Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been having trouble with determining if a function is one-to-one or onto. I found an example and would like to see how to go about this problem. If we have $f(x)=\frac{x}{x^2+1}$ where $f:\mathbb{Q}\rightarrow\mathbb{Q}$, is $f$ one-to-one? Onto? What about if $f:\mathbb{Z}\rightarrow\mathbb{Q}$?

share|cite|improve this question

The function is onto if and only if every rational can be obtained as the image of the function.

Considering for a second the function as a function over all of $\mathbb{R}$, consider $$f'(x) = \frac{(x^2+1) - 2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}.$$ So $f'(x)$ has critical points at $x=1$ and $x=-1$. It is now easy to verify that $f(x)$ has a local maximum at $1$ and a local minimum at $-1$.

Since $f$ is decreasing on $(-\infty,-1)$, increasing on $(-1,1)$, and decreasing on $(1,\infty)$, and $\lim\limits_{x\to \infty} f(x) = \lim\limits_{x\to-\infty}f(x) = 0$, the function is bounded.

Since the function is bounded over $\mathbb{R}$, it is bounded when restricted to $\mathbb{Q}$ and $\mathbb{Z}$, so it cannot be onto.

Restricted to the integers, the function is one-to-one, because it is one to one and positive on $[1,\infty)$, and one-to-one and negative on $(-\infty,-1]$.

A little experimentation to decide if $f$ is one-to-one over $\mathbb{Q}$: assume that $$\frac{q}{q^2+1} = \frac{r}{r^2+1}$$ Then $qr^2 + q = q^2r + r$, hence $$r(qr-1) = q(qr-1).$$ If $qr-1\neq 0$, then this forces $q=r$. But if $qr=1$, i.e., if $r = \frac{1}{q}$, then we have: $$f\left(\frac{1}{q}\right) = \frac{\frac{1}{q}}{\frac{1}{q^2}+1} = \frac{\frac{1}{q}}{\quad\frac{1+q^2}{q^2}\quad} = \frac{q^2}{q(1+q^2)} = \frac{q}{1+q^2} = f(q),$$ so $f(1/2) = f(2)$, hence $f$ is not one-to-one when restricted to the rationals either. That is, in fact we have that $f$ takes every value except $0$, $\frac{1}{2}=f(1)$ and $-\frac{1}{2}=f(-1)$, twice.

share|cite|improve this answer

For this particular problem, it's not hard to solve $\frac{x}{x^2+1} = \frac{y}{y^2+1}$ directly; the equation you get, $y x^2 - (y^2+1)x +y=0$ is a quadratic in $x$, and you already know $x=y$ is a solution, so you can factor that out, giving $y(x-y)(x-1/y) = 0$.

So as a function on the rationals, it fails to be injective, because, for example, $f(2)=f(1/2)$. It is, however, injective on the integers, because $x = 1/y$ implies $x = y = \pm 1$.

share|cite|improve this answer

For ontoness: Given a rational $q$, you want to know if there is a rational $x$ such that ${x \over x^2 + 1} = q$. This translates into the quadratic equation $x^2 + {1 \over q}x + 1 = 0$ (We can ignore $q = 0$ since it is the image of $x = 0$). By the quadratic formula, you'd have to have $x = -{1 \over 2q} + {1 \over 2}\sqrt{{1 \over q^2} - 4}$ or $x = -{1 \over 2q} - {1 \over 2}\sqrt{{1 \over q^2} - 4}$. These don't have to be rational; all you have to do is take some $q$ for which the expression ${1 \over q^2} - 4$ inside the radical is not a square of a rational; say $q = {1 \over 3}$. So the function is not onto.

For one-to-oneness, like the others mention it eventually reduces to $f(x) = f(1/x)$, so it won't be one-to-one either.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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