∩ B')
3. Now recall the distributive law of intersection (Fig.1.14 of section 1.6):
X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z)
• The RHS in this law is similar to the result written in (2)
• So the result in (2) becomes: A ∩ (B ∪ B')
4. So the question becomes:
Prove: A = A ∩ (B ∪ B')
5. Consider (B ∪ B') in the RHS
• (B ∪ B') is U
• So the RHS in (4) becomes: A ∩ U
6. But A ∩ U is A
• So LHS and RHS are equal.

Part (b):
1. We have to prove that A ∪ (B – A) = (A ∪ B)
• Consider (B - A) in the LHS.
• From the results in the previous solved example 1.61, we can write (B ∩ A') in the place of (B - A)
2. So the LHS becomes:
A ∪ (B ∩ A')
3. Now recall the distributive law of union:
X ∪ (Y ∩ Z) = (X ∪ Y) ∩ (X ∪ Z)
• The LHS in this law is similar to the result written in (2)
• So the result in (2) becomes: (A ∪ B) ∩ (A ∪ A')
4. So the question becomes:
Prove: (A ∪ B) ∩ (A ∪ A') = (A ∪ B)
5. Consider (A ∪ A') in the RHS
• (A ∪ A') is U
• So the LHS in (4) becomes: (A ∪ B) ∩ U
6. But (A ∪ B) ∩ U is (A ∪ B)
• So LHS and RHS are equal.

Alternate method using Venn diagrams:
1. Fig.1.29(a) below shows our familiar 'Venn diagram of two sets'.

Fig.1.29

2. Fig.1.29(b) shows the three portions separated from each other. This is also familiar to us.
◼ It is clear that, the union of
    ♦ left crescent portion
    ♦ and the middle portion
    ♦ will give set A 
• This proves part (a): A = (A ∩ B) ∪ (A – B)
3. From part (a), we have: A = (A ∩ B) ∪ (A – B)
• So A ∪ (B – A) = (A ∪ B) is same as: (A ∩ B) ∪ (A – B) ∪ (B – A)
    ♦ This is the union of the three portions in fig.1.29(b)
• The union of those three portions will give (A ∪ B)
• This proves part (b): A ∪ (B – A) = (A ∪ B)   

Solved example 1.63
Using properties of sets, show that
(a) A ∪ ( A ∩ B ) = A
(b) A ∩ ( A ∪ B ) = A.
Solution:
Part (a):
1. We have to prove that A ∪ ( A ∩ B ) = A
• Consider the LHS. Recall the distributive law of union:
X ∪ (Y ∩ Z) = (X ∪ Y) ∩ (X ∪ Z)
2. Applying this law, the LHS becomes:
(A ∪ A) ∩ (A ∪ B)
3. But (A ∪ A) is A
• So the LHS becomes:
A ∩ (A∪B)
4. But A ∩ (A∪B) is A
• So the LHS becomes A
5. Thus we get LHS = RHS

Part (b):
1. We have to prove that A ∩ ( A ∪ B ) = A
• Consider the LHS. Recall the distributive law of intersection (Fig.1.14 of section 1.6):
X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z)
2. Applying this law, the LHS becomes:
(A ∩ A) ∪ (A ∩ B)
3. But (A ∩ A) is A
• So the LHS becomes:
A ∪ (A ∩ B)
4. But in part (a), we proved that A ∪ (A ∩ B) is A.
• So the LHS becomes A.
5. Thus we get LHS = RHS

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In the next section, we will see some relations between three sets.

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Chapter 1.7 - Difference of Sets

In the previous section, we saw intersection of sets. In this section, we will see difference of sets.

Difference of sets

• This can be explained in 9 steps:
1. Let A and B be any two sets.
• We perform an operation called ‘difference of sets’ between A and B.
2. The set which is formed as a result of that operation,
    ♦ will contain only those elements of A which do not belong to B.
• This can be explained further as follows:
    ♦ The resulting set can contain elements of A.
    ♦ The resulting set can not contain any element of B.
    ♦ The resulting set can not contain any element which is common to A and B.
3. The symbol ‘－’ is used to represent difference.
• So the difference of A and B can be represented as A － B.
    ♦ It is read as: A minus B.
4. Let us see some examples:
A = {2, 4, 6, 8, 10} and B = {2, 4, 6, 12, 14}. Find A － B.
Solution:
• A － B can contain elements of A.
• A － B can not contain any element of B.
    ♦ So we can right away discard set B.
• Also, A － B cannot contain any element which is common to both A and B.
    ♦ So we can discard 2, 4 and 6 from A.
• Thus we get: A － B = {8, 10}
• We can represent this difference using Venn diagrams also.
    ♦ Fig.1.16(a) below shows A and B before the difference.
    ♦ Fig.1.16(b) shows the new set which is A － B.

Fig.1.16

5. In the above example, find B － A
Solution:
• B － A can contain elements of B.
• B － A cannot contain any element of A.
    ♦ So we can right away discard set A.
• Also, B － A cannot contain any element which is common to both B and A.
    ♦ So we can discard 2, 4 and 6 from B.
• Thus we get: B － A = {12, 14}
• We can represent this difference using Venn diagrams also.
    ♦ Fig.1.17(a) below shows A and B before the difference.
    ♦ Fig.1.17(b) shows the new set which is B － A.

Fig.1.17

6. The above two examples show that:
    ♦ A － B
    ♦ is not equal to
    ♦ B － A
• So we must be careful about the order while specifying the difference.
• We must clearly examine the two sets:
    ♦ The set on the left side of '－' sign.
    ♦ The set on the right side of '－' sign.
7. Another example:
Let V = {a, e, i, o, u} and B = {a, i, k, u}. Show that V － B ≠ B － V 
Solution:
(i) V － B can contain elements of V.
• V － B cannot contain any element of B.
    ♦ So we can right away discard set B.
• Also, V － B cannot contain any element which is common to both V and B.
    ♦ So we can discard a, i and u from V.
• Thus we get: V － B = {e, o}
(ii) B － V can contain only elements of B.
• B － V cannot contain any element of V.
    ♦ So we can right away discard set V.
• Also, B － V cannot contain any element which is common to both V and B.
    ♦ So we can discard a, i and u from B.
• Thus we get: V － B = {k}
(iii) Thus we see that: V － B ≠ B － V
8. Thus we can write the definition:

Definition 8: The difference of two sets A and B is the set C which consists of all those elements which belong to A but not to B.

9. Let us see the relation between intersection and difference. It can be written in 4 steps:
(i) Consider two sets A and B.
• We can performs the operations of intersection and difference on the two sets.
(ii) As a result of those operations, we get three different sets:
A － B, B － A and A ∩ B
(Recall that, A ∩ B is same as B ∩ A. So the operation of intersection will give only one set)
(iii) Fig.1.18(a) below shows the two sets A and B before the operations.
• Fig.b shows the three resulting sets.

Fig.1.18

(iv) We see that, the three resulting sets do not have any overlapping parts.
◼ Thus we can write:
A － B, B － A and A ∩ B are mutually disjoint. They will never have any common elements.

Now we will see some solved examples:
Solved example 1.26
Find the union of each of the following pairs of sets :
(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = [ a, e, i, o, u} B = {a, b, c}
(iii) A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}
(iv) A = {x : x is a natural number and 1 < x ≤ 6 }
B = {x : x is a natural number and 6 < x < 10 }
(v) A = {1, 2, 3}, B = ɸ
Solution:
(i) X ∪ Y = {1, 2, 3, 5}
(ii) A ∪ B = {a, b, c, e, i, o, u}
(iii) In roster form, A = {3, 6, 9, 12, . . .}
• In roster form, B = {1, 2, 3, 4, 5}
• So A ∪ B = {x : x = 1, 2, 4, 5 or a multiple of 3}
(iv) In roster form, A = {2, 3, 4, 5, 6}
• In roster form, B = {7, 8, 9}
• So A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}
(v) A ∪ B = {1, 2, 3}

Solved example 1.27
Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?
Solution:
• All elements of A are present in B. So A ⊂ B
• A ∪ B = {a, b, c} = B

Solved example 1.28
If A and B are two sets such that A ⊂ B, then what is A ∪ B
Solution:
If A is a subset of B, then A ∪ B will be B

Solved example 1.29
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 } and D = { 7, 8, 9, 10 }; find
(i) A ∪ B  (ii) A ∪ C  (iii) B ∪ C  (iv) B ∪ D  (v) A ∪ B ∪ C  (vi) A ∪ B ∪ D  (vii) B ∪ C ∪ D
Solution:
(i) A ∪ B = {1, 2, 3, 4, 5, 6}
(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(iii) B ∪ C = {3, 4, 5, 6, 7, 8}
(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}
(v) A ∪ B = {1, 2, 3, 4, 5, 6}
   ♦ So A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(vi) A ∪ B = {1, 2, 3, 4, 5, 6}
   ♦ So A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) B ∪ C = {3, 4, 5, 6, 7, 8}
   ♦ So B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

Solved example 1.30
Find the intersection of each pair of sets of example 1.26 above.
Solution:
(i) X ∩ Y = {1, 3}
(ii) A ∩ B = {a}
(iii) In roster form, A = {3, 6, 9, 12, . . .}
• In roster form, B = {1, 2, 3, 4, 5}
• So A ∩ B = {3}
(iv) In roster form, A = {2, 3, 4, 5, 6}
• In roster form, B = {7, 8, 9}
• So A ∩ B = ɸ
(v) A ∩ B = ɸ {1, 2, 3}

Solved example 1.31
If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find
(i) A ∩ B  (ii) B ∩ C  (iii) A ∩ C ∩ D  (iv) A ∩ C  (v) B ∩ D  (vi) A ∩ (B ∪ C) (vii) A ∩ D
(viii) A ∩ (B ∪ D)  (ix) ( A ∩ B ) ∩ ( B ∪ C )  (x) (A ∪ D) ∩ (B ∪ C)
Solution:
(i) A ∩ B = {7, 9, 11}
(ii) B ∩ C = {11, 13}
(iii) A ∩ C = {11}
   ♦ So A ∩ C ∩ D = ɸ
(iv) A ∩ C = {11}
(v) B ∩ D = ɸ
(vi) B ∪ C = {7, 9, 11, 13, 15}
   ♦ So A ∩ (B ∪ C) = {7, 9, 11}
(vii) A ∩ D = ɸ
(viii) B ∪ D = {7, 9, 11, 13, 15, 17}
   ♦ So A ∩ (B ∪ D) = {7, 9, 11}
(ix) A ∩ B = {7, 9, 11}
• B ∪ C = {7, 9, 11, 13, 15}
   ♦ So ( A ∩ B ) ∩ ( B ∪ C )   = {7, 9, 11}
(x) A ∪ D = {3, 5, 7, 9, 11, 15, 17}
• B ∪ C = {7, 9, 11, 13, 15}
   ♦ So ( A ∩ B ) ∩ ( B ∪ C ) = {7, 9, 11, 15}

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The link below gives some solved examples:

Solved examples 1.32 to 1.37

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In the next section, we will see complement of a set.

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