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I am looking for help in regard to a practice question about functions. The question is

Show that a function $f$, defined by $f(x)=x/\sqrt{x^2+1}$ , $x \in \Bbb R$ is a bijection of $\Bbb R$ onto $\{ y: -1<y<1\}$.

So what I know that for it to be a bijection, it must be an injection and also a surjection. So to proof this question, do I just need to prove both of those?

So for injection, when $x_1=x_2$ $f(x_1)=f(x_2)$

To do this I wrote $$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_2^2+1}},$$ squared both sides and expanded to solve that $x_1=x_2$.

Next for a surjection, must show that the range is contained,

so the bottom cannot be $0$ or negative because cannot square root a negative and cannot divide by $0$. I believe in this situation you are supposed to write it as $y=x/\sqrt{x^2+1}$ and solve for $x$ in terms of $y$ but I have trouble doing that. Or can I jsut do it by solving an inequality such as $\sqrt{x^2+1}>0$,

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  • $\begingroup$ You can only square both sides after you check whether or not the sides share the same sign. If you don't do this then you have could prove $-1 = 1$. $\endgroup$ – user157227 Aug 21 '14 at 0:59
  • $\begingroup$ @user157227, How would I go about checking that? $\endgroup$ – Quality Aug 21 '14 at 1:03
  • $\begingroup$ Now that I take a closer look, the quantities are equal so they do share the same sign. Nevermind. You did the injection correctly. $\endgroup$ – user157227 Aug 21 '14 at 1:11
  • $\begingroup$ @LearningMath: If $x_1/\sqrt{x_1^2 + 1} = x_2/\sqrt{x_2^2 + 1}$ then $x_1/x_2 = \sqrt{(x_1^2 + 1)/(x_2^2 + 1)}$. Since the RHS is positive, this shows that $x_1$ and $x_2$ have the same sign, so now it's valid to square both sides. $\endgroup$ – Bungo Aug 21 '14 at 1:11
  • $\begingroup$ @bungo Oh that is smart, makes sense. $\endgroup$ – Quality Aug 21 '14 at 1:12
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Injective: suppose that $$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_2^2+1}}\ .\tag{$*$}$$ Squaring both sides and multiplying out denominators, $$x_1^2(x_2^2+1)=x_2^2(x_1^2+1)\quad\Rightarrow\quad x_1^2=x_2^2\ .$$ Now substituting back into the denominator on the RHS of $(*)$, $$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_1^2+1}}\quad\Rightarrow\quad x_1=x_2\ .$$ Doing it this way avoids having to (explicitly) consider whether or not $x_1$ and $x_2$ have the same sign.

Surjective: we want to solve $$y=\frac{x}{\sqrt{x^2+1}}\tag{$*\!*$}$$ for any $y\in(-1,1)$. Easy algebra gives $$x^2=\frac{y^2}{1-y^2}\ ,$$ and we now have to consider which square root to take in order to get the right $x$. It is not hard to see that $y$ should have the same sign as $x$, so we guess $$x=\frac{y}{\sqrt{1-y^2}}\ .$$ Note the word "guess": if you stop here, the solution is logically backwards and therefore incorrect. We need to actually substitute this expression for $x$ into the RHS of $(**)$ and confirm that it simplifies to $y$. This is easy, so I leave it to you.

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  • $\begingroup$ "substituting back into the denominator..." to avoid the sign check - nice! $\endgroup$ – Bungo Aug 21 '14 at 1:32
  • $\begingroup$ Very good job,David. You see,it's the little things like these kinds of algebraic shorthand that really help in doing actual mathematics. It really helps avoid subtle confusions that can arise when doing computations by brute force. $\endgroup$ – Mathemagician1234 Aug 21 '14 at 1:40
  • $\begingroup$ For the surjective part, plugging x into the RHS of (**) does not seem to simplify so easily. I've tried doing it myself and Wolfram Alpha does not seem to simplify it down to y either. Any tips for doing this? $\endgroup$ – flubsy Sep 6 '15 at 2:35
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    $\begingroup$ @samalamma708 $$\eqalign{x=\frac{y}{\sqrt{1-y^2}}\quad &\Rightarrow\quad x^2+1=\frac{y^2}{1-y^2}+1=\frac{1}{1-y^2}\cr &\Rightarrow\quad \frac{x}{\sqrt{x^2+1}}=\frac{y}{\sqrt{1-y^2}}\bigg/\frac{1}{\sqrt{1-y^2}}=y\ .\cr}$$ $\endgroup$ – David Sep 6 '15 at 5:01
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since $f$ is differentiable $$ \frac{df}{dx} = \frac1{(x^2+1)^{\frac32}} $$ $f$ is monotonic. but it is also continuous...

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  • $\begingroup$ So does that show it is surjective? $\endgroup$ – Quality Aug 21 '14 at 1:10
  • $\begingroup$ have you encountered the intermediate-value theorem? $\endgroup$ – David Holden Aug 21 '14 at 1:15
  • $\begingroup$ Not in analysis, in calculus one though. This seems helpful though. $\endgroup$ – Quality Aug 21 '14 at 1:17
  • $\begingroup$ Clever,David-but may be a bit subtle for students who aren't strong on calculus beyond simple pencil pushing computations with derivatives and integrals. $\endgroup$ – Mathemagician1234 Aug 21 '14 at 1:43
  • $\begingroup$ yes, a good point. i failed to identify and address precisely OP's uncertainty $\endgroup$ – David Holden Aug 21 '14 at 1:48
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Let $y \in (-1, 1)$. Then $y^2 \in (0, 1)$ . That is $ 0 \lt y^2 \lt 1 $. Now take, $$ t = \frac{y^2}{1 - y^2} \implies y^2 = \frac{t}{t + 1} \;\;; \; t\gt 0 \; \;\text{by its definition }$$

This was inspired by trying to write $y^2$ in the form of $ \dfrac{t}{t + 1}$ and then solving for $t$. Now,

$$ y = \pm \sqrt {\dfrac{t}{t + 1} } = \pm \dfrac{\sqrt t}{ \sqrt{(\sqrt t)^2 + 1}} $$

Now take $x = \sqrt t$ or $ x = - \sqrt t $. Such an $x$ exists in $\Bbb R$ since $t \gt 0$. hence we have proven that there is $x \in \Bbb R$ such that $ y = \dfrac{x}{ \sqrt {x^2 +1}} $

$ \mathscr Q.E.D.$

Ultimately your $x$ is equal to $ \pm \sqrt{\dfrac{y^2}{1 - y^2}} $

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