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- Thread starter DrKareem
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In summary, the conversation discussed the Egyptian 2/n-table problem which involves writing 2/N as "egyptian fractions," and the conjecture that the Egyptians had a method for finding these fractions. The conversation also touched on the debate between using inverted integers versus regular fractions and the Erdos-Strauss conjecture. The participants expressed interest in exploring ancient mathematics and finding out more about the problem and its possible solutions.

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What is the conjuncture exactly?

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- #4

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? This is not a conjuncture at all.

[tex] \frac{3}{2} + \frac{1}{2} = 2 [/tex]

Dividing both side by n, we get

[tex] \frac{3}{2n} + \frac{1}{2n} = \frac{2}{n} [/tex]

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

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

[tex] \frac{3}{2} + \frac{1}{2} = 2 [/tex]

Dividing both side by n, we get

[tex] \frac{3}{2n} + \frac{1}{2n} = \frac{2}{n} [/tex]

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

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

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Werg22 said:

[tex] \frac{3}{2} + \frac{1}{2} = 2 [/tex]

Dividing both side by n, we get

[tex] \frac{3}{2n} + \frac{1}{2n} = \frac{2}{n} [/tex]

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

[tex] \frac{1}{x} + \frac{1}{y} = \frac{2}{n} [/tex], 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.

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

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robert Ihnot said:

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...

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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:

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.

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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

all of the angles of a triangle are cast out

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