Proof that sum is rational number

**OReilly** · May 28th 2006, 02:43 PM

I have to prove that
$\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}$
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:

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

I have then discovered that next five fractions are also equal to 1.
$\frac{1}{{\sqrt 4 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 6 }} + \frac{1}{{\sqrt 6 + \sqrt 7 }} + \frac{1}{{\sqrt 7 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt 9 }} = 1$ .

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 $3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99$ whole sum is equal to 9: $\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}=9$

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.

**ThePerfectHacker** · May 28th 2006, 02:48 PM

Originally Posted by OReilly

I have to prove that
$\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}$
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:

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

I have then discovered that next five fractions are also equal to 1.
$\frac{1}{{\sqrt 4 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 6 }} + \frac{1}{{\sqrt 6 + \sqrt 7 }} + \frac{1}{{\sqrt 7 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt 9 }} = 1$ .

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 $3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99$ whole sum is equal to 9: $\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}=9$

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.

One word,
rationalize.

**OReilly** · May 28th 2006, 02:53 PM

Originally Posted by ThePerfectHacker

One word,
rationalize.

Bit more words?

**ThePerfectHacker** · May 28th 2006, 03:16 PM

Originally Posted by OReilly

Bit more words?

Okay whenever you have,
$\frac{1}{\sqrt{x}\pm \sqrt{y}}$ you can change the fraction, called "rationalizing". This is done by mutiplying the numerator and denominator by $\sqrt{x}\mp \sqrt{y}$ . This changes the denominator into whole number because you use the difference of two squares.
Thus,
$\frac{1}{\sqrt{x}\pm \sqrt{y}}\cdot \frac{\sqrt{x}\mp \sqrt{y}}{\sqrt{x}\mp \sqrt{y}}=\frac{\sqrt{x}\mp \sqrt{y}}{x-y}$
------------
Hence you have,
$\frac{1}{\sqrt{1}+ \sqrt{2} }+\frac{1}{\sqrt{2}+ \sqrt{3} }+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{ \sqrt{99}+\sqrt{100}}$
Upon rationalizing,
$\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}$
Simplyfy,
$-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{99}+\sqrt{100}$
Each one cancels each other out (such a series is called telescoping), thus,
$-1+\sqrt{100}=-1+10=9$

**OReilly** · May 28th 2006, 03:21 PM

Thanks!

**ThePerfectHacker** · May 28th 2006, 03:55 PM

Welcomed.
That happens to be a nice problem which I really like. Where did you get it from?

**OReilly** · May 28th 2006, 04:31 PM

Originally Posted by ThePerfectHacker

Welcomed.
That happens to be a nice problem which I really like. Where did you get it from?

From high school book. Author has also wrote book with variety of problems which i find most interesting.

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