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Let $r_1 < r_2$ be two positive rational fractions, both $> 1$ and both in their lowest terms i.e. the numerator and denominator of each fraction have no common factors. If $r_3 = r_1 + r_2$ is in its lowest terms, under what conditions will the denominator of $r_3$ be not divisible by the denominator of $r_1$.

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When a prime factor shared by both denominator is canceled out by the numerator to it's highest power. This also requires denominators to not be both 0.

If $r_1 = \frac{a}{b}$ and $r_2 = \frac{c}{d}$ where $b,d \in Z^{+} \backslash \{1\} $ and $a,c \in Z^{+} $

$$r_3 = \frac{ad+cb}{bd}$$

If there exist $p^k$, such that $p^k \mid ad+cb$ and $p^k \mid bd$ and $p \nmid \frac{bd}{p^k}$

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  • $\begingroup$ Can you give an example of this with $a,b,c,d > 1$ with $(a,b) = 1$ and $(c,d) = 1$? $\endgroup$ – Stacker Oct 7 '18 at 7:35
  • $\begingroup$ Unfortunately no, this is never true when either denominator is 1, as 2+3= 5, they all have a denominator 1 and 1 is divisible by 1. If any 1 denominator is 1, it's a similar case. I.e. 2.5+3 = 5.5. The denominator is 2 in 2.5 and 5.5. $\endgroup$ – Neo Oct 7 '18 at 9:20

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