# Basic inequality question

1. Jun 6, 2009

### hassman

Hi.

Tried to solve first problem in the book "Firfty Challenging Problems in Probability" and solved it although very ugly.

Then I check the answers and see the author use the following inequality:

$$r > \frac{1}{\sqrt{2}-1}b=(\sqrt{2} + 1)b$$

Now correct me if I am wrong, but this implies that

$$\frac{1}{\sqrt{2}-1}b=(\sqrt{2} + 1)b$$

Right? Well, this does not seem right, does it?

2. Jun 6, 2009

### protonchain

Given: $$\frac{1}{\sqrt{2}-1}b=(\sqrt{2} + 1)b$$

Multiply by $$\frac{\sqrt{2}+1}{\sqrt{2}+1}$$

$$\frac{1}{\sqrt{2}-1}*\frac{\sqrt{2}+1}{\sqrt{2}+1}b=(\sqrt{2} + 1)b$$

$$\frac{\sqrt{2}+1}{2-1}b=(\sqrt{2} + 1)b$$

$$\frac{\sqrt{2}+1}{1}b=(\sqrt{2} + 1)b$$

3. Jun 6, 2009

### hassman

sweet mother of god. thanks.

It is always the 1 that is omitted that confuses me. Plus I used sqrt(9) to ease the calculation.

4. Jun 6, 2009

### protonchain

Sure, I just hope you understood the steps I took.

I'll lay it out in English just in case.

Basically you multiply top and bottom by the same thing (aka 1), then when you multiply the denominator, you'll remember that (x + A) (x - A) = x^2 - A^2. So for this we get 2 - 1 = 1 :)

5. Jun 6, 2009

### hassman

yes I understood from the first reply, it's just so simple, hence my amazement.

6. Jun 6, 2009

### protonchain

Ah no worries. I wouldn't have thought to do this step either if this was me several years ago.

Once you see it the first time, it sticks to you. Once you practice it on a couple of problems it becomes natural. So whenever I see square roots like that in the denominator, I automatically turn on simplification mode (since I've seen it so many times).

Such is knowledge and life.