So the goal is to prove that the $\operatorname{lcm}(a,b)$ divides any multiple of of $a$ and $b$. Suppose there is some integer $c$ such that $a|c$ and $b|c$ but I want to prove $\operatorname{lcm}(a,b)|c$ also. I got that $$\operatorname{lcm}(a,b)=\frac{(a\cdot b)}{\gcd(a,b)}$$ and I want to see how we could show $$\frac{(a*b)}{\gcd(a,b)}\bigg| \, c$$ Any help would be appreciated thank you.
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1It's the direction $(\Rightarrow)$ in the equiavlences in this answer in the first linked dupe. See the others for background and motivation.. – Bill Dubuque Apr 27 '20 at 18:08
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Please don't change the question at this point. If you have a related question then pose a new question. – Bill Dubuque Apr 28 '20 at 01:18
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WARNING: This answer is INCORRECT. See comments for what mistakes I am making
Firstly we know that some multiple of $a$ and $b$ can be represented by $kab$ where $k$ is some integer, and thus is divisible by $ab$.
We need to show that $lcm(a,b) \vert ab$. Since $gcd(a,b)$ would be some integer, and that $lcm(a,b) gcd(a,b) = ab$, it is obvious.
Therefore any multiple of $ab$ is divisible by $lcm(a,b)$.
user12986714
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I wonder, why it is not true that $ a \vert c $ and $ b \vert c $ means $ (ab) \vert c$ and therefore $c = kab$ ? (I know that maybe gcd(a,b) is not 1 but c is some multiple of a and b) – user12986714 Apr 27 '20 at 22:39
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1e.g. $,2,4\mid 4,$ but not $,2\cdot 4\mid 4.\ $ Your claim is true $\iff \gcd(a,b) = 1\ \ $ – Bill Dubuque Apr 27 '20 at 22:44
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Thanks. I thought that multiple of a and b means that $c = k a ^m b^n$ where k,m,n are some integers...... – user12986714 Apr 27 '20 at 22:52