Question

How to prove f is onto?
\[f:\mathbb {R} \rightarrow (-1,1)\\x\rightarrow \frac{x}{\sqrt{x^2+1}}\]
Please, help

Answer 1

after solving x w.r.t. y, I have
\[x = \pm \dfrac{y}{\sqrt{1-y^2}}\]
But I don't know how to get rid of minus sign while it is supposed to be.

Answer 2

You don't need the minus. For any \(y\in (-1,1),\) take:$$x=\frac{y}{\sqrt{1-y^2}}.$$This will be a real number since \(y\in (-1,1)\Longrightarrow 1-y^2>0.\) Then a quick computation shows: $$\frac{x}{\sqrt{x^2+1}}=\frac{\frac{y}{\sqrt{1-y^2}}}{\sqrt{\left(\frac{y}{\sqrt{1-y^2}}\right)^2+1}}=\frac{\frac{y}{\sqrt{1-y^2}}}{\sqrt{\frac{y^2}{1-y^2}+1}}=\frac{\frac{y}{\sqrt{1-y^2}}}{\frac{1}{\sqrt{1-y^2}}}=y.$$

Answer 3

Thanks for reply. However, my question is right there, why not minus? If I don't have minus sign, then it turns easy to me.

Answer 4

@tkhunny

Answer 5

This is very subtle. You will have to buy an argument you may never have heard.

When we try to solve \(x^{2} = y^{2}\), for either, we get the odd result \(|x| = |y|\;or\;x = \pm y\) because we do not know anything about x or y. If we know more about them, some of the burdensome notation may be unnecessary. For example, if we KNOW x > 0 and y > 0, then the solution to \(x^{2} = y^{2}\) is simply \(x = y\).

See if you can wedge this principle into your argument, KNOWING that we are only in the 1st and 3rd quadrants (and the Origin).

Answer 6

Thank you.

Answer 7

Part of the problem is the symbol, \(\pm\). Some have enumerated five different interpretations of the symbol. In this case, it might mean "x and y take on the same sign, whichever sign that is". It does NOT mean "both", as it sometimes does - such as the Quadratic Formula.