Task 1

a)

3.) symmetric antisymmetric. A set can also be neither (see for example)

d)

Make sure to state that the function is well-defined and totally-defined!

e)

We don’t need to have a surjection, just an injection!!! Thus we can just define an “almost” trivial reflexive relation which has an extra for every in and then contains the .

Task 2

d) RSA

We cannot assume . If then . But in any case we can factor out as it’s still in , no matter if they’re equal or not.

For b) we can simply use the CRT to get the answer that we find a unique solution.

Task 3

a)

6.) List all zero-divisors of (note not otherwise there wouldn’t be any!). We have not only because each of these can make a multiple of .

7.) cannot be be a zero-divisor in any ring.

d)

2.) The neutral element in a multiplicative group is not , watch out! 3.) I listed the elements wrong, there are 16 of them! Either you correctly list all of them, and don’t forget that in there are 5 possible coefficients! Or you say that it has in total elements, and has two factors which we can multiply by 4 different coefficients and then thus .

Task 4

e) 1.)

If we need to find an interpretation for a formula that looks like this or , then we need to find a formula that replaces or ! This is written as or and not as or !