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

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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} &gt; \frac{r-1}{r+b-1}<br />

<br /> \left(\frac{r}{r+b}\right)^2 &gt; \frac{1}{2} &gt; \left(\frac{r-1}{r+b-1}\right)^2<br />

Thus
\frac{r}{r+b} &gt; \frac{1}{\sqrt{2}} &gt; \frac{r-1}{r+b-1}

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

r &gt; \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
 
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Divide both sides with \sqrt{2}+1 work the product in the denominator to yield one, and you'll see.
 
I would have multiplied both sides by \sqrt{2}-1!:rolleyes:
 
But surely better than to verify, is to understand how to transform things like this in the future. It is rationalizing the denominator.
 
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