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Show that the function $f$ defined by $$f(x):=\frac{x}{\sqrt{x^2+1}}\;,$$ $x$ is an element of the reals, is a bijection of the reals onto $\{y:-1<y<1\}$.

So we need to show that it is 1-1 and onto. I begin by trying to prove that it is 1-1:

$$\frac{x}{\sqrt{x^2+1}} = \frac{y}{\sqrt{y^2+1}}$$

solving, we get $x^2 = y^2$

How can we now prove that $x=y$? I think there must be two solutions but am not sure what to do.

Also, I am not entirely sure where I should begin to prove that it is onto.

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up vote 5 down vote accepted

Once you know $x^2 = y^2$, you know that either $x = y$ or $x = -y$. Plug both into the original equation and you'll know that $x = -y$ is impossible.

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I see, so -y/sqrt((-y)^2+1) does not equal y/sqrt(y^2+1) – Samuel Gregory Aug 23 '12 at 21:17
Do I have to prove why those two do not equal each other? – Samuel Gregory Aug 23 '12 at 21:19
If they are, then they must both be 0 (it's the only quantity that's equal to it's negation), but then that also means $x = 0 = y$. – Tunococ Aug 23 '12 at 21:25
So how does that prove that x=-y is impossible if 0 works? – Samuel Gregory Aug 23 '12 at 21:30
Uh, I thought the question was all settled. I claim that $x = -x$ has only one solution which is $x = 0$. You can find various ways to "prove" it, depending on what axioms you accept. – Tunococ Aug 24 '12 at 3:59

Alternately, you can look at the derivative:

$$f'(x)=\frac{1}{\sqrt{x^2+1}} - \frac{x}{2}(x^2+1)^{-\frac{3}{2}}(2x) = \frac{x^2+1 - x^2}{(x^2+1)^{\frac{3}{2}}} = \frac{1}{(x^2+1)^{\frac{3}{2}}}>0$$

Therefore $f(x)$ is strictly monotone increasing and is therefore one-to-one.

All that you have to do now is show that $f$ is onto: given some $y\in (-1,1)$, we need to find $x$ such that $f(x)=y$, or in other words, $$y \sqrt{x^2+1} = x$$

Just find all values of $x$ that satisfy this equation for a given $y$.

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@SamuelGregory A person who reads your question understands only one thing. That is; you dont have any idea how to start or what to do. And after you receive an answer, saying that you were not looking for an answer with calculus is not polite. – Seyhmus Güngören Aug 23 '12 at 21:42
I said that I was sorry and that I have should have mentioned that. I do not know what else to say? I do know how to start, I set the two equations = to each other and found that x^2 = y^2 but was not sure what to do after that. I have never used this site before, someone just recommended it to me. I would appreciate it if you would not be rude to me as well. Thank you. – Samuel Gregory Aug 23 '12 at 21:44
@JackManey + for the calculus way of the solution ;) – Seyhmus Güngören Aug 23 '12 at 21:46
@Samuel: I have taken the liberty of cleaning up many of the comments here that neither contribute to the question or its answer. As an aside to everyone, please remain civil at all times, especially with first time users who don't know the way of the land. I would think this was a terrible introduction of the site to OP. – mixedmath Aug 24 '12 at 3:49
@mixedmath Seeing that you are a moderator, it would have been more proper to omit the last sentence of your comment. (As an aside, I do not quite understand that you recommend to remain civil at all times especially with first time users... to a first time user.) – Did Aug 24 '12 at 9:44

In the proof that the function is 1-1 you are not using some important information. Suppose you know that $x^2=y^2$. What additional information does $x/\sqrt{x^2+1}=y/\sqrt{y^2+1}$ give you?

To show that the function is onto, unfortunately the supposed range of the function is not correctly displayed. But the range is $(-1,1)$, right? To see this, compute the limits of the functions as $x$ tends to $-\infty$ and to $\infty$ and think about what happens in between. Use the intermediate value theorem.

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Is the additional info just that x^2+1>0? – Samuel Gregory Aug 23 '12 at 21:28
@SamuelGregory - No, it isn't. – user5137 Aug 24 '12 at 6:27

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