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Byju's Answer
Standard XII
Mathematics
Equivalence Relation
Let R be a ...
Question
Let
R
be a relation over the set
N
×
N
and it is defined by
(
a
,
b
)
R
(
c
,
d
)
⇒
a
+
d
=
b
+
c
. Then
R
is
A
Reflexive only
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B
Symmetric only
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C
Transitive only
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D
An equivalence relation
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Solution
The correct option is
C
An equivalence relation
We have
(
a
,
b
)
R
(
a
,
b
)
for all
(
a
,
b
)
∈
N
×
N
. Since
a
+
b
=
b
+
a
.
Hence,
R
is reflexive.
R
is symmetric for we have
(
a
,
b
)
R
(
c
,
d
)
⇒
a
+
d
=
b
+
c
⇒
d
+
a
=
b
+
c
⇒
c
+
b
=
d
+
a
⇒
(
c
,
d
)
R
(
a
,
b
)
Hence
R
is symmetric.
Then by definition of
R
, we have
a
+
d
=
b
+
c
and
c
+
f
=
d
+
e
Hence by addition, we get
a
+
d
+
c
+
f
=
b
+
c
+
d
+
e
or
a
+
f
=
b
+
e
Hence
(
a
,
b
)
R
(
e
,
f
)
Thus
(
a
,
b
)
R
(
c
,
d
)
and
(
c
,
d
)
R
(
e
,
f
)
⇒
(
a
,
b
)
R
(
e
,
f
)
Hence
R
is transitive.
Hence
R
is an equivalence relation
Suggest Corrections
0
Similar questions
Q.
Let R be a relation over the set
N
×
n
and it is defined by (a, b) R (c, d)
⇒
a+ d = b + c. Then, R is
Q.
Let
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be a relation over the set
N
×
N
and it is defined by
(
a
,
b
)
r
(
c
,
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)
⇒
a
+
d
=
b
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, then
r
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Q.
Let R be a relation over the set
N
×
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and it is defined by (a, b) R (c, d)
⇒
a+ d = b + c. Then, R is
Q.
Let
N
denote the set of all natural numbers and
R
be the relation on
N
×
N
defined by
(
a
,
b
)
R
(
c
,
d
)
,
if
a
d
(
b
+
c
)
=
b
c
(
a
+
d
)
,
then show that
R
is an equivalence relation.
Q.
Let
N
denote the set of all natural numbers and
R
be the relation on
N
×
N
defined by
(
a
,
b
)
R
(
c
,
d
)
⟺
a
d
(
b
+
c
)
=
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a
+
d
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. Check weather
R
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