- #1
rwinston
- 36
- 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
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