Set of Rational Numbers whose Numerator Divisible by p is Closed under Addition

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Theorem

Let $p$ be a prime number.

Let $A_p$ be the set of all rational numbers which, when expressed in canonical form has a numerator which is divisible by $p$.

Then $A_p$ is closed under rational addition.


Proof

Let $a, b \in A_p$.


Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:

$n_1, n_2 \in \Z$
$d_1, d_2 \in \Z_{>0}$
$p n_1 \perp d_1, p n_2 \perp d_2$


Then:

\(\ds a + b\) \(=\) \(\ds \frac {p n_1} {d_1} + \frac {p n_2} {d_2}\)
\(\ds \) \(=\) \(\ds \frac {p n_1 d_2 + p n_2 d_1} {d_1 d_2}\) Definition of Rational Addition
\(\ds \) \(=\) \(\ds \frac {p \paren {n_1 d_2 + n_2 d_1} } {d_1 d_2}\)

From Euclid's Lemma for Prime Divisors, if $p \divides d_1 d_2$ then either $p \divides d_1$ or $p \divides d_2$.

But neither of these is the case, so $p \nmid d_1 d_2$.

Hence by Prime not Divisor implies Coprime:

$p \perp d_1 d_2$

where $\perp$ denotes coprimeness.


So when $\dfrac {p \paren {n_1 d_2 + n_2 d_1} } {d_1 d_2}$ is expressed in canonical form, $p$ will still be a divisor of the numerator.

Hence the result.

$\blacksquare$


Sources