Question 11. Let O be the origin . We define a relation between two points P and Q in a plane if OP = OQ . Show that the relation, so defined is an equivalence relation.
Solution:
Let A be the set of points on plane
and let R = {(P, Q): OP = OQ} be a relation on A where O is the origin.
now (i) reflexibility:
let P ∈ A
since, OP = OP
:. P, P ∈ R
so, relation R is reflexive
(ii) symmetry:
let (P, Q) ∈ R for P, Q ∈ A
=>OQ = OP
:. (Q, P) ∈ R
therefore, relation R is symmetric .
Similarly (iii) transitive:
let (P,Q) ∈ R and (Q,S) ∈ R
OP = OS
(P, S) ∈ R
hence, relation R is transitive.
so, relation R is an equivalence relation on A .
Question 12. Let R be the relation defined on set A = {1,2,3,4,5,6,7} by R={(a,b): both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1,3,5,7} are related to each other and all the element of the subset {2,4,6} are related to each other, but no element of the subset {1,3,5,7} is related to any element of the subset {2,4,6}
Solution:
Given, A={1,2,3,4,5,6,7}
R={(a,b): both a and b are either odd or even}
clearly, (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7) ∈ R
so, relation R is reflexive.
Also, for symmetry,
a, b ∈ A such that (a, b) ∈ R
both a and b are either odd or even
both b and a are either odd or even
=>(b, a) ∈ R
so, relation R is also symmetry
Similarly, for transitivity,
let a, b, c ∈ Z such that (a,b) ∈ R, (b,c) ∈ R
both a and b are either odd or even
both b and c are either odd or even
if both a and b are even then,
(b,c) ∈ R => both b and c are even
:. both a and c are even
and if both a and b are odd, then
(b,c) ∈ R =>both b and c are odd
both a and c are odd
Thus, both a and c are even and odd
therefore, (a,c) ∈ R
so, (a,b) ∈ R and (b,c) ∈ R => (a,c) ∈ R
so, relation R is transitive
it proves that R is an equivalence relation.
We observe that {1,3,5,7} are related with each other only and {2,4,6} are related with each other.
Question 13. let set S be a relation on the set of all real numbers defined as S = {(a,b) ∈ R x R: a2 + b2 = 1}
Solution:
Let us observe the following properties of S
(i) Reflexibility: let a be an arbitrary element of R
a ∈ R
=> a2 + a2 ≠1 for all a ∈ R
so, S is not reflexive on R .
(ii) symmetry:
let (a, b) ∈ R
a2 + b2 = 1
b2 + a2 = 1
=> (b, a) ∈ S for all a, b ∈ R
So, S is symmetric on R
(iii) transitivity:
Let (a,b) and (b,c) ∈ S
a2 + b2 =1 and b2 + c2 = 1
Adding both the equations we will get,
a2 + c2 = 2 – 2b2 ≠1 for all a, b, c ∈ R
So, S is not transitive on R
Hence, S is not an equivalence relation on R
Question 14. Let Z be the set of all integers and Z0 be the set of all non – zero integers. Let a relation R on Z x Z0 be defined as (a,b) R (c,d) <=>ad =bc for all (a,b), (c,d) ∈ Z x Z0. Prove that R is an equivalence relation on Z x Z0.
Solution:
Let us observe the properties of R
(i) reflexibility:
let (a,b) be an arbitrary element of Z x Z0
(a,b) ∈ ZxZ0
a,b ∈ Z,Z0
ab = ba
(a,b) ∈ R for all (a,b) ∈ ZxZ0
R is reflexive
(ii) symmetry:
Let (a,b), (c,d) ∈ ZxZ0 such that (a,b) R (c,d)
=> ad = bc
=> cb = da
(c,d) R (a,b)
Thus, (a,b) R (c,d) => (c,d) R (a,b) for all (a,b), (c,d) ∈ ZxZ0
so, R is symmetric.
(iii) transitivity:
Let (a,b), (c,d), (e,f) ∈ NxN0 such that (a,b) R (c,d) and (c,d) R (e,f)
(a,b) R (c,d) => ad = bc
(c,d) R (e,f) => cf = de
from this, we get (ad) (cf) = (bc) (de)
=> af = be
(a,b) R (e,f)
so, R is transitive.
Hence it is proved that relation R is an equivalence relation.
Question 15. If R and S are relations on a set A, then prove that
(i) R and S are symmetric = R ∩ S and R ∪ S are symmetric
(ii) R is reflexive and S is any relation => R∪ S is reflexive.
Solution:
(i) R and S are symmetric relations on the set A
=> R ⊂ A x A and S ⊂ A x A
=> R ∩ S ⊂ A x A
thus, R ∩ S is a relation on A
let a, b ∈ A such that (a,b) ∈ R ∩ S
(a,b) ∈ R ∩ S
=> (a,b) ∈ R and (a,b) ∈ S
=> (b,a) ∈ R and (b,a) ∈ S
thus, (a,b) ∈ R ∩ S
=> (b,a) ∈ R ∩ S for all a, b ∈ A
so, R ∩ S is symmetric on A
Also, let a, b ∈ R such that (a,b) ∈ R ∪ S
=> (a,b) ∈ R or (a,b) ∈ S
= (b,a) ∈ R or (b,a) ∈ S [since R and S are symmetric]
=> (b,a) ∈ R ∪ S
hence, R ∪ S is symmetric on A
(ii) R is reflexive and S is any relation:
Suppose a ∈ A
then, (a,a) ∈ R [since R is reflexive]
=> (a,a) ∈ R ∪ S
=> R ∪ S is reflexive on A .
Question 16. If R and S are transitive relations on a set A, then prove that R ∪ S may not be a transitive relation on A.
Solution:
Let A = {a,b,c} and R and S be two relations on A, given by
R = {(a,a), (a,b), (b,a), (b,b)} and
S = {(b,b), (b,c), (c,b), (c,c)}
Here, the relations R and S are transitive on A
(a,b) ∈ R ∪ S and (b,c) ∈ R ∪ S
but (a,c) ∉ R ∪ S
Hence, R ∪ S is not a transitive relation on A.
Question 17. Let C be the set of all complex numbers and C0 be the set of all non – zero complex numbers. Let a relation R and C0 be defined as Z1 R Z2 <=> z1 – z2 / z1 + z2 is real for all Z1, Z2 ∈ C0. Show that R is an equivalence relation.
Solution:
(i) reflexibility:
since, z1 – z2 / z1 + z2 = 0 which is a real number
so, (z1, z1) ∈ R so,
relation R is reflexive.
(ii)symmetry;
z1 – z2 / z1 + z2 = x, where x is real number
=> – (z1 – z2 / z1 + z2 ) = -x
so, (z2, z1) ∈ R
Hence, R is symmetric.
(iii) transitivity:
Let (z1,z2) ∈ R and (z2,z3) ∈ R
then,
z1-z2 / z1+z2 = x where x is a real number
=> z1 – z2 = xz1 + xz2
=> z1 – xz1 = z2 + xz2
=> z1 (1 – x) = z2 (1+x)
=> z1/z2 = (1+x)/(1-x) …. eqn. (1)
also, z2-z3/z2+z3 = y where y is a real number
=> z2 – z3 = yz2 + yz3
=> z2 – yz2 = z3 + yz3
=> z2(1-y) = z3 (1+y)
=> z2/z3 = (1+y)/(1-y)….eqn.(2)
dividing (1) and (2) we will get
z1/z3 = (1+x / 1-x) X (1-y / 1+y) = z where z is a real number
=> z1-z3 / z1+z3 = z-1/z+1 which is real
=> (z1, z3) ∈ R
Hence, R is transitive
Therefore, it is proved that relation R is an equivalence relation.
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