MHB Why is the implication obvious?

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The discussion centers on proving that the least common multiple (LCM) of two non-zero integers, a and b, shares the same common multiples as a and b themselves. The proof demonstrates that if a divides m and b divides m, then m can be expressed in terms of the LCM, leading to the conclusion that the remainder r must be zero. Consequently, this implies that the LCM divides m, which in turn confirms that both a and b must also divide m. The final implication is considered obvious because the properties of divisibility ensure that if the LCM divides m, then both a and b must divide m as well. The discussion concludes with an affirmation of understanding regarding the implications of these mathematical relationships.
evinda
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Hello! (Smile)

I am looking at the proof of the follwing sentence:

Let $a,b \neq 0$.
The common multiples of $a,b$ are the same as the multiples of $[a,b]$, where $[a,b]$ is the least common multiple of $a \text{ and } b$.

  • Let $a \mid m, b \mid m$

    $$m=q \cdot [a,b]+r , \ 0 \leq r < [a,b] (1) $$

    $$a \mid m, a \mid [a,b] \Rightarrow a \mid r$$

    $$b \mid m, b \mid [a,b] \Rightarrow b \mid r$$

    So, $r$ is a common multiple of $a,b$.

    As $[a,b]$ is the least common multiple of $a,b$, we conclude from the relation $(1)$ that $r=0$,so $m=q \cdot [a,b]$ and so we conclude that $[a,b] \mid m$
  • $$[a,b] \mid m \Rightarrow a \mid m, b \mid m$$

    According to my notes the last implication is obvious. But...why is it like that? :confused: (Thinking)
 
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We know that $a \mid [a,b]$ and $b \mid [a,b]$ since $[a,b]$ is a common multiple of both $a$ and $b$. So if $[a,b] \mid m$, it follows immediately that $a \mid m$ and $b \mid m$. (I.e., if $a \mid b$ and $b \mid c$, then $a \mid c$.)
 
magneto said:
We know that $a \mid [a,b]$ and $b \mid [a,b]$ since $[a,b]$ is a common multiple of both $a$ and $b$. So if $[a,b] \mid m$, it follows immediately that $a \mid m$ and $b \mid m$. (I.e., if $a \mid b$ and $b \mid c$, then $a \mid c$.)

Oh yes,right! Thank you very much! (Nerd) :)
 
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