Ancient Egyptian 2/n-Table: Solved 4k Yrs Ago?

[DrKareem]

One of my former teachers claimed that he had solved an ancient problem (4k y/o) called Egyptian 2/n-table. He told me (a while ago) that he submitted his proof to historia mathematica but i didn't communicate with him since and didn't know what happened next. I'm wondering if anyone could show me where i can find resources on the problem, and if anyone who has subscription to that journal if it is published yet or not. Cheers

 

[Werg22]

What is the conjuncture exactly?

 

[robert Ihnot]

This deals with writing 2/N as "egyptian fractions," i.e., as inverted positive integers. This situation deals with, for example, 2/3=1/2+1/6. See http://www.math.buffalo.edu/mad/Ancient-Africa/mad_ancient_egyptroll2-n.html.

 

[Werg22]

? This is not a conjuncture at all.

\frac{3}{2} + \frac{1}{2} = 2

Dividing both side by n, we get

\frac{3}{2n} + \frac{1}{2n} = \frac{2}{n}

Since 3 | n, \frac{3}{2n} = \frac{1}{x} where x is some positive integer and x = \frac{2n}{3}. Letting 2n = y We have

\frac{1}{x} + \frac{1}{y} = \frac{2}{n}, which is what needed to be shown.

 

[AlephZero]

The conjecture is not the formulas themselves, but trying to discover the method the Egyptians may have used to create them, and therefore to find out what mathematics they knew (not what we know).

 

[d_leet]

? This is not a conjuncture at all.

\frac{3}{2} + \frac{1}{2} = 2

Dividing both side by n, we get

\frac{3}{2n} + \frac{1}{2n} = \frac{2}{n}

Since 3 | n, \frac{3}{2n} = \frac{1}{x} where x is some positive integer and x = \frac{2n}{3}. Letting 2n = y We have

\frac{1}{x} + \frac{1}{y} = \frac{2}{n}, which is what needed to be shown.

Can you explain why you are making the claim that 3|n? That is certainly not true for all n.

 

[Werg22]

For some reason I had understood that it was meant for multiples of 3 only My bad.

 

[DrKareem]

This deals with writing 2/N as "egyptian fractions," i.e., as inverted positive integers. This situation deals with, for example, 2/3=1/2+1/6. See http://www.math.buffalo.edu/mad/Ancient-Africa/mad_ancient_egyptroll2-n.html.

The page that provided states that the conjecture is solved? (however, i still don't fully understand the conjecture yet). Anyways I'm still curious...

 

[robert Ihnot]

Dr.Kareem: The page that provided states that the conjecture is solved? (however, i still don't fully understand the conjecture yet).

My own partial understanding of this is that some people think that the Egyptians had a real method of finding these fractions. In the article itself is stated: There are nearly 30,000 ways of representing the numbers 2/q for 2< q < 102. For those where the formula is used, how was the number a chosen?

However, this matter can be looked at from the other side, some have been amazed that the Egyptians did not seem to understand our ordinary use of fractions such as a/p where 1<a<p, insisting generally upon resolving everything in terms of inverted integers. This view hardly suggests that the Egyptians had any really efficient method of working with fractions.

The other side seems to be that the builders of the pyramids were advanced and must have had some general and efficient method of finding, "Best Fractions," what ever that means, considering the multitude of choices. I assume the conjecture goes along the lines of this view.

There is some argument that in the case, for example, of distributing 5 sacks of grain among 8 people that it would be very difficult to give each person 5/8 of a sack, but it would be easier to give 1/2 sack per person, and then divide 1 sack into 8 parts, giving each person an additional 8th.

The Greeks at first adopted the Egyptian system, just as they did for Plane Geometry, but consider:

2/101 = 1/101 + 1/202 + 1/303 + 1/606. Well if I was to distribute 2 sacks among 101 people, I would just suppose that is about 1/50 a sack per person, and hope to have a little left over for the last person. Or if not, open up another sack to serve the last person.

NOTE: The "Erdos-Strauss conjecture" (ESC) is the statement that for any integer n > 1 there are integers a, b, and c with 4/n = 1/a + 1/b + 1/c; a > 0, b > 0, c > 0.

 

[light_bulb]

i love these systems, i'll have fun finding them all out. for fun i found a new twist on casting out in pi, maybe not new but still interesting.

all of the angles of a triangle are cast out

 

[DrKareem]

Thanks robert Ihnot for this explanation. I guess this is the first time i get exposed to ancient mathematics. I hope i can get the time to fiddle with them before my finals. I'll also try to contact my prof soon and see if his paper about them was published or not.