00:03
In this question, we are given a relation between x 1 y 1 and x 2 y 2 and the relation is given by 3 x 1 plus 2 y 1 is equal to 3 x 2 plus 2 y 2 we need to show that this is an equivalence relation.
00:25
So first is reflexive second is symmetric and the third is transitive.
00:31
So let us take the reflexivity.
00:33
So consider x 1 comma x 1 and x 1 comma x.
00:41
All right.
00:41
So 3 x 1 plus x 1 comma y 1 x 1 comma y 1.
00:47
So 3 x 1 plus 2 y 1 on the left hand side and it will be 3 x 1 plus 2 y 1 on the right hand side and both are equal.
00:56
So trivially true.
00:57
All right, this is true by default.
01:01
So now we will check for symmetry.
01:07
All right.
01:07
So in symmetric we have if x 1 comma y 1 are related to x 2 comma y 2, then we need to check that if x 2 y 2 is related to x 1 y 1.
01:16
So consider x 2 y 2 and x 1 y.
01:22
So 3 x 2 plus 2 y 2 and on the left right hand side, we will be having 3 x 1 plus 2 y 1.
01:34
All right, and we know that if they both are equal by 1 then they are here.
01:40
They are equal here as well.
01:42
So equal by 1 and true due to addition.
01:51
All right.
01:53
So that is if 3 x 1 plus 2 y 1 is equal to 3 x 2 plus 2 y 2, then this must also holds.
02:00
All right.
02:01
Now we will check for transitivity.
02:07
So what do we do in transitivity? that if 3 x 1 plus 2 y 1 is equal to 3 x 2 plus 2 y and we assume the third factor 3 x 2 plus y 2 2 y 2 is equal to 3 x 3 plus 2 y 3.
02:30
All right, we assumed that x 1 y 1 is related to x 2 y 2 and x 2 y 2 is related to x 3 y 3 then we need to show that to show that x 3 y 3 is related to x 1 y 1.
02:52
All right.
02:53
So from here you can see that by comparing the above two equations, we get that 3 x 2 plus 2 y 2 is equal to 3 x 3 plus 2 y 3.
03:13
Also, this is equal to 3 x 1 plus 2 y 1.
03:18
So this implies 3 x 3 plus 2 y 3 is equal to 3 x 1 plus 2 y 1...