# Simple Algebra Question

1. Sep 1, 2007

### rwinston

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:

$$\frac{r}{r+b}\times\frac{r-1}{r+b-1}=\frac{1}{2}$$

Since, for b > 0:
$$\frac{r}{r+b} > \frac{r-1}{r+b-1}$$

$$\left(\frac{r}{r+b}\right)^2 > \frac{1}{2} > \left(\frac{r-1}{r+b-1}\right)^2$$

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 cant 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

2. Sep 1, 2007

### da_willem

Divide both sides with $$\sqrt{2}+1$$ work the product in the denominator to yield one, and you'll see.

3. Sep 1, 2007

### HallsofIvy

I would have multiplied both sides by $\sqrt{2}-1$!

4. Sep 1, 2007

### matt grime

But surely better than to verify, is to understand how to transform things like this in the future. It is rationalizing the denominator.