Question

How to prove there is one to one function from (-1,1) to Real.
Please, help

Answer 1

Oh I remember doing this one earlier in the semester! :)
Start with a function that has an interval for its range like:\[\large\rm f(x)=\arctan(x)\]So we can see that\[\large\rm f:\mathbb R\to (-\pi,\pi)\]

Oh wait wait, I must be thinking in the wrong direction.
I would going to map the reals to (-1,1).
Hmm, thinking... :d

Answer 2

Oh this should work though, right?
If we find a mapping from the reals to (-1,1),
we can just take it's inverse ya? :o

Maybe that's extra work to do so though.
If we just start with tangent:\[\large\rm f(x)=\tan x\]We can see that:\[\large\rm f:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to\mathbb R\]

So then ummm,\[\large\rm g(x)=\tan(\pi x)\]This one is a little bit closer ya?\[\large\rm g:\left(-\frac{1}{2},\frac{1}{2}\right)\to\mathbb R\]

Oh,\[\large\rm h(x)=\tan\left(\frac{\pi x}{2}\right)\]Does that work maybe? :o\[\large\rm h:\left(-1,1\right)\to\mathbb R\]

Answer 3

Oh tangent is not one-to-one though! :)
doh!

Answer 4

Look up at your note, please

Answer 5

? :o

Answer 6

You said you saw it at the beginning of the semester, I assume that you had it. OH, I have students come. Sorry,have to be out

Answer 7

oh ok bye c:

Answer 8

How is tangent restricted to (-1,1) not one to one?

Answer 9

\[
h(x)=\tan\left(\frac{\pi x}{2}\right)\\
\arctan(h)=\frac{\pi x}{2}\\
\frac{2\arctan(h)}{\pi}=x
\]
Image of arctan is (-pi/2,pi/2) so it works right?

Answer 10

I don't think so !! you assume it is bijection, hence, it has inverse. Then you use that inverse to prove it is one to one. That is called circulation proof.

Answer 11

I don't know how to prove something is a bijection but from the graph of \(f(x)=\tan\left(\frac{\pi x}{2}\right)\) it is clear that f is a bijection if restricted to (-1,1).

Answer 12

\[
\tan(x)=\frac{e^{ix}-e^{-ix}}{i\left( e^{ix}+e^{ix}\right)}\\
\]
Find the inverse using this. Then you might able to prove that tangent is a bijection.
https://math.stackexchange.com/questions/165434/how-to-prove-if-a-function-is-bijective