00:01
So solution is for first question 1 .23.
00:11
Example of sets a and b such that mod a minus b is equal to mod a intersection b that is equal to b minus a mod which is equal to 3.
00:33
Now let us take a is equal to 1, 2, 3, 4, 5, 6 and b is equal to 1, 2, 3, 7, 7, 8 and 9.
00:55
Then, a minus b is equal to 4, 5 and 6 and 6 and more a minus b is equal to 4, 5 and 6 and more a minus minus b is equal to 3.
01:11
Again, a intersection b is equal to 1, 2 and 3, which implies modulus a intersection b is equal to 3 and b minus a is equal to 7, 8 and 9 which implies modulus b minus a is equal to 3.
01:37
So we have, we have modulus a minus b is equal to modulus a intersection b that is equal to modulus b minus a which is equal to 3.
01:55
So this is the answer for 1 .23 now for 1 .24 examples of examples of sets a, b and b and such that b belongs to a and v is a subset of c and a intersection c is equal to phi.
02:30
No such example exists with ordinality greater than greater than 1 since if b belongs to a and b is a subset of c then a intersection c must have must have at least one element in common.
03:32
Equal to phi b is equal to phi and c is equal to phi is the only example here b belongs to a subset of c and a intersection c is equal to phi so it is the null set it is the null set...