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I would like to prove the following: if $a_i \mid b_i$ for $i = 1,...,k$, then $lcm(a_1,...,a_k) \mid lcm(b_1,...,b_k)$.

I don't really know how to start. Can someone please help me with this?

Thank you!

Gorid
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2 Answers2

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For each $i$, $a_i \mid b_i$ and $b_i \mid lcm(b_1, \ldots, b_k)$. So that, $a_i \mid lcm(b_1, \ldots, b_k)$, for all $i$. Hence $lcm(a_1, \ldots, a_k) \mid lcm(b_1, \ldots, b_k)$ by definition of $lcm$.

Ali
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  • Thank you for your answer, but I don't quite see why $lcm(a_1,...,a_k) \mid lcm(b_1,...,b_k)$ follow from the definition of the lcm, if $a_i \mid lcm(b_1,...,b_k)$. Can you please elaborate this a little further? – Gorid Jan 08 '17 at 12:56
  • $lcm$ is the least common multiplier. So if $a \mid x$ and $b \mid x$, then $lcm(a,b) \mid x$ too. See for example this link. Does it help you? – Ali Jan 08 '17 at 13:03
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Take any prime divisor of $\text{lcm}(a_1,\dots,a_n)$ and let it be $p$. The exponent of $p$ in $\text{lcm}(a_1,\dots,a_n)$ is $\max\{a_{1_p}, \dots, a_{n_p}\}$, where $a_{i_p}$ is the exponent of $p$ in the prime decomposition of $a_i$. So WLOG let the maximum be $a_{i_p}$.

But now as $a_i \mid b_i$ we have that $b_{i_p} \ge a_{i_p}$ and namely $\max\{b_{1_p}, \dots, b_{n_p}\} \ge b_{i_p} \ge a_{i_p} \ge $ $ \max\{a_{1_p}, \dots, a_{n_p}\}$. Therefore we the exponent of any prime $p$ in the prime decomposition of $\text{lcm}(b_1,\dots,b_n)$ is always greater or equal to the exponent of $p$ in the prime decomposition of $\text{lcm}(a_1,\dots,a_n)$. Hence $\text{lcm}(a_1,\dots,a_n) \mid \text{lcm}(b_1,\dots,b_n) $

Stefan4024
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