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I need help proving the following two formulas:

$$\gcd\left(\frac{a_1}{b_1}, \frac{a_2}{b_2}, \dots, \frac{a_n}{b_n}\right) = \frac{\gcd(a_1, a_2, \dots, a_n)}{\operatorname{lcm}(b_1, b_2, \dots, b_n)},$$

$$\operatorname{lcm}\left(\frac{a_1}{b_1}, \frac{a_2}{b_2}, \dots, \frac{a_n}{b_n}\right) = \frac{\operatorname{lcm}(a_1, a_2, \dots, a_n)}{\gcd(b_1, b_2, \dots, b_n)}.$$

where $\gcd(a_i, b_i) = 1$.

I've managed to prove them for $n = 2$, but can't find the way to do it in more general case.

To be clarify, I assume that $A$ is a divisor of $B$, and that $B$ is a multiple of $A$, if $B/A$ is an integer, where $A$ and $B$ are non-zero rational numbers (e.g. $1/14$ is a divisor of $3/7$, and $3/7$ is a multiple of $1/14$, because $3/7 \div 1/14 = 6$ is an integer).

Bill Dubuque
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  • Use the fact that there exists a unique prime factorisation. When you proofed it for $n=2$, you can also just use induction. – Vincent Batens Jan 14 '24 at 21:23
  • It's obvious for a single prime, so proceed as per the standard argument. – Calvin Lin Jan 14 '24 at 21:35
  • The $n$-ary case follows immediately from the binary case using induction (and associativity of gcd & lcm). Or extend the proof of the binary case to the $n$-ary case (e.g. as here in the 3rd dupe). – Bill Dubuque Jan 15 '24 at 00:18

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