# Egyptian fraction representations of real numbers

I've been looking into Egyptian fractions now, but information on certain topics seems scarce. Can you answer any of these questions that intrigue me:

1) What is known about the Egyptian fraction representation (by the greedy algorithm) of irrational numbers? Are Egyptian fractions known to be interesting in any similar sense as the continued fractions? I have not found any references to these expansions; OEIS lists a few terms for some numbers though.

2) What would be the "average" growth of the n:th term in the sequence? The slowest possible would be 2, 3, 7 etc. but for a random real, what is the expected size of the n:th term?

• You should give some thought toward making Q2 more precise. For instance, what do you mean by "random real"? Perhaps restricting to $(0,1)$ is worthwhile. Note that the growth rate of terms is quite rapid and the expected value of the first term is already infinite (assuming uniform distribution on $(0,1)$), so perhaps a more logarithmic measure of size is warranted. Oct 20, 2013 at 19:11
• If I tried, I would most probably fail. I realize it's not accurate at all but I just want to see if someone has anything relevant to say about the growth of such sequences. Yes, but also there is a probability of 1/2 that the first term is 1/2, assuming uniform distribution on (0,1). I guess what I'd like to ask is, for a such a random real r, 0<r<1, which number is most likely to be the n:th term. This clearly has a solution. Oct 20, 2013 at 19:22

There's a little bit of info here:
H. E. Salzer, The Approximation of Numbers as Sums of Reciprocals, American Mathematical Monthly, Vol. 54, No. 3 (Mar., 1947), pp. 135-142. There are very few "naturally-occcurring" real numbers for which their expansion is known. One example where it is known is for (3-sqrt(5))/2, where the denominators are given by the (2^n)th Fibonacci number.

• so using egyptian fraction ( greedy algorithm ) for irrational numbers means that we can write irrational numbers as sums of unit fraction (rational), but knowing that the sum of rationals is always rational leads to a contradiction , how can an irrational number be equal to a rational number(sum of rational numbers)? Oct 29, 2017 at 20:35
• For irrational numbers you get an infinite sum, not a finite one. Oct 29, 2017 at 21:14
• but yet no matter if it is finite or infinite , the sum of rationals is still rational right? Oct 29, 2017 at 21:20
• No, of course not. Take any irrational number; then its base-10 representation expresses it as the infinite sum of rational numbers. Oct 29, 2017 at 22:07
• Does this mean that irrational numbers are nothing but infinite sums of rationals? Oct 30, 2017 at 8:02

After trying to parse the following painstakingly-simple algorithm into plain English, I gave up all hope:

$$N = \underbrace{\underbrace{\underbrace{\underbrace{\underbrace{\underbrace{[N]}_{A_0}+\frac1{\left[N-A_0\right]}}_{A_1}+\frac1{\left[N-A_1\right]}}_{A_2}+\frac1{\left[N-A_2\right]}}_{A_3}+\frac1{\left[N-A_3\right]}}_{A_4}+\frac1{\left[N-A_4\right]}}_{A_5}+\ldots$$

If the terms are not allowed to oscillate ($\pm$) , then, with the exception of [N] , [...] should stand for upper-rounding. However, if fast convergence is what's desired, [...] should stand for rounding. In any case, continued fractions make for better convergents than the Egyptian ones under the same conditions/restrictions/limitations.

The famous Hungarian—Jewish mathematician Erdos devoted a significant part of his life to the study of Egyptian fractions. See: Greedy algorithm for Egyptian fractions , Odd greedy expansion , Erdos-Graham conjecture , and Erdos-Straus conjecture.

• Your first sentence is much more suggestive of continued fractions, not Egyptian fractions. Oct 20, 2013 at 19:18
• I'm not sure if you read my question at all. Oct 20, 2013 at 19:30
• Take $\pi$. Calculate its integer part. It's 3. Subtract it, then calculate the inverse of its fractional part. It's 7. Now subtract $\tfrac17$ from $\pi-3$. Again, calculate the inverse of its fractional part. It's - 790. Now add $\tfrac1{790}$ to $\pi-3-\tfrac17$ , and calculate the inverse of its fractional part. It's 749896. Subtract $\tfrac1{749896}$ from $\pi-3-\tfrac17+\tfrac1{790}$ . Etc. Oct 20, 2013 at 19:30
• @Lucian I agree with your calculation, but what you wrote (eliminate the integer part) does not really describe the action you perform (subtract $1/7$). The links you gave are relevant to the question, so I'm not accusing you of conflating Egyptian and continued fractions, but this doesn't contribute substantially towards an actual answer. Oct 20, 2013 at 20:01
• I can read questions, but I can't read minds. Seven is the integer part of $\frac1{\pi-3}$. If the terms are not allowed to oscillate, then upper-rounding should replace the integer part. But oscillation makes for a better approximation. Either way, continuous fractions are better than both in this regard. Oct 20, 2013 at 20:11