0
$\begingroup$

I found out that the denominators of two fractions with numerators $1$ add up to a third fraction that has the sum in the numerator and the product in the denominator. For example, ${1\over 5}+{1\over 6}={11\over 30}$ because one, $0.2+0.1\overline6=0.3\overline6$, which convert to the main fractions, and two, the denominators with numerators $1$ add up to a third fraction having a sum in the numerator and evaluates to the product in the denominator. Why does this always happen? This is mind-blown for me!

$\endgroup$
1
  • 2
    $\begingroup$ ${1\over5}={6\cdot1\over6\cdot5}$ and ${1\over6}={5\cdot1\over5\cdot6}$. $\endgroup$ Feb 1, 2015 at 14:49

1 Answer 1

0
$\begingroup$

Let $m, n$ be non-zero integers. Then the sum of the unit fractions $1/m$ and $1/n$ is $$\frac{1}{m} + \frac{1}{n} = \frac{1}{m} \cdot \frac{n}{n} + \frac{1}{n} \cdot \frac{m}{m} = \frac{n}{mn} + \frac{m}{mn} = \frac{m + n}{mn}$$

$\endgroup$
2
  • $\begingroup$ Let me work that out... $\endgroup$ Feb 1, 2015 at 20:31
  • $\begingroup$ Okay, done! Sorry for the wait. $\endgroup$ Feb 1, 2015 at 21:54

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.