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In ancient Egypt, fractions were written as sums of fractions with numerator 1. For instance,$ \frac{3}{5}=\frac{1}{2}+\frac{1}{10}$. Consider the following algorithm for writing a fraction $\frac{m}{n}$ in this form$(1\leq m < n)$: write the fraction $\frac{1}{\lceil n/m\rceil}$ , calculate the fraction $\frac{m}{n}-\frac{1}{\lceil n/m \rceil}$ , and if it is nonzero repeat the same step. Prove that this algorithm always finishes in a finite number of steps.

Note:if $ n\in \mathbf{Z} $ and $n-1<x\leq n , \lceil x\rceil=n$

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    $\begingroup$ Hint: consider the sequence of numerators. $\endgroup$ – Daniel Fischer Aug 2 '13 at 19:50
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    $\begingroup$ Note that because the sum of unit fractions is unbounded, one can hit any positive number in this way, and also miss out any finite initial segment of fractions and still succeed! $\endgroup$ – Mark Bennet Aug 2 '13 at 20:49
  • $\begingroup$ Also, the (greedy) algorithm gives distinct denominators, which requires a little bitty extra proof. Oh, distinct because they are strictly increasing... $\endgroup$ – Will Jagy Aug 3 '13 at 4:25
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We proceed by induction on $m\in \Bbb{N}$.

Base Case: When $m=1$, $\dfrac{m}{n}$ is already in the desired form, so terminate.

Induction Hypothesis: Assume that for all $m \in \{1,...,k\}$ (where $k \geq 1$), the algorithm will terminate when its input is $\dfrac{m}{n}$.

It remains to prove the claim true for $m=k+1$. After one step of the algorithm, we obtain the new fraction: $$ \dfrac{k+1}{n}-\dfrac{1}{\left\lceil \frac{n}{k+1} \right\rceil} = \dfrac{\left\lceil \frac{n}{k+1} \right\rceil(k+1)-n}{\left\lceil \frac{n}{k+1} \right\rceil n} $$

Now observe that: $$ \begin{align*} \left\lceil \frac{n}{k+1} \right\rceil(k+1)-n &= \left( \left\lfloor \frac{n-1}{k+1} \right\rfloor+1 \right)(k+1) -n \\ &\leq \left( \frac{n-1}{k+1} +1 \right)(k+1) -n \\ &= (n-1)+(k+1) -n \\ &= k \end{align*} $$ Hence, by the induction hypothesis, the algorithm must terminate. This completes the induction.

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    $\begingroup$ In my opinion: For homework-type questions, you should give a hint and then wait a while. (Say, 24 hours or until the OP responds.) $\endgroup$ – GEdgar Aug 2 '13 at 20:44
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HINT: Use the division algorithm to write $n=mq+r$, where $q$ and $r$ are non-negative integers, and $0\le r<m$. I’ll leave the case $r=0$ to you. If $r>0$, then $$\frac1{\lceil n/m\rceil}=\frac1{q+1}\;,$$ so

$$\frac{m}n-\frac{1}{\lceil n/m\rceil}=\frac{m}{mq+r}-\frac1{q+1}=\frac{m-r}{(mq+r)(q+1)}\;.$$

Let this fraction be $\dfrac{a}b$ in lowest terms. Then $a\le m-r<m$.

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One can always show that any fraction can be written as a sum of finite fractions, by the factorial rule. That is for $\frac mn$, if $n \mid f!$ then any $\frac mn$ can be written in no more than $f-1$ steps.

For the greedy algorithm, a fraction $\frac mn$, for being greater than $\frac 1a$, will leave a difference $\frac {am-n}{na}$

Suppose you have $\frac 1c \le \frac mn \le \frac 1{c-1}$. Multiply the limits by $\frac mm$ to get the denominators $cm > n > cm-c$, and thence put $n = cm-x$, where $x \lt m$.

Then subtract $\frac 1c$ to get $\frac {cm}{c(cm-x)} - \frac {cm-x}{c(cm-x)} = \frac x{c(cm-x)}$. Since $x<m$, each iteration produces a smaller $x$, which must ultimately divide the denominator.

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