Solve Algebraic Equation: r > b/(sqrt2-1) = (sqrt2+1)b | Easy Explanation

  • Context: Undergrad 
  • Thread starter Thread starter rwinston
  • Start date Start date
  • Tags Tags
    Algebra
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
rwinston
Messages
36
Reaction score
0
Hi

I am working through the superb book "50 challenging problems in probability." I have a slight problem with some of the algebra used in solving one of the problems. The algebra to solve is shown below:

[tex] \frac{r}{r+b}\times\frac{r-1}{r+b-1}=\frac{1}{2}[/tex]

Since, for b > 0:
[tex] \frac{r}{r+b} > \frac{r-1}{r+b-1}[/tex]

[tex] \left(\frac{r}{r+b}\right)^2 > \frac{1}{2} > \left(\frac{r-1}{r+b-1}\right)^2[/tex]

Thus
[tex]\frac{r}{r+b} > \frac{1}{\sqrt{2}} > \frac{r-1}{r+b-1}[/tex]

So (this is the part I have difficulty with):

[tex]r > \frac{b}{\sqrt{2}-1} = (\sqrt{2}+1)b[/tex]

I can't see how the two sides of the = sign can be transformed into each other, or how they are equivalent. Can anyone help with this?

Cheers
 
Physics news on Phys.org
Divide both sides with [tex]\sqrt{2}+1[/tex] work the product in the denominator to yield one, and you'll see.