Egyptian Fraction Representations

It's a well known fact that every positive rational can be written as the sum of distinct Egyptian Fractions, or reciprocals of integers. One way is to apply a greedy algorithm. Start with $\frac {p} {q}$. Then at the $k$th step, if the remaining number is at least $\frac {1} {k}$, subtract $\frac {1} {k}$ to get $\frac {p}{q} - \frac {1} {k} = \frac {pk-q} {qk}$. Note that $pk-q<p$ when $p(k-1)<q$, or when the fraction is less than $\frac {1} {k-1}$. But eventually, because the harmonic series diverges, we'll never reach a state where two unit fractions are subtracted in a row, beucase if $\frac {1} {r}$ was the last non subtracted unit fraction before them we have $\frac {1} {r} < \frac {1} {r+1} + \frac {1} {r+2}$, a contradiction. So eventually whenever we're subtracting a $\frac {1} {k}$ from $\frac {p} {q}$, it will be because $\frac {p} {q} < \frac {1} {k-1}$ and we did not subtract the $\frac {1} {k-1}$. So the numerator will decrease and eventually become $0$.

Another way, which you can figure out yourself, involves creating a binary tree. Start with, say, $2$ as a root (any number at least $2$ will do). Below each number, write $f(x)=x+1$ to the left and $g(x)=x(x+1)$ to the right. You can prove that the numbers in each row are distinct (try it yourself) and have sum of reciprocals $\frac {1} {2}$. From here it shouldn't be too hard to see that every positive half-integer, and thus every positive integer and every positive rational, can be written as the sum of reciprocals of distinct positive integers.