B Algebra Identity I can't figure out

1. Oct 29, 2017

Daniel Sellers

I seem to remember this Algebra identity being covered in one of my classes years ago, but it has cropped back up in studying the relativistic doppler effect for light.

Can anyone please show me the intermediate steps to show that:

(1+x)/(sqrt(1-x^2) = sqrt((1+x)/(1-x))

or similarly

(sqrt(1-x^2)/(1+x) = sqrt((1-x)/(1+x))

I can solve problems well enough by factoring gamma out of these equations but it is bugging me that all the texts I can find keep taking this for granted and I can't see why.

2. Oct 29, 2017

$1-x^2 =(1-x)(1+x)$. Comes from $a^2-b^2=(a-b)(a+b)$. The rest is just things like $\frac{u^1}{u^{1/2}}=u^{1/2}$ etc. where $u^{1/2}=\sqrt{u}$.

3. Oct 29, 2017

Daniel Sellers

I knew it was something obnoxiously simple and obvious! Thanks very much!

4. Nov 1, 2017

Ssnow

Suggestion: to see quickly you can take the square from both side $\frac{(1+x)^2}{1-x^2}\,=\, \frac{1+x}{1-x}$, now it is quite obvious ...
Ssnow

5. Nov 3, 2017

Staff: Mentor

Easy -- just cancel the exponents!
$$\frac{(1 + x)^2}{1 - x^2} = \frac{(1 + x)^{\rlap{/}2}}{1 - x^{\rlap{/}2}} = \frac{1 + x}{1 - x}$$