For Symmetric you just need to provide a counter example, such as 6/2 = 3 but 2/6 = 1/3 which is not an integer.
I was asked to show that the divisibility relation is reflexive, but NOT symmetric.
Here is what I did...Let me know if it seems acceptable.
Assume a|a, for all a e Z. This implies that a = ac, for some c e Z. Abviously if c = 1, this is true. Thus, it is reflexive.
Assume if a|b, then b|a. I don't know what to do now...
REMEMBER: Don't get | confused with \. | = Divides and \ = Divided by