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!
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2$\begingroup$ ${1\over5}={6\cdot1\over6\cdot5}$ and ${1\over6}={5\cdot1\over5\cdot6}$. $\endgroup$David Mitra– David Mitra2015-02-01 14:49:20 +00:00Commented Feb 1, 2015 at 14:49
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1 Answer
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Let $m, n$ be non-zero integers. Then the sum of the unit fractions $1/m$ and $1/n$ is $$\frac{1}{m} + \frac{1}{n} = \frac{1}{m} \cdot \frac{n}{n} + \frac{1}{n} \cdot \frac{m}{m} = \frac{n}{mn} + \frac{m}{mn} = \frac{m + n}{mn}$$
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$\begingroup$ Let me work that out... $\endgroup$ReliableMathBoy– ReliableMathBoy2015-02-01 20:31:23 +00:00Commented Feb 1, 2015 at 20:31
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$\begingroup$ Okay, done! Sorry for the wait. $\endgroup$ReliableMathBoy– ReliableMathBoy2015-02-01 21:54:34 +00:00Commented Feb 1, 2015 at 21:54