Prove that any positive rational number can be expressed as a finite sum
. . of distinct terms of the harmonic series: .$\displaystyle 1,\:\frac{1}{2},\:\frac{1}{3},\:\frac{1}{4},\:\hdo ts,\:\frac{1}{n}$
I created my own Riddle.
I was going to take the following approach....but then I stopped....why?
We will show that $\displaystyle \mathbb{Q}$ is a vector space over $\displaystyle \mathbb{Z}$ spanned by the infinite dimensional set $\displaystyle S=\{1,1/2,1/3,...\}$. But then I stopped there. Why?
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I am not going to answer your riddle because I already knew that, let someone else try.
No one took up the challenge?
This is an oldie from the study of "Eqyptian fractions".
There is no hide/show feature at this site, so here's the solution.
The rational number $\displaystyle \frac{a}{b}$ can be written: .$\displaystyle \underbrace{\frac{1}{b} + \frac{1}{b} + \frac{1}{b} + \cdots + \frac{1}{b}}_{a\text{ terms}}$
We eliminate the duplications by repeatedly using the identity:
. . $\displaystyle \frac{1}{n} \:=\:\frac{1}{n+1} + \frac{1}{n(n+1)} $ .until all denominators are distinct.
Example: $\displaystyle \frac{3}{7}$
$\displaystyle \frac{3}{7} \;=\;\frac{1}{7}\quad\; +\quad \frac{1}{7}\qquad\quad\; +\qquad\quad\; \frac{1}{7}$
. . $\displaystyle = \;\frac{1}{7} + \overbrace{\left(\frac{1}{8} + \frac{1}{56}\right)} +\quad \overbrace{\left(\frac{1}{8}\qquad +\qquad \frac{1}{56}\right)}$
. . $\displaystyle = \;\frac{1}{7} + \left(\frac{1}{8} + \frac{1}{56}\right) + \overbrace{\left(\frac{1}{9} + \frac{1}{72}\right)} + \overbrace{\left(\frac{1}{57} + \frac{1}{3192}\right)} $
Therefore: .$\displaystyle \frac{3}{7}\;=\;\frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{56} + \frac{1}{57} + \frac{1}{72} + \frac{1}{3192}$