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I am trying to prove that a function is bijective and I really am not sure how to go about it. I know that I must show that the function is both injective and surjective for it to be bijective. The function that I am trying to prove is bijective is:

$f:\mathbb{R} \to (-1,1); x \mapsto \frac {x}{\sqrt{1+x^2}}$

So to prove that it is injective, would I begin by letting $x, y \in (-1,1)$ and have:

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

Would this be the right way of going about it? I am not sure on how to prove that it is surjective.

Aaron
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  • I'd square both sides and then multiply both sides by $(1+x^2)(1+y^2)$. – Git Gud Nov 10 '13 at 16:22
  • You've got a fine start: can you make the case that your equation holds if and only if $x = y$? If so, then the function is injective. See @GitGud's suggestion above. But note, we need for $x, y \in \mathbb R$, and $f(x), f(y) \in (-1, 1)$. – amWhy Nov 10 '13 at 16:23
  • Ok so I have $x^2(1+y^2)=y^2(1+x^2)$ so I presume this equation holds iff $x=y$? So this would prove that the function is injective? – Aaron Nov 10 '13 at 16:28
  • @Aaron Well the immediate consequence is merely $x^2=y^2$, so $x=\pm y$, but ... – Hagen von Eitzen Nov 10 '13 at 16:31
  • As Git Gud mentioned in another question, doesn't the starting assumption imply that the signs are the same? – Aaron Nov 10 '13 at 16:33
  • @Aaron That's correct. But do you understand why? – Git Gud Nov 10 '13 at 16:40
  • Not sure... would it simply be due to the fact that $\frac {x}{\sqrt{1+x^2}}$ can be negative or positive? – Aaron Nov 10 '13 at 16:46
  • Suppose $x$ and $y$ have different signs (for instance $x>0$ and $y<0$), if $\frac {x}{\sqrt{1+x^2}} = \frac {y}{\sqrt{1+y^2}}$, then the LHS is positive because $x$ is positive and $\sqrt \cdot$ is positive, however, the RHS is negative over positive, which yields negative. So you get that a positive numbers equals a negative number, which is a contradiction. Therefore $x$ and $y$ can't have different signs. – Git Gud Nov 10 '13 at 16:50
  • Yep :) now for surjectivity, would I follow what N.S. said in regards to setting $y=\frac {x}{\sqrt{1+x^2}}$ – Aaron Nov 10 '13 at 16:54
  • Yes, solve for $y$.You'll need to use the quadratic formula to get $y$ as a function of $x$. For each $y$you'll get two $x$'s, but only one works. To find out which, follow his hint: 'don't forget that $y$ must have the same sign as $x$'. – Git Gud Nov 10 '13 at 16:57

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Hint:

$$\frac {x}{\sqrt{1+x^2}} = \frac {y}{\sqrt{1+y^2}} \Rightarrow x\sqrt{1+y^2}=y\sqrt{1+x^2}$$ and square both side.

For onto, let $y \in (-1,1)$ try to solve

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

This is the same as

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

square again both sides, and don't forget that $y$ must have the same sign as $x$...

P.S. Don't forget to start the problem by showing that $f$ is well defines, that is $f(x) \in (-1,1)$.

N. S.
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  • Squaring yields $x^2=y^2$, so further investigation is necessary. – Michael Hoppe Nov 10 '13 at 16:26
  • @MichaelHoppe That's why i said hint... $x^2=y^2$ is easy to solve, and also easy to check which of the two solutions is wrong/right.... – N. S. Nov 10 '13 at 16:28
  • @MichaelHoppe I know you know this, but I'll reply to help the OP. The starting assumption is that $\frac {x}{\sqrt{1+x^2}} = \frac {y}{\sqrt{1+y^2}}$, among other things, this implies that $x$ and $y$ have the same sign. – Git Gud Nov 10 '13 at 16:28
  • @GitGud I don't doubt -- even in my weirdest dreams -- that you're aware of that extra consideration: but how high do you estimate the beginner's percentage which are aware of that, too? My estimate would be less than a third. – Michael Hoppe Nov 10 '13 at 17:05
  • @MichaelHoppe The fact that I thought that the OP could have some trouble with it is the reason why I explained how to find $x=y$. Meaning I agree with you in principle, but I won't commit to a number $\ddot \smile$ – Git Gud Nov 10 '13 at 17:07
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Hints:

  • for injectivity: Try to find a left inverse
  • for surjectivity: prove that $f$ is increasing and have a look at $\lim_{x\to -\infty} f(x)$ and $\lim_{x\to +\infty}f(x)$
math
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  • Though useful, I think this is too conceptual to the apparent level of the OP. (+1 because of future readers). – Git Gud Nov 10 '13 at 16:25
  • @GitGud You are completely right. I was unsure to post it. But as usual, "the more the better" :) – math Nov 10 '13 at 16:27
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Hint. Show that $f = \sin \circ \arctan$. What can you say about those two functions?

Karolis Juodelė
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