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Byju's Answer
Standard XII
Mathematics
Symmetric Relations
Let Z be the ...
Question
Let Z be the set of all integers and Z
0
be the set of all non-zero integers. Let a relation R on Z × Z
0
be defined as
(a, b) R (c, d) ⇔ ad = bc for all (a, b), (c, d) ∈ Z × Z
0
,
Prove that R is an equivalence relation on Z × Z
0
.
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Solution
We observe the following properties of R.
Reflexivity:
Let
a
,
b
be
an
arbitrary
element
of
Z
×
Z
0
.
Then
,
a
,
b
∈
Z
×
Z
0
⇒
a
,
b
∈
Z
,
Z
0
⇒
a
b
=
b
a
⇒
a
,
b
∈
R
for
all
a
,
b
∈
Z
×
Z
0
So
,
R
is
reflexive
on
Z
×
Z
0
.
Symmetry:
Let
a
,
b
,
c
,
d
∈
Z
×
Z
0
such
that
a
,
b
R
c
,
d
.
Then
,
a
,
b
R
c
,
d
⇒
a
d
=
b
c
⇒
c
b
=
d
a
⇒
c
,
d
R
a
,
b
Thus
,
a
,
b
R
c
,
d
⇒
c
,
d
R
a
,
b
for
all
a
,
b
,
c
,
d
∈
Z
×
Z
0
So
,
R
is
symmetric
on
Z
×
Z
0
.
Transitivity:
Let
a
,
b
,
c
,
d
,
e
,
f
∈
N
×
N
0
such
that
a
,
b
R
c
,
d
and
c
,
d
R
e
,
f
.
Then
,
a
,
b
R
c
,
d
⇒
a
d
=
b
c
c
,
d
R
e
,
f
⇒
c
f
=
d
e
⇒
a
d
c
f
=
b
c
d
e
⇒
a
f
=
b
e
⇒
a
,
b
R
e
,
f
Thus
,
a
,
b
R
c
,
d
a
n
d
c
,
d
R
e
,
f
⇒
a
,
b
R
e
,
f
⇒
a
,
b
R
e
,
f
for
all
values
a
,
b
,
c
,
d
,
e
,
f
∈
N
×
N
0
So
,
R
is
transitive
on
N
×
N
0
.
Suggest Corrections
0
Similar questions
Q.
Let Z be the set of integers. Show that the relation
R = {(a, b) : a, b ∈ Z and a + b is even}
is an equivalence relation on Z.
Q.
Let C be the set of all complex numbers and C
0
be the set of all no-zero complex numbers. Let a relation R on C
0
be defined as
z
1
R
z
2
⇔
z
1
-
z
2
z
1
+
z
2
is real for all
z
1
,
z
2
∈
C
0
.
Show that R is an equivalence relation.
Q.
Let R be a relation on the set Z of all integers defined as (x, y) ∈ R ⇔ x-y is divisible by 2. Then, the equivalence class [1] is _________________.
Q.
Let
Z
be the set of integers. Show that the relation
R
=
{
(
a
,
b
)
:
a
,
b
∈
Z
and a+b is even
}
is an equivalence relation on
Z
.
Q.
Let
S
be the set of all complex numbers
z
satisfying
|
z
−
2
+
i
|
≥
√
5
. If the complex number
z
0
is such that
1
|
z
0
−
1
|
is the maximum of the set
{
1
|
z
−
1
|
:
z
∈
S
}
,then the principal argument of
4
−
z
0
−
¯
¯¯¯
¯
z
0
z
0
−
¯
¯¯¯
¯
z
0
+
2
i
is
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