Proposition 3.1. Let $a,b\in\mathbb{Z}, m_1,m_2\in\mathbb Z^+$ and $\gcd(m_1,m_2)=1$. Then $$\begin{cases} a &\equiv b \pmod{m_1} \\ a &\equiv b \pmod{m_2}\\ \end{cases} \implies a\equiv b \pmod{m_1m_2}$$.
Is there an intuitive way to think about this result? I get the proof, but am not seeing how this is "obviously true". That is, why is it that when I can write some conrguence $a\equiv b$ in two different mods, they are also the same thing in the product of their mods.
For some integers $\ell$ and $k$, by definition of congruence we have that $$\begin{array}{*{20}{c}}\begin{array}{l}a = b + {m_1}k\\a = b + {m_2}\ell \end{array}\end{array}\quad \Rightarrow \quad {m_1}k = {m_2}\ell $$ Both sides of the equation ${m_1}k = {m_2}\ell $ ought to have the same factors. Since $m_{1}$ and $m_{2}$ share no common factors, all of the factors of $m_{2}$ are contained in $k$. In other words, $k = u \cdot {m_2}$ for some integer $u$. Substituting this into the first congruence, $$a = b + u \cdot {m_1}{m_2}\quad \Rightarrow \quad a \equiv b\bmod {m_1}{m_2}$$
