Question

Chứng minh

Answer 1

\(a\equiv b\left(modm\right)\Leftrightarrow a-b=k.m\) với k nguyên

Mặt khác do d là ước chung của a,b,m nên: \(\left\{{}\begin{matrix}a=d.x\\b=d.y\\m=d.z\end{matrix}\right.\) với x;y;z nguyên

\(\Rightarrow x=\dfrac{a}{d};y=\dfrac{b}{d};z=\dfrac{m}{d}\)

\(\Rightarrow\dfrac{a}{d}-\dfrac{b}{d}=\dfrac{a-b}{d}=\dfrac{k.m}{d}=\dfrac{k.d.z}{d}=k.z\)

\(\Rightarrow\dfrac{a}{d}\equiv\dfrac{b}{d}\left(modz\right)\)  hay \(\dfrac{a}{d}\equiv\dfrac{b}{d}\left(mod\dfrac{m}{d}\right)\)