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 $(gf)^{-1}(z)$.