Chapter 2.1 - Relations

In the previous section, we saw Cartesian product. In this section, we will see relations.

We will see the basics about relations using three examples.
Example 1:
This can be written in 7 steps:
1. Consider two sets A and B
   ♦ A is a set of four students: Student 1, Student 2, Student 3, Student 4.
         ✰ A = {1, 2, 3, 4}
   ♦ B is a set of five available courses.
         ✰ B = {Maths, Physics, Chemistry, Biology, Geography}
2. The students have the liberty to learn one or more of the available courses.
• So there are twenty possible combinations. They are:
(1, Maths),           (2, Maths),         (3, Maths),         (4, Maths),
(1, Physics),         (2, Physics),       (3, Physics),        (4, Physics),
(1, Chemistry),    (2, Chemistry),   (3, Chemistry),   (4, Chemistry),
(1, Biology),         (2, Biology),       (3, Biology),        (4, Biology),
(1, Geography),   (2, Geography),  (3, Geography),   (4, Geography).
• They are the 20 ordered pairs.
3. Using the method that we saw in the previous section, the above twenty ordered pairs can be denoted by red dots. This is shown in fig.2.4(a) below:

Fig.2.4

4. The red dots only give us the various possibilities. For example, (3, Biology) indicates that, the student 3 can choose to do Biology.
• So the red dots are useful during the admission processes. They help students and authorities to select and allocate various courses.
• Once the admission processes are complete, the red dots have not much value. At that stage, we will be wanting to know which student chose which course.
5. So we need to pick the appropriate red dots from among the total 20 red dots. This is shown in fig.2.4(b) above. The selected dots are marked with green circles.
• Consider any one green circle. Say the one at the intersection of student 2 line and Chemistry line.
• This green circle indicates that, student 2 chose to do the chemistry course.
6. The ordered pairs with green circle are:
(1, Maths), (2, Maths), (2, Chemistry), (3, Chemistry), (3, Biology), and (4, Geography)
• These pairs give us the following information:
   ♦ Student 1 chose to do Maths.
   ♦ Student 2 chose to do Maths.
   ♦ Student 2 chose to do Chemistry.
   ♦ Student 3 chose to do Chemistry.
   ♦ Student 3 chose to do Biology.
   ♦ Student 4 chose to do Geography.
7. This information can be shown in an arrow diagram also. It is shown in fig.2.4(c) above.
• Note that:
   ♦ The number of green circles in fig.b is 6
   ♦ The number of green arrows in fig.c is also 6
• This is because, both the figs. b and c convey the same information.

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• Let us write this example in terms of sets and ordered pairs. The 8 steps given below will help us:
1. A is the set of students and B is the set of courses.
2. All the ‘possible combinations’ is given by the red dots in fig.2.4(a)
• As we saw in the previous section, all those red dots will be included in the set A × B
3. The green circles in fig.2.4(b) shows the relation between students and their chosen courses.
• We can make a set R which contains all the green circles in fig.2.4(b)
4. All the green circles are obtained from the red dots. So the set R will be a subset of A × B
• We can write: R ⊂ (A × B)
5. In the previous section, we saw that all elements of A × B are ordered pairs.
• Since R is a subset of A × B, all the elements of R will also be ordered pairs.
• In our present case,
R = {(1, Maths), (2, Maths), (2, Chemistry), (3, Chemistry), (3, Biology), and (4, Geography)}
6. The green circles in fig.b can be better visualized using the arrow diagram in fig.c
   ♦ The green circles in fig.b give us the ordered pairs in R.
   ♦ The green arrows in fig.c give us the same ordered pairs in R.
7. The usefulness of the arrow diagram will be clear from the following 3 steps:
(i) Take any ordered pair in R. Look at the corresponding green arrow in the arrow diagram.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
• We can work in the reverse also. It can be written in 4 steps:
(i) Take any green arrow in the arrow diagram. Corresponding to that arrow, there will be an ordered pair in R.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
(iv) All the green arrows must be included as ordered pairs in the set R.
8. In step (5), we wrote R in the roster form. We must be able to write it in the set builder form also. The following 3 steps will enable us to do so:
(i) We know that, set R contains ordered pairs. Let the general form of those ordered pairs be (x,y)
(ii) Then x will be the student and y will be the course chosen by that student.
(iii) So the set builder form will be:
R = {(x,y) : y is the course chosen by student x, x ∈ A, y ∈ B}

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Example 2:
This can be written in 6 steps:
1. Consider two sets A and B
   ♦ A = {5, 6, 7}
   ♦ B = {3, 4, 5}
2. n(A) = 3 and n(B) = 3
• So there are nine possible combinations. They are:
(5, 3),            (6, 3),           (7, 3),
(5, 4),            (6, 4),           (7, 4),
(5, 5),            (6, 5),           (7, 5).
• They are the 9 ordered pairs.
3. Using the method that we saw in the previous section, the above nine ordered pairs can be denoted by red dots. This is shown in fig.2.5(a) below:

Fig.2.5

4. The red dots give us the various possible combinations.
Suppose that, we want only those combinations in which:
   ♦ The element taken from B
   ♦ is 2 less than
   ♦ The element taken from A.
• Then we need to pick the appropriate red dots from among the total 9 red dots.
• The appropriate red dots can be determined using 2 steps:
(i) Let x be the element (which satisfies the relation) from A. Let y be the corresponding element (which satisfies the relation) from B
• Then the algebraic form of the relation is: x - 2 = y
(ii) Let us take each possible value for x from set A:
• When x = 5,
   ♦ y = (x-2) = (5-2) = 3
   ♦ '3' is available in B
   ♦ So the ordered pair (5, 3) satisfies the given relation.
         ✰ Note that (5, 3) is one among the red dots in fig.a
• When x = 6,
   ♦ y = (x-2) = (6-2) = 4
   ♦ '4' is available in B
   ♦ So the ordered pair (6, 4) satisfies the given relation.
         ✰ Note that (6, 4) is one among the red dots in fig.a
• When x = 7,
   ♦ y = (x-2) = (7-2) = 5
   ♦ '5' is available in B
   ♦ So the ordered pair (7, 5) satisfies the given relation.
         ✰ Note that (7, 5) is one among the red dots in fig.a
5. We need to select the above three ordered pairs from among the 9 ordered pairs. This is shown in fig.2.5(b) above. The selected dots are marked with green circles.
• The ordered pairs with green circle are:
(5, 3), (6, 4) and (7, 5)
6. This information can be shown in an arrow diagram also. It is shown in fig.2.5(c) above.
• Note that:
   ♦ The number of green circles in fig.b is 3
   ♦ The number of green arrows in fig.c is also 3
• This is because, both the figs. b and c convey the same information.

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• Let us write this example in terms of sets and ordered pairs. The 8 steps given below will help us:
1. A = {5, 6, 7} and B = {3, 4, 5}
2. All the ‘possible combinations’ is given by the red dots in fig.2.5(a)
• As we saw in the previous section, all those red dots will be included in the set A × B
3. The green circles in fig.2.5(b) shows those ordered pairs which satisfy a particular relation.
• The relation is this:
   ♦ The element taken from B
   ♦ is 2 less than
   ♦ The element taken from A.
• We can make a set R which contains all the green circles in fig.2.4(b)
4. All the green circles are obtained from the red dots. So the set R will be a subset of A × B
• We can write: R ⊂ (A × B)
5. In the previous section, we saw that all elements of A × B are ordered pairs.
• Since R is a subset of A × B, all the elements of R will also be ordered pairs.
• In our present case,
R = {(5, 3), (6, 4), (7, 5)}
6. The green circles in fig.b can be better visualized using the arrow diagram in fig.c
   ♦ The green circles in fig.b give us the ordered pairs in R.
   ♦ The green arrows in fig.c give us the same ordered pairs in R.
7. The usefulness of the arrow diagram will be clear from the following 3 steps:
(i) Take any ordered pair in R. Look at the corresponding green arrow in the arrow diagram.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
• We can work in the reverse also. It can be written in 4 steps:
(i) Take any green arrow in the arrow diagram. Corresponding to that arrow, there will be an ordered pair in R.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
(iv) All the green arrows must be included as ordered pairs in the set R.
8. In step (5), we wrote R in the roster form. We must be able to write it in the set builder form also. The following 3 steps will enable us to do so:
(i) We know that, set R contains ordered pairs. Let the general form of those ordered pairs be (x,y)
(ii) Then x will be the element from A and y will be the element from B.
(iii) So the set builder form will be:
R = {(x,y) : y = x - 2, x ∈ A, y ∈ B}

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Example 3:
This can be written in 6 steps:
1. Consider the set A
   ♦ A = {1, 2, 3, 4, 5, 6}
2. We want the possible combinations of A with itself.
n(A) = 6
• So there are 36 possible combinations. They are:
(1, 1),            (2, 1),           (3, 1),         (4, 1),            (5, 1),           (6, 1),
(1, 2),            (2, 2),           (3, 2),         (4, 2),            (5, 2),           (6, 2),
(1, 3),            (2, 3),           (3, 3),         (4, 3),            (5, 3),           (6, 3),
(1, 4),            (2, 4),           (3, 4),         (4, 4),            (5, 4),           (6, 4),
(1, 5),            (2, 5),           (3, 5),         (4, 5),            (5, 5),           (6, 5),
(1, 6),            (2, 6),           (3, 6),         (4, 6),            (5, 6),           (6, 6).
• They are the 36 ordered pairs.
3. Using the method that we saw in the previous section, the above 36 ordered pairs can be denoted by red dots. This is shown in fig.2.6(a) below:

Fig.2.6

4. The red dots give us the various possible combinations.
Suppose that, we want only those combinations in which:
   ♦ The element taken from B
   ♦ is 1 greater than
   ♦ The element taken from A.
• Then we need to pick the appropriate red dots from among the total 36 red dots.
• The appropriate red dots can be determined using 2 steps:
(i) Let x be the element (which satisfies the relation) from A. Let y be the corresponding element (which satisfies the relation) from B
• Then the algebraic form of the relation is: x + 1 = y
(ii) Let us take each possible value for x from set A:
• When x = 1,
   ♦ y = (x+1) = (1+1) = 2
   ♦ '2' is available in A
   ♦ So the ordered pair (1, 2) satisfies the given relation.
         ✰ Note that (1, 2) is one among the red dots in fig.a
• When x = 2,
   ♦ y = (x+1) = (2+1) = 3
   ♦ '3' is available in A
   ♦ So the ordered pair (2, 3) satisfies the given relation.
         ✰ Note that (2, 3) is one among the red dots in fig.a
• When x = 3,
   ♦ y = (x+1) = (3+1) = 4
   ♦ '4' is available in A
   ♦ So the ordered pair (3, 4) satisfies the given relation.
         ✰ Note that (3, 4) is one among the red dots in fig.a
• When x = 4,
   ♦ y = (x+1) = (4+1) = 5
   ♦ '5' is available in A
   ♦ So the ordered pair (4, 5) satisfies the given relation.
         ✰ Note that (4, 5) is one among the red dots in fig.a
• When x = 5,
   ♦ y = (x+1) = (5+1) = 6
   ♦ '6' is available in A
   ♦ So the ordered pair (5, 6) satisfies the given relation.
         ✰ Note that (5, 6) is one among the red dots in fig.a
• When x = 6,
   ♦ y = (x+1) = (6+1) = 7
   ♦ '7' is not available in A
   ♦ So the ordered pair (6, 7) does not satisfy the given relation.
         ✰ Note that (6, 7) is not among the red dots in fig.a
5. We need to select the above five ordered pairs from among the 36 ordered pairs. This is shown in fig.2.6(b) above. The selected dots are marked with green circles.
• The ordered pairs with green circle are:
(1, 2), (2, 3), (3, 4), (4, 5) and (5, 6)
6. This information can be shown in an arrow diagram also. It is shown in fig.2.6(c) above.
• Note that:
   ♦ The number of green circles in fig.c is 5
   ♦ The number of green arrows in fig.c is also 5
• This is because, both the figs. b and c convey the same information.

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• Let us write this example in terms of sets and ordered pairs. The 8 steps given below will help us:
1. A = {1, 2, 3, 4, 5, 6}
2. All the ‘possible combinations’ from A to A is given by the red dots in fig.2.6(a)
• As we saw in the previous section, all those red dots will be included in the set A × A
3. The green circles in fig.2.6(b) shows those ordered pairs which satisfy a particular relation.
• The relation is this:
   ♦ The element taken from A
   ♦ is 1 greater than
   ♦ The element taken from A.
• We can make a set R which contains all the green circles in fig.2.6(b)
4. All the green circles are obtained from the red dots. So the set R will be a subset of A × A
• We can write: R ⊂ (A × A)
5. In the previous section, we saw that all elements of A × A are ordered pairs.
• Since R is a subset of A × A, all the elements of R will also be ordered pairs.
• In our present case,
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
6. The green circles in fig.b can be better visualized using the arrow diagram in fig.c
   ♦ The green circles in fig.b give us the ordered pairs in R.
   ♦ The green arrows in fig.c give us the same ordered pairs in R.
7. The usefulness of the arrow diagram will be clear from the following 3 steps:
(i) Take any ordered pair in R. Look at the corresponding green arrow in the arrow diagram.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
• We can work in the reverse also. It can be written in 4 steps:
(i) Take any green arrow in the arrow diagram. Corresponding to that arrow, there will be an ordered pair in R.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
(iv) All the green arrows must be included as ordered pairs in the set R.
8. In step (5), we wrote R in the roster form. We must be able to write it in the set builder form also. The following steps will enable us to do so:
(i) We know that, set R contains ordered pairs. Let the general form of those ordered pairs be (x,y)
(ii) Then x will be the element from A and y will be the element from A.
(iii) So the set builder form will be:
R = {(x,y) : y = x + 1, x ∈ A, y ∈ A}

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• The above three examples help us to understand the basics about relations.
• We will now see some important terms involved with relations. The important terms are:
(a) image
(b) domain
(c) range
(d) Codomain

(a) image:
This can be explained in 3 steps:
1. We know that, the set R will contain one or more ordered pairs.
2. Each of those ordered pairs will have two elements.
3. The second element is called the image of the first element.
◼ Consider the ordered pairs in R of our first example. We can write:
    ♦ Maths is the image of student 1
    ♦ Maths is the image of student 2
    ♦ Chemistry is the image of student 2
    ♦ Chemistry is the image of student 3
    ♦ Biology is the image of student 3
    ♦ Geography is the image of student 4
◼ Consider the ordered pairs in R of our second example. We can write:
    ♦ 3 is the image of 5
    ♦ 4 is the image of 6
    ♦ 5 is the image of 7
◼ Consider the ordered pairs in R of our third example. We can write:
    ♦ 2 is the image of 1
    ♦ 3 is the image of 2
    ♦ 4 is the image of 3
    ♦ 5 is the image of 4
    ♦ 6 is the image of 5

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(b) domain:
This can be explained in 5 steps:
1. We know that, the set R will contain one or more ordered pairs.
2. Each of those ordered pairs will have two elements.
3. Pick out all the first elements.
4. Make a set using those first elements.
5. This set is called the domain of the relation R.
◼ Consider the ordered pairs in R of our first example. We can write:
    ♦ The first elements are: 1, 2, 2, 3, 3, 4
    ♦ When we write them as a set, repeating elements should appear only once.
    ♦ So we get: domain = {1, 2, 3, 4}
◼ Consider the ordered pairs in R of our second example. We can write:
    ♦ The first elements are: 5, 6, 7
    ♦ Here there are no repeating elements.
    ♦ So we get: domain = {5, 6, 7}
◼ Consider the ordered pairs in R of our third example. We can write:
    ♦ The first elements are: 1, 2, 3, 4, 5
    ♦ Here there are no repeating elements.
    ♦ So we get: domain = {1, 2, 3, 4, 5}

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(c) range:
This can be explained in 5 steps:
1. We know that, the set R will contain one or more ordered pairs.
2. Each of those ordered pairs will have two elements.
3. Pick out all the second elements.
4. Make a set using those second elements.
5. This set is called the range of the relation R.
◼ Consider the ordered pairs in R of our first example. We can write:
    ♦ The second elements are: maths, maths, chemistry, chemistry, biology, geography.
    ♦ When we write them as a set, repeating elements should appear only once.
    ♦ So we get: range = {maths, chemistry, biology, geography}
◼ Consider the ordered pairs in R of our second example. We can write:
    ♦ The second elements are: 3, 4, 5
    ♦ Here there are no repeating elements.
    ♦ So we get: range = {3, 4, 5}
◼ Consider the ordered pairs in R of our third example. We can write:
    ♦ The second elements are: 2, 3, 4, 5, 6
    ♦ Here there are no repeating elements.
    ♦ So we get: domain = {2, 3, 4, 5, 6}

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(d) codomain:
We know that, the relation R is defined from set A to set B.
• The set B is also known as codomain of the relation R.
◼ Consider the relation R of our first example.
   ♦ The set B for this relation is: {Maths, Physics, Chemistry, Biology, Geography}
   ♦ So codomain of this R is : {Maths, Physics, Chemistry, Biology, Geography}
◼ Consider the relation R of our second example.
   ♦ The set B for this relation is: {3, 4, 5}
   ♦ So codomain of this R is : {3, 4, 5}
◼ Consider the relation R of our third example.
   ♦ The set B for this relation is: {1, 2, 3, 4, 5, 6}
   ♦ So codomain of this R is : {1, 2, 3, 4, 5, 6}

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From the above four definitions, following 4 points can be noted:
(i) domain will contain only those elements belonging to Set A.
    ♦ all elements of A may not be present in domain.
(ii) codomain will contain only those elements belonging to Set B.
    ♦ all elements of B will be present in codomain.
(iii) range will contain only those elements belonging to Set B.
    ♦ all elements of B may not be present in range.
(iv) From (ii) and (iii), it is clear that:
range ⊂ codomain.

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• Once we understand the basics, there will not be any need to write all the lengthy steps. We will be able to obtain the results using minimum steps.
• The solved examples given below will demonstrate the process

Solved example 2.15
Let A = {1, 2, 3,...,14}. Define a relation R from A to A by
R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and
range.
Solution:
1. The relation R is a set which contains ordered pairs of the form (x, y)
   ♦ 'x' should be from set A
   ♦ Since the relation is from A to A, 'y' should also be from set A
• The x and y in each ordered pair in R should satisfy the condition: 3x - y = 0
2. The given condition can be rearranged as: 3x = y
Let us take each possible value for x from set A:
• When x = 1,
   ♦ y = 3x = (3 × 1) = 3
   ♦ '3' is available in A
   ♦ So the first ordered pair in R is (1, 3)
• When x = 2,
   ♦ y = 3x = (3 × 2) = 6
   ♦ '6' is available in A
   ♦ So the second ordered pair in R is (2, 6)
• When x = 3,
   ♦ y = 3x = (3 × 3) = 9
   ♦ '9' is available in A
   ♦ So the third ordered pair in R is (3, 9)
• When x = 4,
   ♦ y = 3x = (3 × 4) = 12
   ♦ '12' is available in A
   ♦ So the fourth ordered pair in R is (4, 12)
• When x = 5,
   ♦ y = 3x = (3 × 5) = 15
   ♦ '15' is not available in A
   ♦ So the ordered pair (5, 15) does not satisfy the given relation.
         ✰ Note that (5, 15) will not be available in A × A also.
3. The set R will contain the four ordered pairs that we determined above. We can write:
R = {(1,3), (2,6), (3,9), (4,12)}
4. Domain is the set containing all the first elements in the ordered pairs of R. So we get:
Domain of R = {1, 2, 3, 4}
5. Codomain is the set from which we take the second elements of the ordered pairs in R. In effect, co domain is the set B.
• In our present case, since the relation is from A to A, we have A in place of B.
• So we get: codomain of R = {1, 2, 3,...,14}
6. Range is the set containing all the second elements in the ordered pairs of R.
• So we get: Range of R = {3, 6, 9, 12}

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• We know that, a relation is defined from one set A to another set B
• But some times, the relation is defined from one set A to the same set A
• In such situations, we can use any one of the two statements below:
   ♦ Relation R from A to A
   ♦ Relation R on A

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More solved examples are given at the link below:

Solved examples 2.16 to 2.25

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In the next section, we will see functions.

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Chapter 2.2 - Functions

In the previous section, we saw relations. In this section, we will see functions.

Some basics about functions can be written in 3 steps:
1. We have seen the method to define a relation from set A to set B. All we need to do is this:
    ♦ Find the ordered pairs which satisfy the relation.
    ♦ Write those ordered pairs as a set.
2. Now, to define a function, we need to check two conditions:
(i) We know that the first elements in R, will be from set A.
• We need to make sure that, every element of A is present as first elements in the R. No element of A should be left out.
(ii) Next we need to make sure that no element in A is present more than once in R.
3. If both the conditions in (2) are satisfied, that relation is a function.

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Let us examine some of the relations that we saw in the previous section.
◼ In Example 1, we have:
R = {(1, Maths), (2, Maths), (2, Chemistry), (3, Chemistry), (3, Biology), and (4, Geography)}
• All the four students are present in R. So it may be a function.
• But student 2 appears more than once. So it is not a function.
(We agree that student 3 also appears more than once. But a single instance is sufficient to confirm that the R is not a function)
◼ In Example 2, we have:
R = {(5, 3), (6, 4), (7, 5)}
• All the three elements in A are present in R. So it may be a function.
• Each of those elements appear only once. So it is a function.
◼ In Example 3, we have:
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
• The element ‘6’ is present in set A. But it is not present in R. So it is not a function.
(If the first condition is not satisfied, we can confirm that it is not a function. There is no need to check the second condition)
◼ In solved example 2.15, we have:
R = {(1,3), (2,6), (3,9), (4,12)}
• The element ‘5’ is present in A. But it is not present in R. So it is not a function.
(We agree that, none of the elements coming after 4, are present in R. But a single instance is sufficient to confirm that the R is not a function.
◼ In solved example 2.16, (file available here) we have:
R = {(1,6), (2,7), (3,8)}
• This relation is on N. But all elements of N are not present in R. So it is not a function.
◼ In solved example 2.17, we have:
R = {(9,3), (9,-3), (4,2), (4,-2), (25,5), (25,-5)}
• All elements of P are present in R. So it may be a function.
• But ‘9’ appears more than once. So it is not a function.
(We agree that ‘4’ and ‘25’ also appear more than once. But a single instance is sufficient to confirm that the R is not a function)
◼ In solved example 2.21, we have:
R = {(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4),
(6,6)}
• All elements of A are present in R. So it may be a function.
• But ‘1’ appears more than once. So it is not a function.
(We agree that ‘2’ and ‘3’ also appear more than once. But a single instance is sufficient to confirm that the R is not a function)
◼ In solved example 2.22, we have:
R = {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)}
• All the elements in the given set are present in R. So it may be a function.
• Each of those elements appear only once. So it is a function.
◼ In solved example 2.25, we have:
R is a relation from Z to Z
• We found out that, domain of R is the set Z
    ♦ That means, all the elements in Z are present as first elements in R
• Since all elements of Z are present, it may be a function.
• But there will be many repetitions of the first element.
    ♦ For example, we can put 1 in the place of ‘a’
    ♦ and put infinite different integers for b.
    ♦ Every result will be an integer.
• Since the first elements in R appear more than once, it is not a function.

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Let us see a solved example:
Solved example 2.26
Let N be the set of natural numbers and the relation R be defined on N such that R = {(x, y) : y = 2x, x, y ∈ N}.
What is the domain, codomain and range of R? Is this relation a function?
Solution:
1. In our present case, the relation R is a set which contains ordered pairs of
the form (x, y)
♦ 'x' should be from set N
♦ Since the relation is on N, 'y' should also be from set N
• The x and y in each ordered pair in R should satisfy the condition:
y = 2x
2. Recall that natural numbers are 1, 2, 3, 4, . . . (Details here)
• Let us take each possible value for ‘x’ from set N:
• Let x = 1
    ♦ Then y = 2x = (2 × 1) = 2
    ♦ So the first ordered pair in R is (1, 2)
• Let x = 2
    ♦ Then y = 2x = (2 × 2) = 4
    ♦ So the second ordered pair in R is (2, 4)
• Let x = 3
    ♦ Then y = 2x = (2 × 3) = 6
    ♦ So the third ordered pair in R is (3, 6)
3. In this way, we can obtain infinite number of ordered pairs in R.
• We will be using all the natural numbers as 'x'. That means, all elements of N will appear in R. So this may be a function
• What happens if we use a natural number (in the place of x) more than once?
Ans: we will be getting the same ordered pair more than once.
• There cannot be repetition of ordered pairs in R. So we will be using every natural number (in the place of x) only once. So this is a function.

Solved example 2.27
Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not?
(i) R = {(2,1),(3,1),(4,2)}
(ii) R = {(2,2),(2,4),(3,3),(4,4)}
(iii) R = {(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)}
Solution:
• Usually, a relation is defined from a set A to set B. Or from a set A to itself.
• But here, we are not given A or B. So we will assume that, all the elements of A are present as first elements in R.
• If all elements are not present, we will be able to straight away say that, they are not functions.
• Thus in all the three questions, we need to check the second condition only.
Part (i):
All the first elements appear only once. So it is a function.
Part (ii):
The first element '2' appear more than once. So it is not a function.
Part (iii):
All the first elements appear only once. So it is a function.

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◼ From the above discussion, it is clear that:
   ♦ All functions are relations.
   ♦ But all relations are not functions.

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• We have seen how to confirm whether a relation is a function or not. Now we will see some technical terms related to functions. Use of technical terms will help us to describe the functions using minimum words. They can be written in 14 steps:
1. If a relation is a function, we use the letter ‘f’ instead of ‘R’.
2. We know that, a relation is defined from one set A to another set B.
• If that relation is a function, we write: f: A→B
3. Some times a relation is defined from one set A to itself.
• If that relation is a function, we write: f: A→A
4. We saw that all functions are relations.
• So the terms domain, codomain, range and image that we saw for relations are applicable to functions also.
• For example,
   ♦ We write domain of a relation.
   ♦ We can write domain of a function also.
5. We saw that, if it is to be a function, every element of A should be present as first elements.
   ♦ The set containing the first elements is the domain.
• So if it is a function, the domain will contain all the elements of A
• Thus it is clear that, if it is a function, the domain will be same as set A.
6. We know that, like R, the f is also a set of ordered pairs.
• If we denote those ordered pairs as (x,y), then:
   ♦ y is called the image of x under f.
   ♦ x is called the image of y under f.
7. Consider a relation that we saw in the previous section:
y = x + 1
• We input various values of x and calculated the corresponding 'y values'. Then we wrote them as ordered pairs in the form (x,y).
8. Note that, the 'y values' are obtained by inputting various 'x values'.
• The input x values are 'processed' according to the rule given by the relation y = x + 1
   ♦ Here, the rule says that, we must add '1' to the input value of x
9. If the relation is a function, we can write:
The 'x values' are 'processed' according to the rule given by the function.
10. Or we can simply write:
The 'x values' are 'processed' according to the function.
11. So we can write:
   ♦ 'y values' are obtained
   ♦ when the 'x values' are processed
   ♦ according to the function.
• This can be schematically represented as in fig.2.9 below:

Fig.2.9

12. We have done this type of 'processing' in our earlier classes. There we denoted the output as 'y'.
• For example, in y = x +1, if we add various values of x to '1', we will get various y values.
    ♦ Here, [addition of '1' to x] is the processing.
    ♦ So it is obvious that y is same as f(x).
    ♦ Then we can write: f(x) = x + 1
13. A function is a relation. So just like R, for f also, there will be a set of ordered pairs.
   ♦ The first values in that ordered pairs will form the domain of that function.
   ♦ The second values in that ordered pairs will form the range of that function.
◼ If all the elements in the range set are real numbers, then that function is called a real valued function.
◼ If in a real valued function, all the elements in the domain set are real numbers, then that function is called a real function.
14. Let us see an example:
Let N be the set of natural numbers. Define a real valued function
f : N→N by f (x) = 2x + 1.
Solution:
(i) Given that, it is a real valued function. That means, all values obtained after processing, must be real values.
• The function can be defined by writing the ordered pairs which satisfy that function.
• So our next task is to find those ordered pairs.
(ii) It is given that, f : N→N
• This indicates that,
   ♦ the first elements of the ordered pairs (input x values) should be taken from the set N.
   ♦ the second elements (resulting y values) must be present in the set N
         ✰ N is a subset of R. So indeed, it will be a real valued function
(iii) The set N is the set of natural numbers. That is., N = {1, 2, 3, 4, . . .}
• Let x = 1
   ♦ This x is processed as follows:
   ♦ f(1) = (2 × 1 + 1) = (2 + 1) = 3
   ♦ f(1) is the 'y value' when 'x value' is 1
   ♦ So the first ordered pair (x,y) is (1,3)
• Let x = 2
   ♦ This x is processed as follows:
   ♦ f(2) = (2 × 2 + 1) = (4 + 1) = 5
   ♦ f(2) is the 'y value' when 'x value' is 2
   ♦ So the second ordered pair (x,y) is (2,5)
• Let x = 3
   ♦ This x is processed as follows:
   ♦ f(3) = (2 × 3 + 1) = (6 + 1) = 7
   ♦ f(3) is the 'y value' when 'x value' is 3
   ♦ So the third ordered pair (x,y) is (3,7)
(iv) Proceeding like this, we will get infinite number of ordered pairs. All those ordered pairs should be included in the set f.
• So we can write: f = {(1,3), (2,5), (3,7), (4,9), (5,11), (6,13), (7,15), . . .}
(v) We can make a table using the x and y values in the set f. Such a table is convenient to draw the graph of the function.

Table 2.1

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Let us see some common functions and their graphs
A. Identity function
This is a real valued function f: R→R defined by y = f (x) = x
Details can be written in 10 steps:
1. Given that, it is a real valued function. That means, all values obtained after processing, must be real values.
• The function can be defined by writing the ordered pairs which satisfy that function.
• So our next task is to find those ordered pairs.
2. It is given that, f : R→R
• This indicates that,
   ♦ the first elements of the ordered pairs (input x values) should be taken from the set R.
   ♦ the second elements (resulting y values) should be present in the set R
3. The set R is the set of real numbers. It will include integers, negative values, positive values, fractions, decimals, recurring decimals, numbers like √2, √5, π etc.,. In short, R will contain every value which can be plotted on a number line. Recall that we plotted √2, √5, π etc., in our previous classes.
• Since different types of numbers are present in R, we will choose some convenient numbers at random.
• Let x = -7
   ♦ This x is processed as follows:
   ♦ f(-7) = x = -7
   ♦ f(-7) is the 'y value' when 'x value' is -7
   ♦ So we get an ordered pair (x,y) as: (-7,-7)
• Let x = -3
   ♦ This x is processed as follows:
   ♦ f(-3) = x = -3
   ♦ f(-3) is the 'y value' when 'x value' is -3
   ♦ So we get another ordered pair (x,y) as: (-3,-3)
• Let x = 1.414
   ♦ This x is processed as follows:
   ♦ f(1.414) = x = 1.414
   ♦ f(1.414) is the 'y value' when 'x value' is 1.414
   ♦ So we get another ordered pair (x,y) as: (1.414,1.414)
• We see that, whatever be the value of x, the value of y will also be the same.
4. Proceeding like this, we will get infinite number of ordered pairs. All those ordered pairs should be included in the set f.
• So we can write: f = {. . . , (-7,-7), (-3,-3), (1.414,1.414), (5,5), . . .}
5. The above set f is written in roster form. But we have to remember an important point. It can be written in 3 steps:
(i) Both elements of the ordered pairs are real numbers.
(ii) Since they are real numbers, there will be integers, negative values, positive values, fractions, decimals, recurring decimals, numbers like √2, √5, π etc.,. We cannot think of a definite sequence to write them.
(iii) So it is better to use set builder form to write f.
6. In the set builder form, we can write:
f = {(x,y) : x ∈ R, y = x}
• That means:
    ♦ The set f contains all ordered pairs such that,
    ♦ x is a real number,
    ♦ y is equal to x.
7. Once we write the set f, we can write the domain and range of f.
(i) First we will write the domain:
• Domain of f is the set containing all the first elements of the ordered pairs in f.
• In our present case, there are infinite number of ordered pairs. So there will be infinite number of first elements.
• We saw that all the first elements are real numbers. Since they are real numbers, there will be integers, negative values, positive values, fractions, decimals, recurring decimals, numbers like √2, √5, π etc.,. We cannot think of a definite sequence to write them. So it is better to use set builder form rather than the roster form.
• We can write:
    ♦ Domain of f = {x : x ∈ R}
• That means:
    ♦ The domain of f is the set of all x such that,
    ♦ x is a real number.
(ii) Next we will write the range:
• Range of f is the set containing all the second elements of the ordered pairs in f.
• In our present case, there are infinite number of ordered pairs. So there will be infinite number of second elements.
• We saw that all the first elements are real numbers. Since the second elements are equal to first elements, they are also real numbers.
• We can write:
    ♦ Range of f = {y : y ∈ R}
• That means:
    ♦ The range of f is the set of all y such that,
    ♦ y is a real number.
8. We can make a table using the x and y values in the set f. Such a table is convenient to draw the graph of the function.
• Note that, to input for x, we choose convenient numbers from the set R.
• It is better not to choose numbers with recurring decimals. They will be difficult to plot.

Table 2.2

9. The red line in fig.2.10(a) below, is the graph of this function.

Fig.2.10

• We can write some peculiarities of this red line. They can be written in 5 steps:
(i) The red line always passes through the origin (0,0)
(ii) If x axis and y axis are drawn to the same scale (Details here), the red line will make 45^o degrees with the x axis.
• In other words, if the two axes are drawn to the same scale, the red line will bisect the angle between the two axes.
(iii) Mark any point on the red line. Note the coordinates of that point.
   ♦ The x coordinate will be same as the y coordinate.
   ♦ This is shown in fig.b
(iv) Mark any point on the x axis. For example, let us mark 3.5.
• Draw a vertical line through that point.
    ♦ Here, it is the green vertical dotted line in fig.b.
• That vertical line will meet the red line at a point.
• Through that meeting point, draw a horizontal line.
    ♦ Here, it is the green horizontal dotted line.
• This horizontal line will meet the y axis at a point which have the same x value (here it is 3.5) from where we started off.
• We will get this result even if the two axes are drawn in different scales.
(v) We see arrows at both ends of the red line.
• The arrow at the top end of the red line indicates that, the line can extend up to the point where x = +∞ and y = +∞.
• The arrow at the bottom end of the red line indicates that, the line can extend up to the point where x = -∞ and y = -∞.
10. The identity function has many applications in science and engineering.

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In the next section, we will see a few more common functions.

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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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