Talk:LCM Divides Common Multiple

From ProofWiki
Jump to navigation Jump to search

Let $ak_1 = s$ and $bk_2 = s$. And let $ap_1 = m$ and $bp_2 = m$.

Then, $ap_1 = bp_2 \implies ap_1k_1 = bp_2k_1 \implies sp_1 = mk_1$.

As $m \nmid p_1$, $m \mid s$.

That last line: "As $m \nmid p_1$, $m \mid s$." Why? What if $s = 6, p_1 = 6, m = 4, k_1 = 9$? Admittedly this can't happen here, but you need more than "As $m \nmid p_1$, $m \mid s$". What am I missing? --prime mover 12:29, 19 May 2011 (CDT)