Number Systems: Problem Sheet 8

The hint for Question 4 may be helpful for Question 2, so I have put it first.

Question 4. This question asks you to find surjective functions g : \mathbb{R} \rightarrow \mathbb{R} and h : \mathbb{R} \rightarrow \mathbb{R} such that the function g+h : \mathbb{R} \rightarrow \mathbb{R} defined by (g+h)(x) = g(x) + h(x) is not surjective. One way to get a function that is not surjective is to take a constant function which takes only one value. For example, the function f : \mathbb{R} \rightarrow \mathbb{R} defined by f(x) = 0 for all x \in \mathbb{R}. If you could arrange that g and h were surjective, and

g(x) + h(x) = 0

for all x \in \mathbb{R}, then you would be done.

Question 2. Your job is to define a function f : X \rightarrow X such that X is neither injective nor surjective. By definition of surjective, f is surjective if for each y \in \{1,2,3\}, there exists at least one x \in \{1,2,3\} such that f(x) = y. So to make sure that f is not surjective, you need to make sure that there is some y \in \{1,2,3\} which is not equal to any f(x) for x \in \{1,2,3\}. Given this guide, try to define f. If you find that with your definition, f is also not injective, you have a suitable example.

Question 6. We haven’t covered relations yet in lectures. But you should be able to make an attempt at this and Question 7 by reading pages 35 and 36 of the lecture notes, issued in Monday’s lecture.

Here is a detailed solution to (6c). We define a relation on the set X of people in a lecture room by x \sim y if x can see the eyes of y. The relation is reflexive if for all x \in X we have x \sim x, i.e. if everyone in the lecture room can see their own eyes. I think we can safely say this is false. (Not everyone can be looking in a mirror at once.) It is symmetric if

x \sim y \Longrightarrow y \sim x,

i.e., if x and see the eyes of y then y can see the eyes of x. This is true (at least if we make the kind assumption that everyone’s eyes are always open in a lecture). Finally, the relation is transitive if

(x \sim y \ \text{and} \ y \sim z) \Longrightarrow x \sim z.

In words, this says that if x can see the eyes of y, and y can see the eyes of z, then x can see the eyes of z. This is false: for example take y to be the lecturer and x and z to be two people facing the lecturer. So \sim is symmetric but not reflexive or transitive.

Question 7. Here are some examples for (a). We have 1 \equiv 4 since 1-4=-3 and

-3 \in \{-6,-3,0,3,6\}.

So you should draw an arrow from 1 to 4. Also 4 \equiv 1, since

4-1 = 3 \in \{-6,-3,0,3,6\},

so there should also be an arrow from 4 to 1. Similarly there should be arrows from 1 to 7, from 7 to 1. There should also be a loop from 1 to 1, since 1 \equiv 1. In the diagram for Example 8.2, I used a single line with two arrow-heads instead of two separate arrows.

Question 8. For (c), you should use the formula for the inverse of a composition given at the top of page 34 of the lecture notes:

(gf)^{-1} = f^{-1} g^{-1}.

Applying this to z \in [0, \infty) we get

(gf)^{-1}(z) = (f^{-1}g^{-1})(z) = f^{-1}(g^{-1}(z)).

You should now use your formula from (b) for g^{-1}(z) and then your formula from (a) for f^{-1}(y) to turn the right-hand side of the equation above into an explicit formula for (gf)^{-1}(z).


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: