Steve asked in Science & MathematicsMathematics · 4 months ago

Is there a method to work out how to express a non-unit fraction as the product of unique unit fractions?

I ve read that the Egyptians didn t have a way to write down non-unit fractions, so they would instead just express them as the sum of unit fractions (eg. instead of writing 11/12, they would write 1/2 + 1/3 + 1/12).

However I m pretty sure they also tried to avoid using the same non-unit fraction more than once, even when it wouldn t be inefficient to do so. So although a simple way to express 2/5 would be to just write 1/5 + 1/5, they might instead express it like 1/3 + 1/15 even though it isn t more efficient in terms of the amount of unit fractions, it is expressed as the sum of 2 unit fractions in both cases. But I guess it is more efficient in terms of using the largest unit fractions possible in 2/5.

I m honestly not entirely sure if it was vital that they avoided using the same fraction more than once, but nevertheless I became interested in trying to express non-unit fractions in this way. But I don t know how to do it in a very efficient way.

The only way I could do it was to find the decimal value of the non-unit fraction with a calculator, and then to just use trial and error to find the largest unit-fraction that would fit into that value. I d then subtract that unit fraction from the decimal value. And then I d find the largest unit fraction that would fit into what s left over, etc. Is there a method though that would allow you to do this without using trial and error?

PS. Could you please explain it in simple terms if that s possible?

Update:

Edit: In the question title, I actually meant to say the SUM of unique unit fractions. Sorry for the confusion.

Relevance
• atsuo
Lv 6
4 months ago

You wrote 11/12 = 1/2 + 1/3 + 1/12 , this is a sum , not a product .

I try to explain "greedy algorithm for Egyptian fractions" in simple terms .

Let given non-unit fraction be 0 < n/d < 1 .

Take a positive number k so that 1/k < n/d < 1/(k-1) .

n/d - 1/k = (nk - d)/(dk) , so n/d = 1/k + (nk - d)/(dk) .

1/k < n/d so (nk - d)/(dk) is positive , and n/d < 1/(k-1) so

n(k-1) < d

nk - n < d

nk - d < n

Therefore , we can divide n/d into an unit fraction 1/k and another fraction

(nk - d)/(dk) with a smaller numerator . And (nk - d)/(dk) may be able to reduce .

If (nk - d)/(dk) becomes an unit fraction then this process is finished .

If not , then let (nk - d)/(dk) be new given non-unit fraction and

repeat this process .

The numerator decreases at every step , so this process is always finished

after finite steps .

For example , let n/d = 5/11 .

1/3 < 5/11 < 1/2 , so k = 3 .

5/11 = 1/3 + 4/33 , so let new n/d = 4/33 .

1/9 < 4/33 < 1/8 , so k = 9 .

4/33 = 1/9 + 3/297 = 1/9 + 1/99 , so this process is finished .

That is , 5/11 = 1/3 + 1/9 + 1/99 .

But this method may give very complicated result .

• Steve4 months agoReport

Thank you, this is perfect! I really appreciate you writing this all out, it was just what I was looking for!

• 4 months ago

Do you want do express as a PRODUCT, like your question says, or as a SUM? Product is simple: for example 7/13 is = (7/1) * ( 1/13). If you want a sum, then the result is not always unique. Wikipedia has one method, but it often gives weird results.

• Steve4 months agoReport

Oh yeah, sorry. I meant the sum alright. I'll change it now.