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It is said that every positive rational number can be represented by infinitely many Egyptian fractions (defined as the sum of distinct unit fractions).

I am struggling to understand in a formal way, what algebraic structure such a set of Egyptian fractions of a positive rational is, and of what algebraic properties?

Thanks in advance and references are also welcome.

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1 Answer 1

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In fact, the rational number are rational because they are a fractions of two numbers. it may be different with positive integers or the sum of egyptians fractions which are link by a rule.

$1/2 = 1/3 + 1/6$

$1/3 = 1/4 + 1/12$

$1/4 = 1/5 + 1/20$

$1/5 = 1/6 + 1/30$

$1/6 = 1/7 + 1/42$

so we can say $1/U(1)=1/U(0)-1/U(0)U(1)$

$U(1)/U(0)-U(1)/U(0)U(1)=1 $ if $ U(1)=U(0)+1$

it is in french, but you can study the egyptian fraction (and the code) in clouds: http://jeux-et-mathematiques.davalan.org/arit/egy/index.html#table2

There are the same question here. A question about rational.

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  • $\begingroup$ There must be a misunderstanding. This is definitely not what I am asking for. The question is about the algebraic structure as in the context of abstract algebra (and possibly group theory). $\endgroup$
    – al-Hwarizmi
    Commented Aug 23, 2013 at 10:37
  • $\begingroup$ you are looking for something like that?en.wikipedia.org/wiki/Jordan_algebra $\endgroup$
    – user52413
    Commented Aug 23, 2013 at 10:51
  • $\begingroup$ I look for a solution in this direction: en.wikipedia.org/wiki/Algebra#Abstract_algebra $\endgroup$
    – al-Hwarizmi
    Commented Aug 23, 2013 at 11:20
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    $\begingroup$ (+1) Interesting, if one define a product $(x,y) \mapsto \rho(x,y) = \frac{xy}{x+y}$, then $\rho$ does satisfy the axiom of Jordan algebra as you say: $$\rho(x,y) = \rho(y,x)\quad\text{ and }\quad\rho(\rho(x,y),\rho(x,x)) = \rho(x,\rho(y,\rho(x,x)))$$ The set of fractions representable as Egyptian fractions is the reciprocal of those number span by $\mathbb{Z}_{+}$ as a Jordan algebra using product $\rho$. $\endgroup$
    – achille hui
    Commented Aug 23, 2013 at 11:25
  • $\begingroup$ @user52413 if you (or anybody else) would elaborate in your question rigorousely towards your hint to Jordan Algebra I would love to accept it as answer. $\endgroup$
    – al-Hwarizmi
    Commented Aug 28, 2013 at 6:45

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