rwinston
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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:
<br /> \frac{r}{r+b}\times\frac{r-1}{r+b-1}=\frac{1}{2}<br />
Since, for b > 0:
<br /> \frac{r}{r+b} > \frac{r-1}{r+b-1}<br />
<br /> \left(\frac{r}{r+b}\right)^2 > \frac{1}{2} > \left(\frac{r-1}{r+b-1}\right)^2<br />
Thus
\frac{r}{r+b} > \frac{1}{\sqrt{2}} > \frac{r-1}{r+b-1}
So (this is the part I have difficulty with):
r > \frac{b}{\sqrt{2}-1} = (\sqrt{2}+1)b
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:
<br /> \frac{r}{r+b}\times\frac{r-1}{r+b-1}=\frac{1}{2}<br />
Since, for b > 0:
<br /> \frac{r}{r+b} > \frac{r-1}{r+b-1}<br />
<br /> \left(\frac{r}{r+b}\right)^2 > \frac{1}{2} > \left(\frac{r-1}{r+b-1}\right)^2<br />
Thus
\frac{r}{r+b} > \frac{1}{\sqrt{2}} > \frac{r-1}{r+b-1}
So (this is the part I have difficulty with):
r > \frac{b}{\sqrt{2}-1} = (\sqrt{2}+1)b
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