Equivalence
Relation
A relation R, defined on a set P is
an equivalence relation if
and only if
(i) R is reflexive, i.e. p R p for all p ∈ P.
(ii) R is symmetric, i.e. p R q ⇒ q R p for all p, q ∈ P.
(iii) R is transitive, i.e. p R q and q R r ⇒ p R r for all p, q, r ∈ P.
The relation defined by “x is equal to y” on the set P of
Integers is an equivalence relation.
Solved examples on equivalence relation
Q 1. Let
P be a set of triangles in a plane. The
relation R is defined as “m is
similar to n, m, n ∈ P”.
Solution:
We see that R is;
(i) Reflexive,
for, every triangle is similar to
itself.
(ii) Symmetric,
for, if m is similar to n, then n
is also similar to m.
(iii) Transitive,
for, if m be similar to n and n be
similar to r, then m is also
similar to r.
Hence R is an equivalence
relation.
Q 2. A
relation R is defined on the set Z by “p
R q if p – q is divisible by 2” for
p, q ∈ Z. Examine if R is an equivalence
relation on Z.
Solution:
(i) Let p ∈ Z. Then p
– p is divisible by 2.
Therefore p
R p holds for all p in Z and R is
reflexive.
(ii) Let p, q ∈ Z and pRq
hold. Then p – q is
divisible by 2 and
therefore
q – p is also divisible by 2.
Thus, p R q ⇒ q R p
and therefore R is
symmetric.
(iii) Let p, q, r ∈ Z and p R
q, q R r both hold. Then p – q and q – r are both divisible by 2.
Therefore p – r = (p – q) + (q – r) is divisible by 2.
Thus, p R q and q R r ⇒ p R r and therefore R is transitive.
Since R is reflexive, symmetric and transitive so, R is an
equivalence relation on Z.
Q.3. Let
x be a positive integer. A relation R is
defined on the set Z by “a R b
if
and only if p – q is divisible by x” for p, q ∈ Z.
Show that R is an
equivalence relation on set Z.
Solution:
(i) Let p ∈ Z. Then p
– p = 0, which is divisible by x
Therefore, p R p holds for all p ∈ Z.
Hence, R is reflexive.
(ii) Let p, q ∈ Z and p R
q holds. Then p – q is divisible by x and
therefore, q – p is also divisible by x.
Thus, p R q ⇒ q R p.
Hence, R is symmetric.
(iii) Let p, q, r ∈ Z and p R
q, q R r both hold. Then p – q is divisible by x and q – r is also divisible by
x. Therefore, p – r = (p – q) + (q – r) is divisible by x.
Thus, p R q and q R r ⇒ p R r
Therefore, R is transitive.
Hence R is an equivalence relation on set Z
Q.4. Let
S be the set of all lines in 3 dimensional
space. A relation ρ is
defined
on S by “l ρ m if and only if l lies
on the plane of
m” for l, m ∈ S.
Examine if ρ is (i) reflexive, (ii) symmetric,
(iii)
transitive
Solution:
(i) Reflexive: Let l ∈ S. Then l is coplanar with itself.
Therefore, lρl holds for all l in S.
Hence, ρ is reflexive
(ii) Symmetric: Let l, m ∈ S and l ρ m holds. Then l lies on the plane
of m.
Therefore, m lies on the plane of l. Thus, l ρ m ⇒ m ρ l and therefore ρ is symmetric.
(iii) Transitive: Let l, m, p ∈ S and l ρ m, m ρ p both hold. Then l lies on the plane of m and m lies on the plane
of p. This does not always implies that l lies on the plane of p.
That is, l ρ m and m ρ p do not necessarily imply l ρ p.
Therefore, ρ is not transitive.
Since, R is reflexive and symmetric but not transitive so, R is
not an equivalence relation on set Z
Wonderful
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