Chapter 2 - Relations And Functions

In the previous section, we completed the discussion on sets. In this chapter, we will see Relations and Functions.

First we will see some basics about Cartesian product of sets. It can be written in 7 steps:
1. Consider two sets A and B
   ♦ A is a set of two colors.
         ✰ A = {red, blue}
   ♦ B is a set of three objects.
         ✰ B = {bag, coat, shirt}
2. We can assign any of the two colors (from set A) to any of the three objects (in set B).
• So there is a possibility of making six different colored objects. They are:
(red, bag), (red, coat), (red, shirt), (blue, bag), (blue, coat), (blue, shirt)
3. No two pairs among the above six, are identical.
• In other words, each pair is distinct from others.
4. The formation of the pairs can be shown diagrammatically as in fig.2.1(a) below:
• The horizontal and vertical lines intersect at various points.
• A red dot is placed at each of those intersecting points.

Fig.2.1

5. The six red dots represent the six pairs that we wrote in (2)
• For example:
   ♦ The intersection of red-line and coat-line will give the pair (red, coat)
   ♦ The intersection of blue-line and shirt-line will give the pair (blue, shirt)
6. We see that, no two red dots overlap. This is additional proof that, the pairs that we wrote in (2), are distinct.
7. Each of the six pairs that we wrote in (2), is an ordered pair. That means, the order inside the pair should not be changed. For example, (blue, bag) should not be written as (bag, blue)

Let us see another case. It can be written in 6 steps:
1. Consider two sets A and B
   ♦ A is a set of three locations.
         ✰ A = {DL, MP, KA}
(DL represents Delhi, MP represents Madhya Pradesh and KA represents Karnataka) 
   ♦ B is a set of three codes for license plates of vehicles.
         ✰ B = {01, 02, 03}
2. Using the two sets, we can make several combinations. The fig.2.1(b) above will help us to make those combinations.
• Consider the first vertical line. When we move from bottom to top along that line, we get: (DL, 01), (DL, 02), (DL, 03)
• Consider the second vertical line. When we move from bottom to top along that line, we get: (MP, 01), (MP, 02), (MP, 03)
• Consider the third vertical line. When we move from bottom to top along that line, we get: (KA, 01), (KA, 02), (KA, 03)
• So in total, there are nine pairs.
3. No two pairs among the above nine, are identical.
• In other words, each pair is distinct from others.
4. Red dots are used to mark the pairs.
• As before, those red dots are at the intersecting points.
• There is a total of nine red dots.
5. We see that, no two red dots overlap. This is additional proof that, the pairs that we wrote in (2), are distinct.
6. Each of the nine pairs that we wrote in (2), is an ordered pair. That means, the order inside the pair should not be changed. For example, (MP, 03) should not be written as (03, MP)

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Now we can write the definition for Cartesian product A⨯B. It can be written in 9 steps:
1. Cartesian product A⨯B is a set.
2. A⨯B is formed from two sets A and B.
3. A⨯B contains all possible combinations between A and B.
4. The combinations are written as ordered pairs.
5. So A⨯B is a set and the ordered pairs are the elements of that set.
6. The first member of all ordered pairs should be from A. The second member of all ordered pairs should be from B.
7. If it is B⨯A, the order also reverses:
• The first member of all ordered pairs should be from B. The second member of all ordered pairs should be from A.
8.Since A⨯B is a set, we can write it in the set builder form:
A⨯B = { (a,b) : a∈A, b∈B}
• It is read as:
Set of all ordered pairs (a,b) such that a is an element of A and b is an element of B
9. Similarly we can write B⨯A also:
B⨯A = { (b,a) : a∈A, b∈B}
• It is read as:
Set of all ordered pairs (b,a) such that b is an element of B and a is an element of A 
(In the examples that we saw above, the ordered pairs represented by the red dots, are elements of the set A⨯B.)

Let us see one more example:
• Let A ={a₁, a₂} and B = {b₁, b₂, b₃, b₄}
• All possible combinations can be obtained using fig.2.1(c) above. We get:
A × B = {(a₁, b₁), (a₁, b₂), (a₁, b₃), (a₁, b₄), (a₂, b₁), (a₂, b₂), (a₂, b₃), (a₂, b₄)}
• Note that in each ordered pair,
   ♦ The first member is from set A.
   ♦ The second member is from set B.

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◼ Once we become familiar with the process, we will not need to draw diagrams like the ones shown in fig.2.1 above. All we need to do is:
• Take the first element of A. Combine it with each element of B. This is shown by the red arrows in fig.2.2(a) below.
    ♦ From the red arrows, we get:
    ♦ (a₁, b₁), (a₁, b₂), (a₁, b₃), (a₁, b₄)

Fig.2.2

• Take the second element of A. Combine it with each element of B. This is shown by the green arrows in fig.2.2(b) above.
    ♦ From the green arrows, we get:
    ♦ (a₂, b₁), (a₂, b₂), (a₂, b₃), (a₂, b₄)

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• In the above examples, we used sets which contain colors, objects, places etc.
• Let us see the peculiarities when sets contain only real numbers. It can be written in 3 steps:
1. Suppose that two sets P and Q contain only real numbers.
2. We have seen that the set P⨯Q will contain some ordered pairs.
• If both P and Q contain only real numbers, those ordered pairs will also contain only real numbers.
3. We know that, ordered pairs containing real numbers can be plotted on a plane.
• For example, (1,2), (-1,3), (2,√5) etc., can be plotted on the x-y plane.
• When we plot such ordered pairs, we see that, they never overlap. That is., each ordered pair has it’s own unique position in the plane.
    ♦ The ordered pair (p₁,q₁)
    ♦ Can never occupy the position of
    ♦ The ordered pair (p₂,q₂)
• If they do occupy the same position, then it is obvious that:
    ♦ P₁ is equal to P₂
    ♦ q₁ is equal to q₂

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Based on the discussion so far, we can write four remarks about the Cartesian product PQ:
Remark 1
Two ordered pairs are equal if and only if:
    ♦ Corresponding first elements are equal.
    ♦ And
    ♦ Corresponding second elements are equal.
• Let us see an example:
If (x + 1, y – 2) = (3,1), find the values of x and y.
Solution:
• The two ordered pairs are equal. So the corresponding elements must be equal.
• Equating the first elements, we get: x + 1 = 3
    ♦ This gives x = (3 - 1) = 2
• Equating the second elements, we get: y - 2 = 1
    ♦ This gives y = (2 + 1) = 3
Remark 2
If there are n₁ elements in A and n₂ elements in B, then there will be n₁n₂ elements in A × B.
• This can be explained in two steps:
(i) Number of elements in each set:
    ♦ There are n₁ elements in A. So we write: n(A) = n₁
    ♦ There are n₂ elements in B. So we write: n(B) = n₂
(ii) We want the number of elements in the set A × B.
• The number of elements in A × B is denoted as n(A × B)
• We will always get: n(A × B) = n₁n₂
◼ The examples that we saw earlier, will give enough proof:
• In our first example in fig.2.1(a) above, we have:
    ♦ n(A) = n₁ = 2 and n(B) = n₂ = 3
    ♦ In A × B, we obtained 6 red dots. That is., 6 ordered pairs.
    ♦ We see that: n₁n₂ = 2 × 3 = 6  
• In our second example in fig.2.1(b) above, we have:
    ♦ n(A) = n₁ = 3 and n(B) = n₂ = 3
    ♦ In A × B, we obtained 9 red dots. That is., 9 ordered pairs.
    ♦ We see that: n₁n₂ = 3 × 3 = 9  
• In our third example in fig.2.1(c) above, we have:
    ♦ n(A) = n₁ = 2 and n(B) = n₂ = 4
    ♦ In A × B, we obtained 8 red dots. That is., 8 ordered pairs.
    ♦ We see that: n₁n₂ = 2 × 4 = 8
Remark 3
If A and B are non-empty sets and either A or B is an infinite set, then A × B is an infinite set.
This can be explained in 4 steps:
(i) Given that A is a non-empty set. So if n(A) = n₁, then n₁ is a non zero number.
(ii) Given that B is also a non-empty set. So if n(B) = n₂, then n₂ is a non zero number.
(iii) Suppose that B is an infinite set. Then n₂ = ∞
(iv) Then we will get: n(A × B) = n₁n₂ = (n₁ × ∞) = ∞
Remark 4
A × A × A is a set. It can be written in set builder form as:
A × A × A = {(a,b,c) : a, b, c ∈ A}
• That means, the set A × A × A will contain elements of the form (a,b,c)
a, b and c must be elements of the set A
• We named (a,b) as: ordered pair.
    ♦ Similarly, (a,b,c) is named as: ordered triplet.
• The order in (a,b,c) is important. The following example will make this clear.
◼ If P = {1, 2}, form the set P × P × P.
Solution:
1. In fig.2.1, we used horizontal and vertical lines. But here we use horizontal and vertical planes. This is shown in fig.2.3 below:

Fig.2.3

• The two horizontal planes are pink in color.
    ♦ One is for ‘1’ and the other is for ‘2’
• Two vertical planes are green in color.
    ♦ One is for ‘1’ and the other is for ‘2’
• The other two vertical planes are blue in color.
    ♦ One is for ‘1’ and the other is for ‘2’
(Blue planes are perpendicular to green planes)
2. In fig.2.1, we marked the intersection of lines. But here, we mark the intersection of planes.
• Any two non parallel planes will intersect along a line.
• But three non parallel planes can intersect only at a unique single point.
• Let us find those intersection points. We will take the order: (Pink, Green, Blue)
3. Consider the pink plane ‘1’.
    ♦ The green plane ‘1’ will cut this pink plane along a line.
    ♦ The blue plane ‘1’ will cut this line at a point.
        ✰ We can name this point as (1,1,1)
4. Consider the pink plane ‘1’.
    ♦ The green plane ‘1’ will cut this pink plane along a line.
    ♦ The blue plane ‘2’ will cut this line at a point.
        ✰ We can name this point as (1,1,2)     
5. Consider the pink plane ‘1’.
    ♦ The green plane ‘2’ will cut this pink plane along a line.
    ♦ The blue plane ‘1’ will cut this line at a point.
        ✰ We can name this point as (1,2,1)     
Consider the pink plane ‘1’.
    ♦ The green plane ‘2’ will cut this pink plane along a line.
    ♦ The blue plane ‘2’ will cut this line at a point.
        ✰ We can name this point as (1,2,2)
6. Proceeding like this, we will get eight ordered triplets. They are:
(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)
7. The above process is shown only for understanding the 3D aspect of the problem. While solving this type of problems, we adopt the following two steps:
(i) First find A × A:
We get: {(1,1), (1,2), (2,1), (2,2)}
(ii) Then find (A × A) × A
• That is: {(1,1), (1,2), (2,1), (2,2)} × {1,2}
• We get:{(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)}

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The link below gives some solved examples based on Cartesian Products.

Solved examples 2.1 to 2.14

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In the next section, we will see Relations.

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