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The answer at Show that $x \equiv 0$ says that $a|x$ and $b|x$ implies $\operatorname{lcm}(a,b)|x$. I can see that if $a=\prod p_i^{e_i}$ and $b=\prod p_i^{f_i}$ where $p_i$ is the i-th prime, and $\operatorname{lcm}(a,b)=\prod p_i^{max(e_i,f_i)}$ then this kind of shows the result because $x$ has to have prime powers at least $\max(e_i,f_i)$. Is there a way to more directly and/or explicitly motivate this conclusion?

gt6989b
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Anna Naden
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  • By definition of $\text{lcm}(a, b) = L$, there are $k, r$ such that $ka = L = rb$. Also since $a|x $ and $b| x$, there are $s, t$ such that $as = x = bt$. But by the definition of $L$ as the least number with such a property, it must be the case that $L|x$. – Daniel Akech Thiong Aug 27 '20 at 03:10
  • @akech: lcm= least common multiple which means "the smallest common multiple". So your "it must be the case,,," needs a proof and that is what OP asked. For the same reason, the question which is designated as duplicate does not have proper answers for this question. – markvs Aug 27 '20 at 03:18
  • You have a point. – Daniel Akech Thiong Aug 27 '20 at 03:23
  • @JCAA Given that my answer in the dupe is essentially the same as yours here, your claim that none of them are "proper answers" is puzzling. Did you even read the answers? – Bill Dubuque Aug 27 '20 at 03:27
  • @Gone; Your answer is 28 lines long and contains a lot of irrelevant information. Are you sure this is the same? – markvs Aug 27 '20 at 03:35
  • @JCAA The proof is one line longer by design - to highlight the Euclidean descent (in simpler subtractive form). The rest adds insight lacking in your answer (where the proof is pulled out of a hat like magic). Since when is a "longer" (more insightful) answer not a "proper" answer? – Bill Dubuque Aug 27 '20 at 03:36
  • @Gone: You claimed that your answer is essentially the same as mine. That is false. Euclidean descent has nothing to do with OP. – markvs Aug 27 '20 at 03:39
  • @JCAA It most certainly does (as you'd learn if you bothered to read the remark in my answer - assuming you know some ideal theory). – Bill Dubuque Aug 27 '20 at 03:40
  • @Gone: I know "some ideal theory". I just do not like bragging about it in answers to elementary questions. – markvs Aug 27 '20 at 03:43
  • @JCAA If you seriously (wrongly) believe that it is "bragging" to make conceptual remarks about how proofs naturally generalize then our viewpoints about pedagogy are so disparate that there is no point in continuing. – Bill Dubuque Aug 27 '20 at 03:48

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Recall: $lcm(a,b)$ is the smallest number divisible by $a$ and $b$. If $lcm(a,b)$ does not divide $x$ we can divide with a remainder: $x=p\cdot lcm(a,b)+r$ where $r<lcm(a,b)$. But then both $a$ and $b$ divide $r$, so $r\ge lcm(a,b)$, a contradiction.

markvs
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