(3,3) doesnt count,if we consider 3 as a and 5 as b,for a relation to be transitive (a,b) should be an element of R (b,c) also element of R and (a,c) should be element of R ,here we got only a and b and not c so we only get (a,a) as an element.hope anyone who has doubt in that understood
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5:10 reflexive aavan ella elementsintethum mirror eyuthano?
8:15 anyone please explain this step😊
3b is 7b-4b and 4a is 7a-3a. Then group 7a and 7b. Take - common and group 4b and 3a. Rearranging the terms.
How 3b becomes 7b-4b?
@@kiddo.x23 7-4 = 3
@@kiddo.x23 We need to bring it in the 7n format so as to relate. Hence we are writing like that.
6:00 if (1,3) and (3,1) then there should be (1,1) also which means its not transitive.
That will be reflexive
8:30 multiple of 7 ayaal symmetric engane?
3:50 isn't it transitive? there is (3,5),(5,3) and also (3,3) in the set
mam that part is little bitt confusing
For (1,3)(3,5) (1,5) is not in the set. Hence not transitive.
ath sheri aanlo appo entha cheyya
Transitive alla enn mention cheytha mathi
(3,3) doesnt count,if we consider 3 as a and 5 as b,for a relation to be transitive (a,b) should be an element of R (b,c) also element of R and (a,c) should be element of R ,here we got only a and b and not c so we only get (a,a) as an element.hope anyone who has doubt in that understood
please elaborate 10:32 reflexive part
5:27 (1,1) (2,2) okke symmetric alle
Yes
Thnkuu miss💗
Please justify the steps from 8:11 . Im having a hard time understanding! This is making me not watch your videos if you want me to be honest.
3b+4b=7b => 3b = 7b-4b
3a+4a=7a => 4a = 7a-3a
substitute value of 3b and 4a in eqn 3b+4a
hence we get (7b-4b)+(7a-4a)
really helped this
❤
no much expanation not clear
Kindly ask your doubts
You're a💎
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