How do we conclude that $[m_1, m_2] \mid a-b$??

Asked Apr 29 '14 at 19:26

Active Apr 29 '14 at 19:26

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$(m_1,m_2)=1, a \equiv b \pmod {m_1} , a \equiv b \pmod {m_2} \Leftrightarrow a \equiv b \pmod {m_1 \cdot m_2}$

The proof of the direction $" \Rightarrow "$ is the following:

$a \equiv b \pmod {m_1} , a \equiv b \pmod {m_2}$

Then $m_1 \mid a-b, m_2 \mid a-b$

So $[m_1, m_2] \mid a-b$

But $[m_1, m_2]=\frac{m_1 \cdot m_2}{(m_1, m_2)}=m_1 \cdot m_2$

So $m_1 \cdot m_2 \mid a-b$, so $a \equiv b \pmod {m_1 \cdot m_2}$

$$$$ Could you explain me how we conclude from $"m_1 \mid a-b, m_2 \mid a-b "$ that $"[m_1, m_2] \mid a-b"$??

asked Apr 29 '14 at 19:26

Mary Star

2

$m,n\mid c,\Rightarrow, {\rm lcm}(m,n)\mid c,\ $ e.g. see here. – Bill Dubuque Apr 29 '14 at 19:32
1

You might also skip a couple of steps. Since $(m_1,m_2)=1$, all prime factors in $m_1$ are not in $m_2$ and vice versa. Since all those prime factors are in $a-b$, it follows that $m_1 \cdot m_2 | a-b$. – Klaas van Aarsen Apr 29 '14 at 20:50
I got it...Thank you very much!!! :-) – Mary Star May 12 '14 at 22:19

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