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Let us call a fraction whose denominator is odd 'odd fraction'. Also, let us call an odd fraction whose numerator is $1$ 'odd unit fraction'.

Then, here is my question.

Question : Is the following true?

"Any odd fraction can be represented as the finite sum of different odd unit fractions."

I can neither find any counterexample nor prove that this is true. Can anyone help? If you have any helpful information, please let me know it.

Motivation : I've been interested in Egyptian Fraction. By using computer, I reached this expectation.

Examples : $$\frac25=\frac13+\frac1{15},\frac35=\frac13+\frac15+\frac1{15},\frac45=\frac13+\frac15+\frac17+\frac19+\frac1{105}+\frac1{315},$$ $$\frac27=\frac15+\frac1{13}+\frac1{115}+\frac1{10465},\frac37=\frac13+\frac1{11}+\frac1{231},$$ $$1=\frac13+\frac15+\frac17+\frac19+\frac1{15}+\frac1{21}+\frac1{25}+\frac1{27}+\frac1{105}+\frac1{135}+\frac1{225}.$$ We have several expressions in some cases : $$1=\frac13+\frac15+\frac17+\frac19+\frac1{15}+\frac1{21}+\frac1{25}+\frac1{35}+\frac1{49}+\frac1{147}+\frac1{441}+\frac1{3675}+\frac1{11025},$$ $$1=\frac13+\frac15+\frac17+\frac1{11}+\frac1{13}+\frac1{15}+\frac1{33}+\frac1{35}+\frac1{77}+\frac1{105}+\frac1{143}+\frac1{1365}+\frac1{5005},$$ $$1=\frac13+\frac15+\frac17+\frac1{11}+\frac1{13}+\frac1{15}+\frac1{21}+\frac1{33}+\frac1{91}+\frac1{3003}+\frac1{15015}.$$

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Can we express $2/3$, which is an odd fraction in your sense, as a sum of distinct odd unit fractions? –  hardmath Oct 2 '13 at 15:14
    
$\frac{2}{3}=\frac{1}{1}+\frac{1}{-3}$ Don't know if it is a licit decomposicition. –  Pocho la pantera Oct 2 '13 at 15:23
4  
$2/3 = 1/3 + 1/5 + 1/9 + 1/45$ –  Robert Israel Oct 2 '13 at 15:25

3 Answers 3

up vote 1 down vote accepted

I've just been able to prove that any odd fraction can be represented as the finite sum of different odd unit fractions.

First, note that it is sufficient to prove the following $(\star)$ :

$(\star)\ \ $ Any natural number $N$ can be represented as the finite sum of different odd unit fractions.

(If we get $(\star)$, then all we need is to divide $N$ by an odd which is the very denominator.)

Now, let $p_i$ be the $i$-th odd prime number arranged in ascending order. ($p_1=3, p_2=5, p_3=7, \cdots$) Here, we use the following two famous facts :

$(1) \ \ p_{i+1}\lt 2p_i$ (by Chebyshev)

$(2) \ \frac1{p_1}+\frac1{p_2}+\cdots$ diverges. (by Euler)

By $(2)$, we can take $n\ge 5$ such that $$\frac1{p_1}+\frac1{p_2}+\cdots+\frac1{p_n}\gt N+1.$$ Let's fix one of such $n$. Then, letting $P=p_1p_2\cdots p_n$, we know that $P\gt 3$ and that $$\left(1+\frac1{p_1}\right)\cdots\left(1+\frac1{p_n}\right)\gt 1+\left(\frac1{p_1}+\cdots+\frac1{p_n}\right)\gt 1+(N+1)\gt N+1+\frac 3P.$$

Now, let $q_j$ be the $j$-th positive divisor of $P$ arranged in ascending order: ($q_1, q_2,\cdots,q_m$ where $m=2^n$) $$q_1=1, q_2=3,q_3=5,q_4=7, q_5=11, q_6=13, q_7=15, \cdots, q_n=P.$$

Lemma 1 : $q_{i+1}\lt 2q_i$ when $i=2,\cdots,m-2$.

(Note that this lacks $i=1,m-1$ because $\frac{q_2}{q_1}=\frac{q_m}{q_{m-1}}=3\gt 2$.)

Lemma 2 : Each of $3,4,\cdots, q_1+q_3+q_4+\cdots+q_i$ can be represented as the sum of some different elements in $q_1,q_2,\cdots,q_i$ when $i=3,\cdots,m-1$.

(For example, when $i=3$, we get $3=q_2, 4=q_1+q_2, 5=q_3, 6=q_1+q_3$. When $i=4$, we get $7=q_4, 8=q_1+q_4,9=q_1+q_2+q_3, 10=q_2+q_4, 11=q_1+q_2+q_4, 12=q_3+q_4, 13=q_1+q_3+q_4.)$

(I'll prove these lemmas at the last.) Let's prove that $(\star)$ follows these lemmas. To do this, let $R=q_1+q_3+q_4+\cdots+q_{m-1}$.(Note that this lacks both of $q_2=3$ and $q_m=P$) Then, we get $$R=(q_1+q_2+\cdots+q_m)-(3+P)=(p_1+1)\cdots(p_n+1)-(3+P).$$ Hence, since we get $$\frac RP=\frac{(p_1+1)\cdots (p_n+1)}{P}-\left(\frac 3P+1\right)=\left(1+\frac1{p_1}\right)\cdots\left(1+\frac1{p_n}\right)-\left(\frac 3P+1\right)\gt\left(N+1+\frac 3P\right)-\left(\frac 3P+1\right)=N,$$ we know that $3\le NP\lt R$. By lemma 2, we can represent $NP$ as $$NP=q_{a_1}+q_{a_2}+\cdots+q_{a_k} \ \ (1\le a_1\lt aa_2\lt \cdots\lt a_k\le m-1).$$ Hence, we get $$N=\frac{q_{a_1}}{P}+\cdots+\frac{q_{a_k}}{P}.$$ Here, since each of $q_{a_i}$ is a positive divisor of $P$, each of $\frac{q_{a_i}}{P}$ is an odd unit fraction. Now the proof for $(\star)$ is completed.

In the following, we are going to prove the above lemmas.

Proof for lemma 1 : Let $_kr_i$ be the $i$-th number in ascending order which can be represented the product of $k\ (k=0,\cdots,n)$ different elements in $p_1,\cdots,p_n$.

For example, $$_kr_1=p_1p_2\cdots p_k,\ _kr_{\binom{n}{k}}=p_np_{n+1}\cdots p_{n-k+1},\ _0r_1=1, \ _nr_1=P.$$ Then, by the fact $(1)$, we know that we can get $_kr_{j+1}\lt 2 _kr_j$.

Now, let's prove that $q_{i+1}\lt 2q_i$ for $i=2,\cdots, m-2$. Note that $q_i=_kr_j$ for some $k,j$ where $1\le k\le n-1$. First, if $j\lt \binom nk$, then we know that $$q_{i+1}\le _kr_{j+1}\lt 2_kr_j=2q_i.$$ Next, if $j=\binom nk$, then we get $$q_i=p_n\times p_{n-1}\times \cdots\times p_{n-k+1}$$ where $i\le m-2$ leads $k\le n-2$ and $n-k+1\ge 3$. In the $n=5$ case, by $p_n=p_5=13$ and $p_1\times p_2=3\times 5=15$, we know $$q_{i+1}\le p_1\times p_2\times p_{n-k+1}\times p_{n-k+2}\times\cdots\times p_{n-1}=\frac{15}{13}p_{n-k+1}\times p_{n-k+2}\cdots\times p_{n-1}\times p_n=\frac{15}{13}q_i\lt 2q_i.$$ In the $n\ge 6$ cases, by $p_n\ge p_6=17\gt 15=p_1\times p_2$, we get $$_{k+1}r_1\lt _kr_{\binom nk}\lt _{k+1}r_{\binom{n}{k+1}}.$$ Hence, since we can tak $l$ such that $$_{k+1}r_{l}\lt _kr_{\binom nk}\lt _{k+1}r_{l+1},$$ we know that $$q_{i+1}\le _{k+1}r_{l+1}\lt 2 _{k+1}r_{l}\lt 2q_i$$ for $q_i= _kr_{\binom nk}.$ Now the proof for lemma 1 is completed.

Proof for lemma 2 : The $i=3$ case is obvious. We treat $l\ge 4$ in the following. First, by induction on $i$, we are going to prove $$(\star\star)\ \ 3+q_{i+1}\le q_1+q_3+q_4+\cdots+q_i+1$$ when $4\le i\le m-2$.

In $i=4$ case, we get $$LHS=3+q_5=3+11=14, RHS=q_1+q_3+q_4+1=1+5+7+1=14.$$ Now, supposing that $(\star\star)$ is true when $i=k\ (4\le k\le m-3)$, in $i=k+1$ case, by lemma 1 and inductive supposition, $LHS=3+q_{k+2}\lt 3+2q_{k+1}=(3+q_{k+1})+q_{k+1}\lt (q_1+q_3+q_4+\cdots+q_k+1)+q_{k+1}=RHS.$ Hence, the proof for $(\star\star)$ is completed.

Now, let's prove lemma 2 by induction on $i$. The $i=4$ case has been already proved, so let's suppose that lemma 2 is true when $i=k\ (4\le k\le m-2)$. We know that the numbers from $3+q_{k+1}$ to $q_1+q_3+q_4+\cdots+q_{k+1}$ can be represented as the sum of some different elements in $q_1,q_2, \cdots, q_{k+1}.$ By the way, by $(\star\star)$, since $$3+q_{k+1}\le q_1+q_3+q_4+\cdots+q_k+1,$$ as a result we know that the numbers from $3$ to $q_1+q_3+q_4+\cdots+q_{k+1}$ can be represented as the sum of some different elements in $q_1,q_2,\cdots,q_{k+1}$. So lemma 2 is true when $i=k+1$. Hence we know that lemma 2 is true for $4\le i\le m-1$. Now the proof for lemma 2 is completed.

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For Lemma 1, do you have a typo. Do you really mean: $q_{i+1} < 2q_i$? Also, is $q_i$ the same as $1$ followed by the list of odd primes? If so, why not call it out with $p$ instead of $q$. I am not clear on the advantage of calling it $q$. –  Larry Freeman Oct 5 '13 at 16:20
    
@Larry Freeman : I had a typo as you said. Thank you for pointing it out. $q_{i+1}=2q_i$ was wrong. $q_{i+1}\lt 2q_i$ is correct where $q_i$ is a positive divisor of $P$. (I forgot this) On the other hand, $p_i$ is an odd prime number, so you can see the difference; $q_7=15\not =p_7=19$. –  mathlove Oct 6 '13 at 4:07

Hint:

$$ \frac{a}{b}-\frac{1}{n}=\frac{a\,n-b}{bn} $$ Then if we can choose $n$ such that $-a< a\,n-b<a$ we are done.

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I don't get what you mean. Could you please explain more? –  mathlove Oct 2 '13 at 15:46
    
The absolute value of the numerator is decreasing. $x_1=\frac{a}{b}$ and $x_2=\frac{an-b}{bn}$ with $|an-b|<|a|$. Then in a finite number of steps you can decompose the fraction. –  Pocho la pantera Oct 2 '13 at 15:57
    
This idea works if we allow negative integers. –  Pocho la pantera Oct 2 '13 at 16:04
    
Thank you. I understood what you mean. But sadly negative integers are not allowed. –  mathlove Oct 2 '13 at 16:06

The MathWorld page you linked to gives a reference:

Each fraction $x/y$ with $y$ odd has an Egyptian fraction in which each denominator is odd (Breusch 1954; Guy 1994, p. 160).

I could not find Breusch's original paper but the Wikipedia page on Odd greedy expansion states that (note that this is not the Odd greedy expansion, which has not been proved to terminate for each $x/y$, but an example of a simpler method)

it is known that whenever $y$ is odd, every fraction $x/y$ has a representation of this type in which all the unit fractions are different from each other. For instance, such a representation can be found by replacing the fraction $x/y$ by $Ax/Ay$ where $A$ is a number of the form $35\times3^i$ for a sufficiently large $i$, and then expanding $Ax$ as a sum of divisors of $Ay$ (Breusch 1954; Stewart 1954).

Apparently, Wikipedia leaves finding the correct $i$ and the proof that this method works as an exercise.

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What I want is a proof. Anyway, thanks for nice info. –  mathlove Oct 5 '13 at 7:21

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