Page 5 - mathsvol1ch1to3ans

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(i) reflexive (ii) symmetric (iii) transitive (iv) equivalence
Solution: Since (c, c) is not in R, it is not reflexive. It is enough to include (c, c) to make it
reflexive. (a, c) ∈ R but (c, a) /∈ R. It is enough to include (c, a) to make it symmetric. After
including the above mentioned elements, R = {(a, a), (b, b), (c, c), (a, c), (c, a)} we see that R is
transitive. Hence the minimum number of ordered pairs to be included to R to make it equivalence
relation is TWO.

4. Let P be the set of all triangles in a plane and R be the relation defined on P as aRb if a is similar
to b. Prove that R is an equivalence relation.
Solution: Each triangle is similar to itself. Hence R is reflexive. a is similar to b ⇒ b is similar to
a. Hence R is symmetric. Let a is similar to b and b is similar to c. Then a is similar to c. Hence
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R is transitive. Since R is reflexive, symmetric and transitive, it is an equivalence relation.
5. On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down
the relation by listing all the pairs. Check whether it is
(i) reflexive (ii) symmetric (iii) transitive (iv) equivalence
Solution: R = {(3, 8), (6, 6), (9, 4), (12, 2)}. R is not reflexive, not symmetric, transitive (How?).
Hence it is not an equivalence relation.
Note: The first elements viz, 3, 6, 9, 12 and second elements viz 2, 4, 6, 8 are in Arithmetic
Progression. The values of first elements are in the form 2n + 1 (odd!) while the values of second
elements are in the form 2(5 − n)(even!). To make a total even, both terms should be even. Hence
second term is multiplied by even. But it is interesting to note that first term is multiplied by odd
while the second term is multiplied by even.
6. Prove that the relation ”friendship” is not an equivalence relation on the set of all people in
Chennai.

7. On the set of natural numbers let R be the relation defined by aRb if a + b ≤ 6. Write down the
relation by listing all the pairs. Check whether it is
(i) reflexive (ii) symmetric (iii) transitive (iv) equivalence
Solution:
R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)}
The relation is not reflexive since (4, 4), (5, 5) /∈ R. The relation is symmetric. It is not transitive
since (4,1),(1,4) belongs to R but (4, 4) /∈ R.
Note: Interesting to note that even when b 6= c whenever (a, b), (b, c) ∈ R,we find that (a, c) ∈ R
both a and c are same.
8. Let A = {a, b, c}. What is the equivalence relation of smallest cardinality on A? What is the
equivalence relation of largest cardinality on A?
Solution:
The smallest set of equivalence relation on A is = {(a, a), (b, b), (c, c)}. The largest set of relation
on A is = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}.

9. In the set Z of integers, define mRn if m − n is divisible by 7.Prove that R is an equivalence
relation.
Solution:
m − n = 7k where k is any integer. We see that mRm since zero is divisible by 7. Hence R is
reflexive.
mRn ⇒ nRm. Hence R is symmetric.
Let mRn and nRp. That is, m − n = 7k and n − p = 7t for some integers k, t. Now, (m − n) +
(n − p) = m − p = 7(k + t) ⇒ mRp. Hence R is transitive.
Hence R is an equivalence relation.