Find a bijection between the set of real numbers and the interval $(−1, 1) ≡ \{\,x ∈ \Bbb R\mid − 1 < x < 1\,\}$.
Hi am I trying to revise for an an exam and I came across this question which I can not figure out. I tried looking online but I kept just finding graphs.
Thank You
Answer
You have to find bijection map
$g:\mathbb{R} \to \left( { - 1,1} \right)$
Then, first write the any bijection map from $\mathbb{R}$ to any open interval.
And then make its range (co-domain) by some adjustment to required open interval.
Example:
$f\left( x \right) = {\tan ^{ - 1}}x$
then,
$\begin{gathered}
Dom\left( f \right) = \mathbb{R} \\
Range\left( f \right) = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right) \\
\end{gathered} $
then, your job is to arrange your range (co-domain) as
$\left( { - 1,1} \right)$
then, for this multiply your function f(x) by
$\frac{2}{\pi }$,
$\frac{2}{\pi }f\left( x \right) = \frac{2}{\pi }{\tan ^{ - 1}}x$
then take,
$\begin{gathered}
g\left( x \right) = \frac{2}{\pi }f\left( x \right) \\
= \frac{2}{\pi }{\tan ^{ - 1}}x \\
\end{gathered} $
And then,
$\begin{gathered}
Dom\left( g \right) = \mathbb{R} \\
Range\left( g \right) = \frac{2}{\pi }\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right) \\
= \left( { - 1,1} \right) \\
\end{gathered} $
Hence, the required bijection map $g:\mathbb{R} \to \left( { - 1,1} \right)$ is,
$g\left( x \right) = \frac{2}{\pi }{\tan ^{ - 1}}x$
I hope this will fulfill you required bijection map.
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