In the previous section,
we saw equivalence relation. We saw one solved example also. In this section , we will see a few more solved examples. Later in this section, we will see equivalence classes.
Solved example 17.3
Let L be the set of all lines in a plane. R is a relation in L.
R = {(L1,L2) : L1 is perpendicular to L2}. Show that R is symmetric but neither reflexive nor transitive.
Solution:
•
Given that, L is the set of all lines in a plane. There will be
infinite number of lines. Let us number them as: 1,2,3,4, . . .
All those lines will be present in L.
•
R is a relation in L. So we must consider L × L
•
(1,1), (1,2), (1,3), (1,4), . . . , (2,1), (2,2), (2,3), . . . are all elements of L × L.
There will be infinite elements in L × L
1. First we check whether R is reflexive. It can be written in 3 steps:
(i)
(1,1) is not eligible to be included in R. This is because, the first
line is not perpendicular to itself. In fact, no line is perpendicular to
itself.
(ii) In this way, (2,2), (3,3), (4,4), . . . are not eligible to be included in R.
(iii) Therefore, R is not a reflexive function.
2. Now we check whether R is symmetric. It can be written in 3 steps:
(i) Suppose that (3,9) is eligible to be included in R.
• Then it means that, the third line is perpendicular to the ninth triangle.
(ii) Now, (9,3) is an element of L × L.
• The element (9,3) is eligible to be included in R. This is because:
If third line is perpendicular to the ninth line, then ninth line will be perpendicular to the third.
(iii) So in general, if (L1, L2) is an element of R, then (L2,L1) will also be an element of R
• Therefore, R is a symmetric relation.
3. Now we check whether R is transitive. It can be written in 4 steps:
(i) Suppose that (4,7) is eligible to be included in R.
• Then it means that, the fourth line is perpendicular to the seventh line.
(ii) Also suppose that (7,10) is eligible to be included in R.
• Then it means that, the seventh line is perpendicular to the tenth line.
This situation is shown in fig.17.1 below:
 |
| Fig.17.1 |
(iii) It is clear that, fourth line is not perpendicular to the tenth line. In fact, fourth line is parallel to the tenth line.
So (4,10) is not eligible to be included in R.
(iv) In general, if both (L1,L2) and (L2,L3) are elements of R, then (L1,L3) will not be an element of R.
Therefore R is not a transitive relation.
4. We see that:
R is symmetric but neither reflexive nor transitive.
•
So R is not an equivalence relation.
Solved example 17.4
Show
that the relation R in the set {1,2,3} given by R = {(1,1), (2,2),
(3,3), (1,2), (2,3)} is reflexive but neither symmetric not transitive.
Solution:
1. Let us check whether R is reflexive:
•
There are three elements in the original set.
•
All those three elements are present in reflexive form in the set R.
They are: (1,1), (2,2) and (3,3)
•
Therefore, R is reflexive.
2. Let us check whether R is symmetric:
•
(1,2) is present in R. But (2,1) is not present.
Therefore, R is not symmetric.
3. Let us check whether R is transitive:
•
(1,2) and (2,3) are present in R. But (1,3) is not present.
Therefore, R is not transitive.
Solved example 17.5
Show that the relation R in the set
Z of integers given by:
$\rm{R = \{(a,b) : 2 ~\text{divides}~ a-b\}}$
is an equivalence relation.
Solution:
•
We have to consider the set Z. It is the set of integers. Integers are those numbers which do not have fraction or decimal parts. They can be +ve or -ve. Zero is also an integer. There are
infinite integers. All of them will be present in Z.
•
R is a relation in Z. So we must consider Z × Z.
•
. . . (-1,0), (-1,1), (-1,2), (-1,3), (-1,4), . . . , (2,0), (2,1), (2,2), (2,3), . . . are all elements of Z × Z.
There will be infinite elements in Z × Z
1. First we check whether R is reflexive. It can be written in 6 steps:
(i) (-1,0) is not eligible to be included in R. This is because, (-1 + 0) is not divisible by 2.
$\frac{-1 + 0}{2} = \frac{-1}{2}$. The quotient $\frac{-1}{2}$ is not an integer.
(ii) (-1,1) is eligible to be included in R. This is because, (-1 + 1) is divisible by 2.
$\frac{-1 + 1}{2} = 0$. The quotient zero is an integer.
(iii) (-1,2) is not eligible to be included in R. This is because, (-1 + 2) is not divisible by 2.
$\frac{-1 + 2}{2} = \frac{1}{2}$. The quotient $\frac{1}{2}$ is not an integer.
(iv) By doing a few calculations like these, we get to know some facts:
•
For (a,b) to be eligible,
♦ Both a and b must be odd
♦ OR both a and b must be even.
♦ The sign of a and b does not matter.
(v) Based on this information, we can easily write some eligible pairs:
(-7, -3), (-4, -10), (-8, 12), (3, -7), (11,13), etc,.
(vi) It is clear that all elements of the form (a,a) are eligible. This is because, the difference (a-a) will be always zero.
•
Some examples are: (-17,-17), (2,2) (5,5) etc.,
•
Since all (a,a) are eligible, we can say that R is reflexive.
2. Now we check whether R is symmetric. It can be written in 3 steps:
(i) Consider an example: (-5,9).
•
This pair is eligible.
(ii) Then (9,-5) is also eligible. We are just interchanging the positions. Both will be the same odd numbers. Or both will be the same even numbers. When (a-b) is calculated after the interchange, only the sign will change. Divisibility by 2 will not be affected.
(iii) We can write:
If (a,b) is an element of R, then (b,a) will also be an element of R.
• Therefore, R is a symmetric relation.
3. Now we check whether R is transitive. It can be written in 3 steps:
(i) Consider an example: (-5,9) and (9,11)
Both pairs are eligible.
(ii) (-5,11) is also eligible. This can be explained as follows:
The first pair consists of odd numbers.
‘9’ is common in both pairs.
So the second pair will also consist of odd numbers.
(iii) We can write:
If (a,b) and (b,c) are eligible, then (a,c) will be eligible.
•
Therefore R is a transitive relation.
4. We see that:
R is reflexive, symmetric and transitive.
•
So R is an equivalence relation.
5. In this problem, we must note an important point. It can be written in 2 steps:
(i) Consider the condition:
Both coordinates in the pair must be even numbers.
•
For such pairs, one of the coordinates can be zero. So (0,2) (0,8), (0, -12), (16,0), (-24,0) etc., are acceptable.
•
We can never use ‘1’ as one of the two coordinates. Because ‘1’ is an odd number.
• The smallest eligible pair is (0,2). We can go on adding 2. All numbers thus obtained can be used to form eligible pairs. We get:
0, 2, 4, 6, 8, . . .
• The smallest eligible pair is (0,2). We can go on subtracting 2. All
numbers thus obtained can be used to form eligible pairs. We get:
0, -2, -4, -6, -8, . . .
•
Combining the two, we get the set:
{. . . -8, -6, -4, -2, 0, 2, 4, . . . }
•
This is the set of even numbers. We can denote it by the letter 'E'. Take any two elements from E. The ordered pair formed by those two numbers will be eligible.
(ii) Consider the condition:
Both coordinates in the pair must be odd numbers.
•
For such pairs, one of the coordinates can be ‘1’. So (1,3) (1,9), (1, -11), (17,1), (-23,1) etc., are acceptable.
•
We can never use ‘0’ as one of the two coordinates. Because ‘0’ is an even number.
• The smallest eligible pair is (1,3). We can go on adding 2. All
numbers thus obtained can be used to form eligible pairs. We get:
1, 3, 5, 7, 9, . . .
• The smallest eligible pair is (1,3). We can go on subtracting 2. All
numbers thus obtained can be used to form eligible pairs. We get:
1, -1, -3, -5, -7, -9, . . .
•
Combining the two, we get the set:
{. . . -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, . . . }
•
This is the set of odd numbers. We can denote it by the letter 'O'. Take any two elements from O. The ordered pair formed by those two numbers will be eligible.
Based on the above solved example 17.5, we can learn some new terms about relations. It can be written in steps:
1. Recall the basic definition about relations:
If (a,b) is an element in set R, then a will be related to b.
◼ For example, a is the sister of b.
We write: R = {(a,b) : a ∈ A, b ∈ B, a is the sister of b}
♦ A is the set of all students in class 11.
♦ B is the set of all students in class 12.
•
Here, (a,b) pairs are picked from A × B
•
The relation R is from A to B.
•
However, R in this case is not an equivalence relation.
◼ Another example: a is the sister of b.
We write: R = {(a,b) : a, b ∈ A, a is the sister of b}
♦ A is the set of all students in the school.
•
Here, (a,b) pairs are picked from A × A
•
The relation R is on A.
•
However, in this case also, R is not an equivalence relation.
◼ Another example: a is 3 greater than b
We write R = {(a,b) : a, b ∈ Z, a = b + 3}
♦ Z is the set of all integers.
•
Here, (a,b) pairs are picked from Z × Z
•
The relation R is on Z.
•
However, in this case also, R is not an equivalence relation.
2. In the solved example 17.5 that we saw above, R is an equivalence relation. Let us see some interesting points:
(i) Consider the condition: a and b must be even.
•
We saw that, to satisfy this condition, we can make ordered pairs by taking any two elements from E.
•
So any even number ‘a’ is related to any other even number ‘b’ by the relation: 2 divides (a-b)
(ii) Consider the condition: a and b must be odd.
•
We saw that, to satisfy this condition, we can make ordered pairs by taking any two elements from O.
• So any odd number ‘a’ is related to any other odd number ‘b’ by the relation: 2 divides (a-b)
3. Based on the above step (2), we can write:
(i) The relation R partitions Z into E and O.
(ii) E can give ordered pairs which are eligible to be included in R.
(iii) O can also give ordered pairs which are eligible to be included in R.
(iv) E is said to be an equivalence class of Z. It is a subset of Z.
(v) O is also said to be an equivalence class of Z. It is also a subset of Z.
4. We can use this as a general result. It can be written in 5 steps:
(i) R is a relation in set X
(ii) R is an equivalence relation.
(iii) Then R will partition X into various sets: A1, A2, A3, . . .
♦ These sets are called equivalence classes.
(iv) Union of all equivalence classes will give set X
(v) Take any two equivalence classes. There will not be any common elements between them. So intersection of any two equivalence classes will give a null set.
Let us see an example:
Show that the relation R in the set
Z of integers given by:
$\rm{R = \{(a,b) : 3 ~\text{divides}~ a-b\}}$
is an equivalence relation. Hence find the equivalence classes.
Solution:
•
We have to consider the set Z. It is the set of integers. Integers are
those numbers which do not have fraction or decimal parts. They can be
+ve or -ve. Zero is also an integer. There are
infinite integers. All of them will be present in Z.
•
R is a relation in Z. So we must consider Z × Z.
•
. . . (-1,0), (-1,1), (-1,2), (-1,3), (-1,4), . . . , (2,0), (2,1), (2,2), (2,3), . . . are all elements of Z × Z.
There will be infinite elements in Z × Z
1. First we check whether R is reflexive. It can be written in 2 steps:
(i) Ordered pairs like (-1,-1), (0,0), (7,7) etc., are eligible because, $\frac{a-b}{3}$ will give zero.
(ii) Since all (a,a) are eligible, we can say that R is reflexive.
2. Now we check whether R is symmetric. It can be written in 2 steps:
(i) Pairs like (0,3), (3,6), (-6,-9) etc., are eligible.
(ii) Even if we interchange 'a' and 'b', they will be eligible.
(iii) We can write:
If (a,b) is eligible, then (b,a) is also eligible.
• Therefore, R is a symmetric relation.
3. Now we check whether R is transitive. It can be written in 3 steps:
(i) Consider an example: (-7,-1) and (-1,8)
Both pairs are eligible.
(ii) (-5,8) is also eligible.
We can check any two eligible pairs like this. The third pair formed will be eligible.
•
The proof can be written as follows:
♦ (a-b) is divisible by 3. So (a-b) = 3k, where k is an integer.
♦ (b-c) is divisible by 3. So (b-c) = 3m, where k is an integer.
♦ Now, (a-c) = [(a-b) + (b-c)] = [3k + 3m] = 3(k+m).
♦ So (a-c) is divisible by 3.
(iii) We can write:
If (a,b) and (b,c) are eligible, then (a,c) will be eligible.
•
Therefore R is a transitive relation.
4. We see that:
R is reflexive, symmetric and transitive.
•
So R is an equivalence relation
5. Since R is an equivalence relation, we must find the equivalence classes. It can be done in four steps:
(i) The smallest eligible pair will contain zero. It is (0,0).
•
So we put zero in the middle.
♦ Then we add 3 on the right side in a repeating manner.
♦ Also,we subtract 3 on the left side in a repeating manner.
•
We get: . . . -12, -9, -6, -3, 0, 3, 6, 9, . . .
(ii) The next smallest eligible pair will contain 1. It is (1,-2).
•
So we put 1 in the middle.
♦ Then we add 3 on the right side in a repeating manner.
♦ Also,we subtract 3 on the left side in a repeating manner.
•
We get: . . . -11, -8, -5, -2, 1, 4, 7, 10, . . .
(iii) The next smallest eligible pair will contain -1. It is (-1,-4).
•
So we put -1 in the middle.
♦ Then we add 3 on the right side in a repeating manner.
♦ Also,we subtract 3 on the left side in a repeating manner.
•
We get: . . . -13, -10, -7, -4, -1, 2, 5, 8, . . .
(iv) So the equivalence classes are:
A1 = {. . . -12, -9, -6, -3, 0, 3, 6, 9, . . .}
A2 = {. . . -11, -8, -5, -2, 1, 4, 7, 10, . . .}
A3 = {. . . -13, -10, -7, -4, -1, 2, 5, 8, . . .}
•
There are no common elements between the above three sets. That means, they are disjoint sets.
•
Also the union of the three sets will give Z.
Solved example 17.6
Show that the relation R in the set P={1,2,3,4,5,6,7} given by:
$\rm{R = \{(a,b) : ~\text{both a and b are either odd or even}\}}$
is an equivalence relation. Hence find the equivalence classes.
Solution:
•
R is a relation in P. So we must consider P × P.
•
(1,1), (1,2), 1,3), . . . (2,1), (2,2), (2,3), . . . are all elements of P × P.
There will be (7 × 7) = 49 elements in P × P
1. First we check whether R is reflexive. It can be written in 2 steps:
(i) Ordered pairs (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) and (7,7) are eligible to be included in R. This is because, both a and b are either odd or even.
(ii) Since all (a,a) are eligible, we can say that R is reflexive.
2. Now we check whether R is symmetric. It can be written in 2 steps:
(i) Pairs like (1,3), (2,6), (5,7) etc., are eligible.
(ii) Even if we interchange 'a' and 'b', they will be eligible. This is because, both a and b are either odd or even.
(iii) We can write:
If (a,b) is eligible, then (b,a) is also eligible.
• Therefore, R is a symmetric relation.
3. Now we check whether R is transitive. It can be written in 3 steps:
(i) Consider an example: (1,3) and (3,7)
Both pairs are eligible.
(ii) (1,7) is also eligible.
We can check any two eligible pairs like this. The third pair formed will be eligible.
•
The proof can be written as follows:
If two pairs are eligible then it means that all four coordinates in them are either odd or even. So the new pair formed will also be either odd or even.
(iii) We can write:
If (a,b) and (b,c) are eligible, then (a,c) will be eligible.
•
Therefore R is a transitive relation.
4. We see that:
R is reflexive, symmetric and transitive.
•
So R is an equivalence relation
5. Since R is an equivalence relation, we must find the equivalence classes. It can be done in four steps:
(i) The set of all odd numbers in P is: {1,3,5,7}
(ii) The set of all even numbers in P is: {2,4,6}
(iii) So the equivalence classes are:
A1 = {1,3,5,7}
A2 = {2,4,6}
•
There are no common elements between the above two sets. That means, they are disjoint sets.
•
Also the union of the two sets will give P.
Link to a few more solved examples is given below:
In the next section, we will see types of functions.
