Can every rational number be represented as a finite sum of reciprocal numbers? You are only allowed to use each reciprocal number one time per expression (So for example 3/2 cannot be 1/2+1/2+1/2).
You could then express these numbers in a kind of "binary-reciprocal" where a 1 in the nth place from the right denotes adding 1/n. E.g 17/10 could be 10011 as it is 1+1/2+1/5.
Thanks in advance!
Yes, and there is a simple algorithm to achieve this. First, we subtract successive reciprocal numbers $(1/1,\ 1/2,\ 1/3,\ \ldots)$ until the remainder is less than the next reciprocal number. This is always possible because the harmonic series is divergent.
If the remainder is zero then we are done, so let us assume that $$0 < \frac{a}{b} < 1.$$ There exists a unique positive integer $m$, greater than any of the preceding denominators, so that $$\frac{1}{m} \le \frac{a}{b} < \frac{1}{m-1}.$$ Therefore, $$\frac{a}{b} - \frac{1}{m} = \frac{am-b}{bm} \ge 0.$$
But $$\frac{a}{b} < \frac{1}{m-1} \implies am-a < b \implies am-b < a,$$ so the numerator has decreased.
If the process is repeated, then the remainder will eventually be zero, since the numerators cannot decrease forever. Therefore, every positive rational number can be represented as a sum of distinct reciprocal numbers.
Example: To write 11/4 as a sum of distinct unit fractions, we subtract as many successive unit fractions as possible, starting with 1.
$$\frac{11}{4} - \frac11 - \frac12 - \frac13 - \frac14 - \frac15 - \frac16 - \frac17 - \frac18 = \frac9{280}.$$
The remainder lies between $\frac1{32}$ and $\frac1{31}$, so we subtract $\frac1{32}$.
$$\frac{9}{280} - \frac{1}{32} = \frac{1}{1120}.$$
Since the result is a unit fraction, we are done. $$\frac{11}4 = \frac11 + \frac12 + \frac13 + \frac14 + \frac15 + \frac16 + \frac17 + \frac18 + \frac1{32} + \frac1{1120}.$$ Thanks to Barry Cipra for pointing out an error in a previous version of this post.
Every positive number is a subseries of the harmonic series (maybe infinite). You do the following: if $x$ is your number - add numbers till the next step is over $x$, Then don't add the following till adding again will give you something less or equal. Proceed in this manner.
For positive rationals there is a method to achive them in a finite manner - using the following replacement $1/k+1/k=1/k+1/(k+1)+1/k(k+1)$ - Botts, Truman (1967), "A chain reaction process in number theory", Mathematics Magazine, 40 (2): 55–65
