$\Bbb R$ stands for real numbers.

$ f(x) = \begin{cases} 2-x, & \text{if $x \le 1 \qquad \text{is one to one but not onto } \Bbb R $ } \\ \frac{1}{x} , & \text{if $x >1$ } \end{cases}$

I know how to prove that this is one to one by saying that an element in the domain maps too exactly one element in the range set.

$x\le1 \;\Rightarrow \; f(x) = f(y) \Rightarrow \; 2-x =2-y \Rightarrow x=y$

$x>1\Rightarrow$ $\frac{1}{x} = \frac{1}{y} \Rightarrow x=y$

We can say from this that this is one-one. I am having trouble understanding why this is not onto. I know that $0 \notin \Bbb x$. This is where I get lost. I know that a function is onto if every element in the range set has a preimage is the domain set. I am just not too sure what that means in applying to this problem.

  • 2
    $\begingroup$ For example, it is I think easy to see that there is no $x$ such that $f(x)=-10$. By the way, your one-to-one is incomplete. You also need to check there is no $x$ and $y$ with $x\le 1$ and $y\gt 1$ such that $f(x)=f(y)$. $\endgroup$ – André Nicolas Jul 16 '16 at 22:10
  • $\begingroup$ The mapping $f\colon?\to ?$ is defined by $$f(x) = \begin{cases} 2-x, & \text{if $x \le 1$}\\ \frac{1}{x} , & \text{if $x >1$ } \end{cases}$$ Those question marks become important. Presumably they are $\mathbb{R}$ and $\mathbb{R}$, but are they? $\endgroup$ – Daniel W. Farlow Jul 16 '16 at 22:11
  • 1
    $\begingroup$ "I know that $0\notin\mathbb{R}$". This is not correct. Did you mean that $0$ is not in the image of $\mathbb{R}$, i.e. $0\notin f(\mathbb{R})$? $\endgroup$ – smcc Jul 16 '16 at 22:21
  • $\begingroup$ What I mean is that $0 \notin x$ $\endgroup$ – Jon Jul 16 '16 at 22:22
  • $\begingroup$ What do you mean by $0\not\in x$? $\endgroup$ – user84413 Jul 16 '16 at 22:23

It is not onto because the value of the function is always above zero. Thus, for example, $0\in\mathbb{R}$ but there is no $x\in\mathbb{R}$ such that $f(x)=0$.

To elaborate on Andre's comment, you need to add this to complete your proof that $f$ is one-to-one: For any $x\leq 1$ and $y>1$, $$f(x)=2-x\geq 1>\frac{1}{y}=f(y).$$

  • $\begingroup$ Thanks. I misread the function. See edit. $\endgroup$ – smcc Jul 16 '16 at 22:19

In showing this is one-to-one, you should also consider the case where $x>1$ and $y\le 1$ and show that in that case you cannot have $f(x)=f(y)$. What you've done takes care of the case where both are $>1$ or both are $\le 1$.

It's not onto $\mathbb R$ because no value of this function is negative or $0$.

  • $\begingroup$ It cannot be all real numbers in other words it is not onto. $\endgroup$ – Jon Jul 16 '16 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.