I have these two functions, in which I have to prove or disprove they are injective and surjective:
- $f:[0,\infty) \to (0,\infty)$ by $f(x) = \frac{1}{x+1}$.
- $h:\mathrm R \to \mathrm R$ by $h(x,y) = xy$
For the first, I have that it is surjective, as an arbitrary element y in $(0,\infty)$, x = $\frac{1}{y} – 1$, and $\frac{1}{y} – 1$ is in $[0,\infty)$ for any y in $(0,\infty)$. It is also injective as we assume $f(a)$ = $f(b)$, so $\frac{1}{a+1} = \frac{1}{b+1}$ for a,b, in $[0,\infty)$. This means that b+1=a+1 $\to$ b+1-a-1=0 $\to$ b-a=0 $\to$ b=a, so it is injective.
For the second, I have a feeling that it is onto, but I'm not sure how to prove it. Could I say that since x and y are in $\mathrm R$, it follows that xy is in $\mathrm R$ as well? For injectivity, I have that it is not injective, as we can take an example h(2,2) = 4 = h(4,1), but h(2,2) does not equal h(4,1).
Is this valid work?
Best Answer
$f$ is not surjective. For a function $f: A \to B$ to be surjective means that for every $y \in B$, there is some $x \in A$ such that $f(x) = y$. So, for example, try solving $f(x) = 2$. What happens? Is $x \in [0, \infty)$? Is is, however, injective. To see this, we put $$f(x) = f(y) \iff \frac{1}{x+1} = \frac{1}{y+1} \iff x+1 = y+1 \iff x = y.$$
As for $g$, it is surjective. Take $c \in \mathbb{R}$. Then we can make $h(x,y) = c$ by taking $x = c$ and $y = 1$. A function $f$ is injective if $f(x) = f(y) \implies x = y$. You showed that $h(2,2) = h(4,1)$, but $(2,2) \neq (4,1)$, so it is not injective.