Is it true that $$a\equiv b\pmod{m}\implies\frac{a}{n}\equiv\frac{b}{n} \pmod{\frac{m}{n}},$$ where $a, b, m, n, \frac{a}{n}, \frac{b}{n}, \frac{m}{n}\in\mathbb{N}$? If so, how do I prove it?
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2Note that this is an essential property for solving modulus equations. For instance $6x\equiv 4 \pmod{10}$. Since $6$ is not invertible modulo $10$ we are stuck. But dividing by $2$ we get $3x\equiv 2\pmod 5$ and now $3$ is invertible and we get $x\equiv 4\mod 5$. – zwim Jan 19 '18 at 05:34
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If $a\equiv b \pmod{m}$ then there is $q\in\mathbb{Z}$ such that $a=b+qm$ which implies that, for $n\not=0$, $$\frac{a}{n}=\frac{b}{n}+q\frac{m}{n}.$$ So, when $\frac{a}{n},\frac{b}{n},\frac{m}{n}\in\mathbb{Z}$, we may conclude that $\frac{a}{n}\equiv\frac{b}{n}\pmod{\frac{m}{n}}$.
Robert Z
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$$a\equiv b\pmod m$$ means $$\frac{a-b}m\in\Bbb Z.$$ $$\frac{a}n\equiv \frac bn\pmod{\frac mn}$$ means $$\frac{a/n-b/n}{m/n}\in\Bbb Z.$$
Angina Seng
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From the definition of modulo:
$$a \equiv b \pmod m \implies a=km+b \tag{1}$$ $$\frac{a}{n} \equiv {\frac{b}{n}} \pmod {\frac{m}{n}} \implies \frac{a}{n}=k \frac{m}{n} + \frac{b}{n} \tag{2}$$
Observe $(2)$.
\begin{align} \frac{a}{n}&=k \frac{m}{n} + \frac{b}{n} \tag {2} \\ \frac{a}{n} \cdot n&=k \frac{m}{n} \cdot n + \frac{b}{n} \cdot n \\ a & = km+b \tag{1} \end{align}
By multiplying $(2)$ by $n$, we get $(1)$, so the expressions are the same. Hence $a \equiv b \pmod m \implies \frac{a}{n} \equiv \frac{b}{n} \pmod{ \frac{m}{n}}$
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