1
$\begingroup$

If I have two unity fractions, like $\frac{1}{12} + \frac{1}{180}$, for instance. These two fractions can be re-writen as $\frac{1}{15} + \frac{1}{45}$ or even $\frac{1}{18} + \frac{1}{30}$, which satisfies the Egyptian Fraction concept of maximising the value of the smallest fraction. Is there a formula or an algorithm that when followed, one can find out all the possible equal pairs of any unity fraction pair?

$\endgroup$

2 Answers 2

1
$\begingroup$

This might be worth something.

Let $x=p/q, p<q.$

Then $1<q/p=[q/p]+z, z<1$, where [] is the greatest integer function or floor.

Now $[q/p]<q/p\implies p/q <1/[q/p]$, but we need to add additional fractions $1/[p/q]$ to get $p/q$. This is not possible since the first term already exceeds the fraction. So we have to add 1 to the denominator as we go.

So $p/q=\frac{1}{1+[q/p]}+u/v$.

$u/v=p/q-\frac{1}{1+[q/p]}<1$ and the next term is $\frac{1}{1+[v/u]}$

$p/q=\frac{1}{1+[q/p]}+\frac{1}{1+[v/u]}+...$

This iterative process generates the Egyptian Fractions for a given number.

$\endgroup$
1
$\begingroup$

For a given rational number $\dfrac{c}{d}$, we wish to find all pairs of positive integers $(a,b)$ such that $$\dfrac{c}{d} = \dfrac{1}{a}+\dfrac{1}{b}.$$

We can manipulate the equation as follows: $$cab = da+db$$ $$cab - da-db = 0$$ $$c^2ab-cda-cdb = 0$$ $$c^2ab-cda-cdb+d^2 = d^2$$ $$(ca-d)(cb-d) = d^2.$$

Hence, $ca-d$ and $cb-d$ must be complementary factors of $d^2$. So the pairs of positive integers $(a,b)$ such that $\dfrac{c}{d} = \dfrac{1}{a}+\dfrac{1}{b}$ are all of the form $$(a,b) = \left(\dfrac{f_1+d}{c},\dfrac{f_2+d}{c}\right)$$ for all integers $f_1,f_2$ such that $d^2 = f_1f_2$ and both $\dfrac{f_1+d}{c}$ and $\dfrac{f_2+d}{c}$ are integers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.