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
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.