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I want to show that the function $f(x) := \frac{x}{\sqrt{x^2+1}}$, $x\in \mathbb{R}$, is the bijection of $\mathbb{R}$ onto $B:={y:0<y<1}$

Firstly, I use the horizontal line test. Taking ${x_1}$ and ${x_2}$ in $\mathbb{R}$ and assume that $f({x_1})$ = $f({x_2})$,

$\frac{x_1}{\sqrt{x^2_1+1}}$=$\frac{x_2}{\sqrt{x^2_2+1}}$

$x^2_1 = x^2_2$

$x_1 = \pm x_2 $

$x_2 = \pm x_1 $

which proved that the function is not injective.

Did I do something wrong?

user356512
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  • Yes... how did you get from the first line to the second? You lost something there -- the sign. – Clement C. Jul 27 '16 at 04:44
  • (From a more high-level point of view: you didn't proceed by equivalence, but by implication. Your conclusion is not equivalent to the original hypothesis...) – Clement C. Jul 27 '16 at 04:46

2 Answers2

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There is something wrong in the reasoning.

The square operation does not guarantee the sign of the RHS and LHS.

To show injection, one should try to assume that $x_1,x_2 \geq 0$ or $x_1 , x_2 \leq 0$. Since $f(x) = \frac{x}{\sqrt{x^2+1}}$ shows that the sign of left about zero and right about zero is different which means cases $x_1 \leq 0, x_2 \geq 0$ or $x_1 \geq 0, x_2 \leq 0$ can be ignored. So one should delve into these two cases.

Zau
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  • Thanks. I think the problem the question who does not restrict the sign of {x_1} and {x_2}. Hence, there is possible for ${x_1}$ and ${x_2}$ to be in different signs. For example, by the result of calculation,$ {x_1^2} = {x_2^2}$, it is possible that ${(-2)^2} = {(2)^2}$ which means ${x_1 \ne x_2}$. However, if I cannot prove the injection, the proof of bijection cannot be vaild – user356512 Jul 27 '16 at 04:57
  • Yes that's why I said square operation doesn't ensure the sign of RHS and LHS. For example $2 \neq -2$, after squaring, $4 = 4$. So that's not ensure the sign. – Zau Jul 27 '16 at 04:59
  • Indeed, I feel very strange on the question cause wolfram math calculator said that the function is injection but not bijection
    https://www.wolframalpha.com/input/?i=is+y%3Dx%2F(x%5E2%2B1)%5E1%2F2+a+one-to-one+function%3F     – user356512 Jul 27 '16 at 05:03
  • @user356512 wolframalpha is right because there is no surjectiveness of the function. To be surjective, a function should map to all the elements in the codomain. It is easy to see this property is invalidate. Pick $f(x) = 2$ and you will get a quadratic equation. Finally it has no root in Real. (i.e. there is no such element in domain can be mapped to the $2$ which is a real number in the codomain) – Zau Jul 27 '16 at 05:09
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If $f(x)=f(y)$ then $$\frac{x^2}{x^2+1}=\frac{y^2}{y^2+1}$$so that $x^2y^2+x^2=x^2y^2+y^2$, or $x^2=y^2$. Hence, if $f(x)=f(y)$, then $x=y$ or $x=-y$. On the other hand, if $x\neq 0$, by inspection $f(x)\neq 0 $ and $f(-x)=-f(x)$, so that $f(x) \neq f(-x)$.

It follows that whenever $f(x)=f(y)$ we have $x=y$, and thus $f$ is injective.

Fimpellizzeri
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