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
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 2.4$: The rational numbers and some finite fields: Exercise $6$