00:01
In this problem of relation and function, we have to show that relation r is equivalence relation and relation r is given in the set a, a is a set in which x belongs to z and such that x is a value between 0 and 12.
00:20
So this is the set a and relation r is given y.
00:25
Say relation r is the first, so relation r is given y.
00:31
R is a relation between ordered pair a and b such that modulars of a minus b is multiple of 4.
00:43
Is multiple of 4.
00:49
So this is the relation.
00:51
Now we have to show that this is a equivalence relation.
00:55
So here, a is the value between 0 to 12.
01:00
So we can write as a is a set in which the values are from 0 to 12, like 0 ,1 ,000.
01:07
2, 3 up to this is 12 and now we can write r so r would be say this should be multiple of 4 so first we are taking 4 and 0 the first pair would be 4 and 0 so we can write it 4 and 0 and also we can take 0 and 4 so this would be 0 and 4 and now 1 5 so this would be 1 5 and 5 1 and 5 1 and then 2662, 26 and 62.
01:49
Similarly, this would be up to 129912 here 129 and 912 and then this would be here 84448 8448 and in the end this would be 12 minus 12 or we can say 12 12.
02:16
So this is the r.
02:19
This relation is said to be reflexive if ordered pair like x and x should also be there.
02:26
So which belongs to the given r.
02:29
So when we write x is equal to 0 2, say we are seeing 4, 5, 6, 7, 8, 9, 10.
02:36
So there would be always such pair such as this is like 7 minus 7 would be equals to 0.
02:44
So this is a multiple of 4, 8 minus 8, modules of 8 minus 8 is 0.
02:48
So we can say that relation r is reflexive.
02:56
And now we have to check for symmetric relation.
02:59
So suppose the ordered pair, a, b is in r, then ordered pair b and a should also be there.
03:09
So as you can know that here, 4 -0 is there, that means 0 -4 is also there.
03:13
1 and 5 is there, 5 and 1 is there, 2 and 6 are there, and 6 and 2 are there...