In the previous section, we completed appendix B. With that, we have completed all topics in class 11. In this chapter, we will see Relations and Functions which is the first chapter in class 12.
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In class 11, we saw some basic details about relations and functions. See chapter 2.
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Let us recall five important points:
(i) A and B are two sets.
(ii) A × B is a new set formed from A and B. This new set will contain all possible ordered pairs between the elements of A and B.
(iii) The ordered pairs are written in the form (a,b).
♦ a is from set A
♦ b is from set B
(iv) We pick out some of those ordered pairs and form a new set R.
(v) Take any ordered pair from R. There will be a definite relation between the a and the b of that ordered pair.
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For example,
♦ a is the brother of b.
♦ a is 2 less than b.
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We write this as a R b.
In this section, we will learn more details about relations and functions.
Types of relations
First we will see empty relation. It can be explained in 4 steps:
1. We have seen the relation from set A to itself. It is written as R in A.
2. Relation in A will be a subset of A × A
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But Φ (empty set) is a subset of any set. So there exists a relation R in A such that R is a empty set.
3. Let us see an example. It can be written in 4 steps:
(i) Let A = {1,2,3,4}
(ii) R in A in set builder form is:
R = {(a,b): a-b = 10}
(iii) a-b = 10 is same as b = a-10
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Take the first element 1. Ten subtracted from 1 is -9. This -9 is not present in A. So when a = 1, we cannot write b. In other words, no ordered pair in R can have a = 1.
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Take the second element 2. Ten subtracted from 2 is -8. This -8 is not present in A. So when a = 2, we cannot write b. In other words, no ordered pair in R can have a = 2.
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Similarly, we will find that:
♦ No ordered pair in R can have a = 3
♦ No ordered pair in R can have a = 4
(iv) None of the elements in A can become 'a' of the ordered pairs in R. So there will be no element in R. In other words, R is an empty set.
4. If R = Φ, then that relation is called an empty relation.
Now we will see universal relation. It can be explained in 4 steps:
1. We have seen the relation from set A to itself. It is written as R in A.
2. Relation in A will be a subset of A × A
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But a set itself is a subset of any set. So there exists a relation R in A such that R is A × A.
3. Let us see an example. It can be written in 4 steps:
(i) Let A = {1,2,3,4}
(ii) R' in A in set builder form is:
R' = {(a,b): |a-b| ≥ 0}
(iii) A has four elements. So A × A will have 16 elements.
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Take the first element (1,1). Here a = 1 and b = 1
|a-b| = |1-1| = |0| = 0
So (1,1) is eligible to be included in the set R'.
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Take the second element (1,2). Here a = 1 and b = 2
|a-b| = |1-2| = |-1| = 1 ≥ 0
So (1,2) is eligible to be included in the set R'.
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Take the third element (1,3). Here a = 1 and b = 3
|a-b| = |1-3| = |-2| = 2 ≥ 0
So (1,3) is eligible to be included in the set R'.
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Take any one of the 16 elements, say (4,1). Here a = 4 and b = 1
|a-b| = |4-1| = |3| = 3 ≥ 0
So (4,1) is eligible to be included in the set R'
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In this way we can check all the 16 elements. We will see that, all of them are eligible to be included in the set R'.
(iv)
We can write: R' = A × A
4. If R' = A × A, then that relation is called a universal relation.
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Here every element in A has the relation R' with every other element of A.
Both the empty relation and the universal relation are some times called trivial relations.
Solved example 17.1
A is the set of all students of a boys school. Show that the relation R in A given by R={(a,b) : a is sister of b} is the empty relation and R’ = {(a,b) : the difference between heights of a and b is less than 3 meters} is the universal relation.
Solution:
Part (i):
1. We can try to write the elements in R. Those elements will be in the form (a,b)
‘a’ will be from set A. ‘b’ will also be from set A.
2. Set A contains only boys.
So ‘a’ can never be the sister of ‘b’.
3. As a consequence, we will not be able to write a single element of the form (a,b). That means, R is an empty set.
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So the relation R is an empty relation.
Part (ii):
1. Heights greater than 2 m are very rare. We can safely assume that, the tallest student has a height of 2.5 m.
2. Heights smaller than 1.5 are very rare. We can safely assume that, the shortest student has a height of 1 m.
3. So the maximum possible difference between the heights is (2.5 – 1) = 1.5 m.
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None of the “differences” will be greater than 1.5 m.
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Consequently, none of the “differences” will be greater than 3 m.
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In other words, all differences will be less than 3 m.
4. We can take any (a,b) from A × A.
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The difference in that pair will be less than 3 m.
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So all elements in A × A are eligible to be included in R`
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Thus we get R` = A × A.
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That means, R` is a universal relation.
Next we will see reflexive relation. It can be explained in 3 steps:
1. Let A = {1,2,3,4}
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We know that there will be 16 elements in A × A.
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(1,1), (2,2), (3,3) and (4,4) will be among those 16 elements.
2. Suppose that, there is a relation R in A in such a way that, (1,1), (2,2), (3,3) and (4,4) are elements of R.
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Then that relation is called a reflexive relation.
3. We can write the definition in two steps:
(i) R is a relation in A
(ii) For every a ∈ A, if (a,a) ∈ R, then R is a reflexive relation.
Next we will see symmetric relation. It can be explained in 3 steps:
1. Let A = {1,2,3,4}
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We know that there will be 16 elements in A × A.
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(1,2), (2,1), (1,3), (3,1) etc., will be among those 16 elements.
2. Suppose that, there is a relation R in A which satisfies the following conditions:
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If (1,2) is an element of R, then (2,1) is also an element of R.
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If (1,3) is an element of R, then (3,1) is also an element of R.
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• If (2,4) is an element of R, then (4,2) is also an element of R.
so on . . .
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If this is true for all such pairs, then that relation is called a symmetric relation.
3. We can write the definition in two steps:
(i) R is a relation in A
(ii) For all a1, a2 ∈ A, if (a1,a2) ∈ R ⇒ (a2,a1) ∈ R, then R is a symmetric relation.
Next we will see transitive relation. It can be explained in 3 steps:
1. Let A = {1,2,3,4}
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We know that there will be 16 elements in A × A.
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(1,2), (2,3), (1,3), (3,4) etc., will be among those 16 elements.
2. Suppose that, there is a relation R in A which satisfies the following conditions:
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If (1,2) and (2,3) are elements of R, then (1,3) is also an element of R.
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If (1,3) and (3,4) are elements of R, then (1,4) is also an element of R.
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• If (2,4) and (4,3) are elements of R, then (2,3) is also an element of R.
so on . . .
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If this is true for all such pairs, then that relation is called a transitive relation.
3. We can write the definition in two steps:
(i) R is a relation in A
(ii) For all a1, a2, a3∈ A,
if (a1,a2) ∈ R and (a2,a3) ∈ R ⇒ (a1,a3) ∈ R, then R is a transitive relation.
Now we can write the definition of an equivalence relation. It can be written in 2 steps:
1. R is a relation in A.
2. This R is an equivalence relation if all three conditions below are satisfied.
(i) R is a reflexive relation.
(ii) R is a symmetric relation.
(iii) R is a transitive relation.
Solved example 17.2
Let T be the set of all triangles in a plane. R is a relation in T.
R = {(T1,T2) : T1 is congruent to T2}. Show that R is an equivalence relation.
Solution:
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Congruent triangles are those triangles which have the same sides and same corresponding angles.
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Given that, T is the set of all triangles in a plane. There will be infinite number of triangles in that set. Let us number them as: 1, 2, 3, 4, . . .
All those triangles will be present in T.
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R is a relation in T. So we must consider T × T
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(1,1), (1,2), (1,3), (1,4), . . . , (2,1), (2,2), (2,3), . . . are all elements of T × T.
There will be infinite elements in T × T
1. First we check whether R is reflexive. It can be written in 3 steps:
(i) (1,1) is eligible to be included in R. This is because, the first triangle is congruent to itself. In fact, any triangle is congruent to itself.
(ii) In this way, (2,2), (3,3), (4,4), . . . are eligible to be included in R.
(iii) Therefore, R is 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 triangle is congruent to the ninth triangle.
(ii) Now, (9,3) is an element of T × T.
• The element (9,3) is eligible to be included in R. This is because:
If third triangle is congruent to the ninth triangle, then ninth triangle will be congruent to the third.
(iii) So in general, if (T1, T2) is an element of R, then (T2,T1) 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 triangle is congruent to the seventh triangle.
(ii) Also suppose that (7,10) is eligible to be included in R.
• Then it means that, the seventh triangle is congruent to the tenth triangle.
(iii) It is clear that, fourth, seventh and tenth triangles are congruent.
Then fourth triangle is congruent to the tenth triangle.
So (4,10) is eligible to be included in R.
(iv) In general, if both (T1,T2) and (T2,T3) are elements of R, then (T1,T3) will also be an element of R.
Therefore R is a transitive relation.
4. We see that:
R is reflexive, symmetric and transitive. So R is an equivalence relation.
In the next section, we will see a few more solved examples.
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