Here are some hints for Problem Sheet 9, as given in office hours or by email.
Question 2. In (b) we want to find all such that . Clearly we can’t get far unless we use the definition of . Substituting this ‘deflates’ the problem, by turning a statement involving the unfamiliar relation into a more concrete problem about arguments. The principal argument of is , so, by definition of , we have
Now try to find all such that . It will useful to put into exponential form. (This should seem like a sensible thing to do, since was given in exponential form.)
Once you’ve done (b) you should find (c) is very similar.
Question 3. For (a), you need to find a relation on that is reflexive, symmetric but not transitive. So there must exist such that , but . You could start by making sure this is the case, by defining so that and , but . You will now need to make some further pairs relate in order to satisfy the reflexive and symmetric conditions.
Note that there is no need to define your relation by some sort of ‘natural’ condition, such as ‘ is even’. A diagram, as in Example 8.2, or a list of pairs such that is also entirely acceptable as a definition.
Question 4. For (b) you need to find some pairs such that , and . Substituting the definition of , this means that . Try putting : for some values of you will get a suitable
Note that in Question 4, and (read ‘ prime’ or ‘ dash’) will usually be different numbers. We use primes to show that while these numbers are different, they play a similar role in the equation.
Question 5. The congruence relation is defined in Definition 9.1. We haven’t covered this in lectures yet, but it is the same relation defined in Example 8.7. So for instance if and only if is divisible by . Clearly is divisible by (since ), so for 5(a) you can take .
Question 6. It might help to look again at Example 8.7. The Part C Slides on Moodle have the specific case of this example where . In this question you should take . For example, to find , use that mod if and only if is divisible by . So
Now divides if and only if for some . So
Note that we can obtain from by subtracting from each element of . This might help you to check your answers to the other parts.