Originally Posted by

**OReilly** I have to prove that

$\displaystyle \frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}$

is rational number.

I have one idea, I hope it's not crazy!

I have spoted that the sum of first three fractions are equal to 1:

$\displaystyle \frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }}=1$

I have then discovered that next five fractions are also equal to 1.

$\displaystyle \frac{1}{{\sqrt 4 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 6 }} + \frac{1}{{\sqrt 6 + \sqrt 7 }} + \frac{1}{{\sqrt 7 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt 9 }} = 1$.

Continuing, we have that every odd number of fractions are equal to 1.

Considering that there are 99 fractions in that sum we get that because of $\displaystyle 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99$ whole sum is equal to 9: $\displaystyle \frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}=9$

I don't know how to prove that but it seems to me that there is a rule in that sum which I have described.