CBSE Class 12 > Class 12 Videos > Example Equivalence Relation-2 (Part - 16) - Relations and Functions, Maths, Class 12
Example Equivalence Relation-2 (Part - 16) - Relations and Functions, Maths,
FAQs on Example Equivalence Relation-2 (Part - 16) - Relations and Functions, Maths, Class 12
| 1. What is an equivalence relation? |  |
Ans. An equivalence relation is a binary relation that is reflexive, symmetric, and transitive. In simpler terms, it is a relation that satisfies three properties: every element is related to itself (reflexive property), if element A is related to element B, then element B is also related to element A (symmetric property), and if element A is related to element B, and element B is related to element C, then element A is also related to element C (transitive property).
| 2. How is an equivalence relation different from other relations? | |
Ans. An equivalence relation is different from other relations because it satisfies three specific properties: reflexivity, symmetry, and transitivity. These properties ensure that the relation partitions the set into equivalent classes, where each class consists of elements that are related to each other in the same way. Other relations may not have these properties and may not create such partitions.
| 3. Can you give an example of an equivalence relation? | |
Ans. Yes, an example of an equivalence relation is the relation "is congruent to" on the set of integers. Two integers are considered equivalent if their difference is divisible by a given integer. This relation is reflexive because every integer is congruent to itself, symmetric because if A is congruent to B, then B is congruent to A, and transitive because if A is congruent to B and B is congruent to C, then A is congruent to C.
| 4. How are equivalence classes defined in an equivalence relation? | |
Ans. Equivalence classes in an equivalence relation are defined as sets of elements that are related to each other in the same way. Each equivalence class consists of elements that are related to each other reflexively, symmetrically, and transitively. For example, if we consider the relation "is congruent to" on the set of integers, the equivalence class of an integer would be the set of all integers that are congruent to it.
| 5. What are some applications of equivalence relations in mathematics? | |
Ans. Equivalence relations have various applications in mathematics. Some examples include: - In algebra, equivalence relations are used to define quotient structures, such as quotient groups, quotient rings, and quotient spaces. - In graph theory, equivalence relations are used to define equivalence classes of vertices, which can help in understanding the connectivity and structure of a graph. - In geometry, equivalence relations are used to define congruence of geometric figures, which is essential for proving geometric theorems. - In set theory, equivalence relations are used to define equivalence classes, which are used in constructing mathematical objects such as quotient sets and partitions. - In number theory, equivalence relations are used to define modular arithmetic, which is essential in solving problems involving remainders and divisibility.
Text Transcript from Video
hello friends this video on relations
and function part 16 is brought to you
by example calm no more fear from exam
before watching this video please make
sure that you have watched part 1 to
part 50 from example this is an example
of equivalence class if I remembered I
have told that in equivalence relation
it divided that set into even into III
and sets and all are disjoint sets and
all these are holding equivalence
relations so you see here there is a set
a 1 2 3 or 5 where we have this
equivalence relation it is stored is
equal installation where a minus B mod
is even you have to do that all the
elements 1 3 5 & 2 4 are like into each
other but there is no element common so
what we can do is if you see this it's
already told the equivalence a minus B
is even swastik never do here is we have
to prove the physical installation for
the relation to be equivalent it has to
be reflects a mental event transitive so
for relation to be reflexive a comma a
should be member of this set so if you
say a comma a that weighs a magazine
should divide should be even a minus a
is nothing but 0 0 is even yes so this
relation is reflexive none for symmetry
I am saying a comma B is member this
should imply DiPalma is member so if a
comma B is member I am saying that a
minus B mod is even and a minus B 1 and
B minus a 1 both are same because we
have taken 1 so I'll pause it should be
a minus B mod is even B minus a bond is
also even be - a mod is even that means
B minus a is member of this set so this
is symmetry possum
norm comes from the
but positive I told if a comma B is
member B gamma C is membered this should
imply a commis is number that means a
minus B is even told B minus c ZN old so
you add these two actually so this
becomes canceled I get a minus C modest
even so a minus C mod is even this
implies that a comma C is a member of
the same if a gamma C if ever the
sentence transitive also said prove that
this relation is reflexive symmetric and
transitive
that means this relation is an
equivalence relation first part is done
since it is equal the equivalence
relation it will break this into
multiple equivalence class that we know
so if you see here is nothing but 1 2 3
4 5 and this has been you can do 1 3 5
done stir all the limits are there not
so it is 2 class is even and this is
able to equivalence last so if you see a
1 all the members satisfy this condition
if you see B minus a is even if you take
any number 1 - 3 mod that is minus 2
more even so if you take all this said 1
3 1 5 9 1 1 3 3
each five and five three five five all
these neighbors if you see all these
members satisfy the condition a minus B
is even 1 minus 3 minus 2 is even minus
4 is again 0 is even 3 and one I think a
trillion won Q is even then I have 3 + 3
0 a 7 3 + 5 -2 is even 5 5 is even 5 one
is even 5-1 seen and five minus three is
even or even correct all its value
satisfy this relation similarly if you
take 2 4 then you have to do 4 and 4
these value also if you see 2 - 2 0 even
2 minus 4 is 0 even for - 4 0 even and
for comment for - to who even so if you
see these satisfies or the condition is
the other nation correct so they
independently our equivalence relation
but there is no continuity between this
side and this there is no common element
between said even and you do so what we
have done we have broke at the same a
into even plus e2 and they are disjoint
sets we'll take our example this is one
example of the balance class where you
break the bigger relation in class you
bring into smaller subsets so here the
question says we have to show that this
relation defined by set a is given by E
minus v1 is multiple of 4 that is going
to be the two oceans is an equivalence
relation and also we have to find a set
of elements now start with to prove it
the equivalence relation we have to
prove their this reflexive symmetric and
transitive
so no one trying to solve push in part
one now
so prove it is reflects as I told into
that a gamma it is a member of this a
comma is a member of this mod of e minus
a that is zero is a multicolored form
that is true so this is reflexive like a
listeria reflects in history for
symmetric
I told if a gamma V is no word of this
relationship this will lie that become
is also member of this said so
a gamma BZ member this means a minus B
mod is multiple of hope now am I will be
mod is nothing but B minus m1 same thing
because any view to him on the same
thing I wrote it up a minus V I am
saying B minus a multiple of one CP
minus m1 multiple of four that means V
gamma is part of this set so this
symmetric also but rather day I told a
comma B and V gamma C a member and this
should imply that a comma C set so I am
saying that a minus B mod is multiple of
four and saying B minus C mod is
multiple of four right is multiple of
four you add both of these you get a
minus c1 is what people of what all
right doesn't send tell you you can see
that a minus B is something good for get
some key key one writes in your medieval
book similarly B minus C is also some 14
you add both this is it cancel get a
minus Sigma B but for or k1 plus k2 this
a percent that is a minus C model to
them since E - E is multiple of four I
can say that a comma C is part of this
set since a comma C is part of this set
I can say that it is positive also
correct now since it is reflexive
symmetry
if I can say it is an equivalence
relation one part is done
second is we will find the set of all
elements so you have element in this
from 0 to 12 all right so the number
here is 0 1 4 5 6 7 8 9 10 11 and so out
of these values only I have to find out
super sets or subjects so you see if I
take 0 as 1 number and say 0 has one
number the next time I possible for
these 4 s 4 minus 0 is 4 that is
multiple of 4 so next number before
right who is known and the next number
will be 8 because 8 minus 0 is also
going to cut off four and eight minus
cosmorly / / take any other number I
want to again check that in the next
number will be 12 only phone number in
this set C if you take 0 and if you take
1 0 minus 1 is not material 4 0 and 6 if
you take is one particular 4 so 0 4 8 12
or in numbers similarly the next set
you'll get you take one if you take one
as the one number the next number will
get is five plus five is one spot so one
fine when you get 9 minus 1 is 8 and
then you get 30 that will be not there
solely three numbers right 1 5 9
next you can take is do we take the next
possible number is 6 2 goes to minus 6
minus 4 and then 10 so use extend our
number uses step then you have III III
gram 3 and then you have 7-eleven 3
7-eleven has a number so if you see this
what we have done we grow this a into 4
pot the hose it makes it a that is
having on some
well is now broken into a 1 plus a 2
plus a 3 plus a 4 which is 3 and the
other question you can solve on your own
same thing same you know set we have and
we have to find a is going to be said
thank you
visit exam viacom to watch free
education videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more thanks again
and function part 16 is brought to you
by example calm no more fear from exam
before watching this video please make
sure that you have watched part 1 to
part 50 from example this is an example
of equivalence class if I remembered I
have told that in equivalence relation
it divided that set into even into III
and sets and all are disjoint sets and
all these are holding equivalence
relations so you see here there is a set
a 1 2 3 or 5 where we have this
equivalence relation it is stored is
equal installation where a minus B mod
is even you have to do that all the
elements 1 3 5 & 2 4 are like into each
other but there is no element common so
what we can do is if you see this it's
already told the equivalence a minus B
is even swastik never do here is we have
to prove the physical installation for
the relation to be equivalent it has to
be reflects a mental event transitive so
for relation to be reflexive a comma a
should be member of this set so if you
say a comma a that weighs a magazine
should divide should be even a minus a
is nothing but 0 0 is even yes so this
relation is reflexive none for symmetry
I am saying a comma B is member this
should imply DiPalma is member so if a
comma B is member I am saying that a
minus B mod is even and a minus B 1 and
B minus a 1 both are same because we
have taken 1 so I'll pause it should be
a minus B mod is even B minus a bond is
also even be - a mod is even that means
B minus a is member of this set so this
is symmetry possum
norm comes from the
but positive I told if a comma B is
member B gamma C is membered this should
imply a commis is number that means a
minus B is even told B minus c ZN old so
you add these two actually so this
becomes canceled I get a minus C modest
even so a minus C mod is even this
implies that a comma C is a member of
the same if a gamma C if ever the
sentence transitive also said prove that
this relation is reflexive symmetric and
transitive
that means this relation is an
equivalence relation first part is done
since it is equal the equivalence
relation it will break this into
multiple equivalence class that we know
so if you see here is nothing but 1 2 3
4 5 and this has been you can do 1 3 5
done stir all the limits are there not
so it is 2 class is even and this is
able to equivalence last so if you see a
1 all the members satisfy this condition
if you see B minus a is even if you take
any number 1 - 3 mod that is minus 2
more even so if you take all this said 1
3 1 5 9 1 1 3 3
each five and five three five five all
these neighbors if you see all these
members satisfy the condition a minus B
is even 1 minus 3 minus 2 is even minus
4 is again 0 is even 3 and one I think a
trillion won Q is even then I have 3 + 3
0 a 7 3 + 5 -2 is even 5 5 is even 5 one
is even 5-1 seen and five minus three is
even or even correct all its value
satisfy this relation similarly if you
take 2 4 then you have to do 4 and 4
these value also if you see 2 - 2 0 even
2 minus 4 is 0 even for - 4 0 even and
for comment for - to who even so if you
see these satisfies or the condition is
the other nation correct so they
independently our equivalence relation
but there is no continuity between this
side and this there is no common element
between said even and you do so what we
have done we have broke at the same a
into even plus e2 and they are disjoint
sets we'll take our example this is one
example of the balance class where you
break the bigger relation in class you
bring into smaller subsets so here the
question says we have to show that this
relation defined by set a is given by E
minus v1 is multiple of 4 that is going
to be the two oceans is an equivalence
relation and also we have to find a set
of elements now start with to prove it
the equivalence relation we have to
prove their this reflexive symmetric and
transitive
so no one trying to solve push in part
one now
so prove it is reflects as I told into
that a gamma it is a member of this a
comma is a member of this mod of e minus
a that is zero is a multicolored form
that is true so this is reflexive like a
listeria reflects in history for
symmetric
I told if a gamma V is no word of this
relationship this will lie that become
is also member of this said so
a gamma BZ member this means a minus B
mod is multiple of hope now am I will be
mod is nothing but B minus m1 same thing
because any view to him on the same
thing I wrote it up a minus V I am
saying B minus a multiple of one CP
minus m1 multiple of four that means V
gamma is part of this set so this
symmetric also but rather day I told a
comma B and V gamma C a member and this
should imply that a comma C set so I am
saying that a minus B mod is multiple of
four and saying B minus C mod is
multiple of four right is multiple of
four you add both of these you get a
minus c1 is what people of what all
right doesn't send tell you you can see
that a minus B is something good for get
some key key one writes in your medieval
book similarly B minus C is also some 14
you add both this is it cancel get a
minus Sigma B but for or k1 plus k2 this
a percent that is a minus C model to
them since E - E is multiple of four I
can say that a comma C is part of this
set since a comma C is part of this set
I can say that it is positive also
correct now since it is reflexive
symmetry
if I can say it is an equivalence
relation one part is done
second is we will find the set of all
elements so you have element in this
from 0 to 12 all right so the number
here is 0 1 4 5 6 7 8 9 10 11 and so out
of these values only I have to find out
super sets or subjects so you see if I
take 0 as 1 number and say 0 has one
number the next time I possible for
these 4 s 4 minus 0 is 4 that is
multiple of 4 so next number before
right who is known and the next number
will be 8 because 8 minus 0 is also
going to cut off four and eight minus
cosmorly / / take any other number I
want to again check that in the next
number will be 12 only phone number in
this set C if you take 0 and if you take
1 0 minus 1 is not material 4 0 and 6 if
you take is one particular 4 so 0 4 8 12
or in numbers similarly the next set
you'll get you take one if you take one
as the one number the next number will
get is five plus five is one spot so one
fine when you get 9 minus 1 is 8 and
then you get 30 that will be not there
solely three numbers right 1 5 9
next you can take is do we take the next
possible number is 6 2 goes to minus 6
minus 4 and then 10 so use extend our
number uses step then you have III III
gram 3 and then you have 7-eleven 3
7-eleven has a number so if you see this
what we have done we grow this a into 4
pot the hose it makes it a that is
having on some
well is now broken into a 1 plus a 2
plus a 3 plus a 4 which is 3 and the
other question you can solve on your own
same thing same you know set we have and
we have to find a is going to be said
thank you
visit exam viacom to watch free
education videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more thanks again
About this Video
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Apr 20, 2026 Last updated
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