L5 : Equal set - SETS, Mathematics, Class 11 Video Lecture

hello friends this video sets part 5 is
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fear from exam before watching this
video you should have watched sets part
1 - part 4 now let's talk about equal
sets what are equal sets two sets a and
B two sets a and B are safe to be equal
in they have exactly same element and
then we say a is equal to B as we live
otherwise a is not equal to B let me
again repeat the definition two sets a
and B are equal if they have exactly the
same element then we say both the sets
are equals let's take some example he is
one two three volt B is 3 1 2 4 then we
say a is equal to B why because we know
that in rows perform or it does not
matter and if you see first let's
compare the number of elements number of
elements in a is equal to 1 2 3 4 the
sting element and number of elements in
B is equal to 4 this is a step 1 if the
number of elements are same then there
is a chance that they're equal then we
will go for step 2 the number of
elements are not same that means it is
not a person we can ignore it
now since the number of sets the same
now we'll go to the step 2 so this is
step 1 and this is step 2 so in step 2
what we do we compare the dimension so
we see element 1 is here in set a
element B is also in set a element a has
two he also has two element a has 3 B
also has three element a has 4 we also
have scored similarly we'll do the other
way round mapping also B has 3 a has 3 1
B has won a has 1 B has two a has to be
has 4 s
we see in both the sides mapping is
correct that means all the elements
which are in a car also elements in V
similarly all relevant which are in B
are also in and since the number of
elements in a and B are the same we can
say that both are equal let's repeat
what I have told once again let's repeat
what I have said two sets a and B are
equal if they have exactly same elements
then we say is equal to the else a is
not equal to B step 1 you check the
number of elements the number of
elements in both assets are same then
you for step 2 in step 2 you check the
elements actual elements all the remains
which are in a should be part of B and
all their eventual B should be part of a
in that is to both the sets are equal to
clear concepts let's find the pair of
equal sets so these are sets given said
ABCD efg we have three four five six
seven six and we have to find them
equals x to do that let's first convert
everything in post reform first is
convert to roster form so we'll say e is
equal to XY minus 2x minus 15 so when we
solve this equation what we get X square
minus 5x plus 3x minus 15 to 0 X into X
minus 5 plus 3 is equal so X is going to
minus 3 and 5 so y right set a has
element minus 3 comma 5
similarly for said we it sets except the
letter of alloy so we will say B has
element e
l.o y similarly for C is the set of
letter loyal L o y L is already there so
we will not consider
d is it says X is greater than 5 less
than 5 greater than 50 so there is no
element actually because if it is
greater than 15 so we have something
called 16 17 18 but again we are seeing
less than 5 so there is no element so B
is equal to 5
there is no element in B similarly e X
minus 5 is equal to 0 so that means X is
equal to 5 so he has element Phi for F
it says X square for 25 that means X is
equal to plus minus Phi because Phi
squares for it 25 so f will have element
minus 5 comma 5 and G is already in
those two form has the element 0 now we
see then we observe that B and C then
you see a let us say e is minus C Phi is
there any other set which is minus D Phi
no so is not equal to any other self be
B has element alo why is there any other
element with any other side which is
element only alo Y so when we CCC has
element loi and that is alo my and
element count is also same so we will
say B is equal to C also will c d e f
none of these are equal to B so B and G
is done now let's take D D has 0 element
5
so now let's see is there any other set
which has zero element no G has element
or 0 it is not having zero element so
there is no other set which is empty let
us take E is having 5 is there any set
which is fine I think I did a mistake
here see that's what non will do X is a
positive integer so when we say X is
positive minus 3 is gone - 3 is gone
because minus 3 is not
so X is if Health's has only one element
and E also has only one element so we
can say e is equal to e and this set is
equal to PI and this set is equal to a n
mu by cover separately basis I hope you
got the point here is going to find it
by mistake because a is a positive
integer I ignored that and now we have
counted that also so EA has 5k and E has
both Phi and then other sets are equal
so the answer is B and C are equal
because both have elo Y and a and E are
equal because both have one element that
is Phi
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