## Number Systems: FInal Comments on Sheet 8

No-one has asked for any hints on Sheet 9 yet, so here instead are some final thoughts on Sheet 8. Let $f : X \rightarrow Y$ be a function. This means that for every $x \in X$, there is an associated element $f(x) \in Y$, which we call the image of $x$ under $f$. So if you are defining a function, it is your responsibility to ensure that $f(x)$ is defined for each $x$ in the domain $X$.

(Incidentally, some people confused this with the definition of injectivity, by writing something like ‘for each $x \in X$ there exists some $y \in Y$‘. This doesn’t really make sense as it stands, because the function $f$ is never mentioned. If we change it to ‘for each $x \in X$ there exists an element $f(x) \in Y$‘ then it’s just the defining property of a function. Injectivity is something different: for each $y \in Y$ there exists at most one $x \in X$ such that $f(x) = y$.)

I found it a little surprising how few people tried to use the method suggested in lectures to show that a function $f : X \rightarrow Y$ is injective. To repeat it one more time: suppose that $f(x) = f(x')$ and deduce, using whatever you know about $f$ that $x=x'$.

Please see the answers on Moodle for examples of the approved method, especially for Question 5(a). Of course any correct proof is acceptable, and there were several people who gave correct, if sometimes a little long-winded, proofs of 5(b) using proof by contradiction.

Question 2 asked for a function $f : \{1,2,3\} \rightarrow \{1,2,3\}$ that was neither injective nor surjective. Several people tried something like $f(x) = (x-2)^2$. This is not a bad attempt, but there is one problem: we have $f(2) = 0$, and $0$ is not in the codomain $\{1,2,3\}$ of $f$. You could get round this by taking $f(x) = (x-2)^2 + 1$. However there is no need to specify $f$ by a quadratic expression in this way: when the domain is a finite set it is usually easier just to list the images. So you could just say: let $f(1) = 2$, $f(2) = 1$ and $f(3) = 2$.