-1
$\begingroup$

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!

2 Answers 2

1

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 $lcm$.

  • 0
    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? – 2017-01-08
  • 0
    $lcm$ 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](http://math.stackexchange.com/questions/322136/if-a-b-are-integers-such-that-a-mid-x-and-b-mid-x-must-mathrmlcma?rq=1). Does it help you? – 2017-01-08
0

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) $