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Let $a_1, a_2, b_1, b_2$ be positive integers such that $\gcd(a_1, b_1) = \gcd(a_2, b_2) = 1$.

It appears to me based on some intuition and numerical evidence that the (reduced) denominator of $\displaystyle \frac{a_1}{b_1} + \frac{a_2}{b_2} = \frac{a_1b_2 + a_2b_1}{b_1b_2}$ is bounded below by $\displaystyle\frac{\operatorname{lcm}(b_1,b_2)}{\gcd(b_1,b_2)}$. I reckon that there's a simple way to prove this, but I don't see it immediately.

The bound on the other side is clear to me: the (reduced) denominator of $\displaystyle \frac{a_1}{b_1} + \frac{a_2}{b_2} = \frac{a_1b_2 + a_2b_1}{b_1b_2}$ is bounded above by $\operatorname{lcm}(b_1,b_2)$, because

  • $b_1b_2 = \operatorname{lcm}(b_1,b_2)\gcd(b_1,b_2)$, and
  • $\gcd(b_1,b_2)$ divides $a_1b_2 + a_2b_1$ since it divides both $b_1$ and $b_2$,

so $$ \frac{a_1b_2 + a_2b_1}{b_1b_2} = \frac{a_1b_2 + a_2b_1}{\operatorname{lcm}(b_1,b_2)\gcd(b_1,b_2)} = \frac{a_1b'_2 + a_2b'_1}{\operatorname{lcm}(b_1,b_2)}, $$ where $b'_1 = \frac{b_1}{\gcd(b_1,b_2)}$ and $b'_2 = \frac{b_2}{\gcd(b_1,b_2)}$.

Peter Kagey
  • 5,052
  • The reduced denominator is not bounded above by $\mathrm{lcm}(b_1,b_2)$, it is bounded below by it. That's why it's sometimes called the "lowest common denominator" too. – anon Nov 09 '20 at 22:48
  • $\frac 12 + \frac 12$ has a denominator of $1$, and $1 < \operatorname{lcm}(2,2) = 2$, so the LCM appears to an upper bound. – Peter Kagey Nov 09 '20 at 22:54

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