What is the Inequality Used in Solving Probability Problems?

[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


<br /> <br /> \frac{1}{\sqrt{2}-1}b=(\sqrt{2} + 1)b<br /> <br />

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

 

[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

 

[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.

 

[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 :)

 

[hassman]

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

 

[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.