Congruence of Quotient

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Theorem

Let $a, b \in \Z$ and $n \in \N$.

Let $a$ be congruent to $b$ modulo $n$, that is $a \equiv b \pmod n$.

Let $d \in \Z: d > 0$ such that $d$ is a common divisor of $a, b$ and $n$.


Then:

$\dfrac a d \equiv \dfrac b d \pmod {n / d}$


Proof 1

By definition of congruence modulo $n$:

$a = b + k n$

Dividing through by $d$ (which you can do because $d$ divides all three terms), we get:

$\dfrac a d = \dfrac b d + k \dfrac n d$

from which the result follows directly.

$\blacksquare$


Proof 2

From Congruence by Product of Moduli, we have that:

$\dfrac a d \equiv \dfrac b d \pmod {n / d} \iff d \dfrac a d \equiv d \dfrac b d \pmod {n / d} \iff a \equiv b \pmod n$

$\blacksquare$


Examples

Example: $8 \equiv 18 \pmod {10}$

\(\ds 8\) \(\equiv\) \(\ds 18\) \(\ds \pmod {10}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 8 2\) \(\equiv\) \(\ds \dfrac {18} 2\) \(\ds \pmod {10 / 2}\)
\(\ds \leadsto \ \ \) \(\ds 4\) \(\equiv\) \(\ds 9\) \(\ds \pmod 5\)


Example: $6 \equiv 36 \pmod {15}$

\(\ds 6\) \(\equiv\) \(\ds 36\) \(\ds \pmod {15}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 6 3\) \(\equiv\) \(\ds \dfrac {36} 3\) \(\ds \pmod {15 / 3}\)
\(\ds \leadsto \ \ \) \(\ds 2\) \(\equiv\) \(\ds 12\) \(\ds \pmod 5\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(\equiv\) \(\ds 6\) \(\ds \pmod 5\)


Warning

Note that in general it is not the case that:

\(\ds a\) \(\equiv\) \(\ds b\) \(\ds \pmod m\)
\(\ds c\) \(\equiv\) \(\ds d\) \(\ds \pmod m\)
\(\ds \leadsto \ \ \) \(\ds \dfrac a c\) \(\equiv\) \(\ds \dfrac b d\)


For example:

\(\ds 8\) \(\equiv\) \(\ds 18\) \(\ds \pmod m\)
\(\ds 27\) \(\equiv\) \(\ds 7\) \(\ds \pmod m\)

But we may not conclude that:

$\dfrac 8 {27} \pmod {10} = \dfrac {18} 7 \pmod {10}$

We may not even conclude that:

$\dfrac 8 2 \pmod {10} = \dfrac {18} 2 \pmod {10}$

because:

\(\ds \dfrac 8 2 \pmod {10}\) \(\equiv\) \(\ds 4 \pmod {10}\)
\(\ds \dfrac {18} 2 \pmod {10}\) \(\equiv\) \(\ds 9 \pmod {10}\)
\(\ds \) \(\not \equiv\) \(\ds 4 \pmod {10}\)


But we do have:

\(\ds \dfrac 8 2 \pmod {10 / 2}\) \(\equiv\) \(\ds 4 \pmod 5\)
\(\ds \dfrac {18} 2 \pmod {10 / 2}\) \(\equiv\) \(\ds 9 \pmod 5\)
\(\ds \) \(\equiv\) \(\ds 4 \pmod 5\)


Sources

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