Existence of Canonical Form of Rational Number

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Theorem

Let $r \in \Q$.

Then:

$\exists p \in \Z, q \in \Z_{>0}: r = \dfrac p q, p \perp q$

That is, every rational number can be expressed in its canonical form.


Proof

We have that the set of rational numbers is the quotient field of the set of integers.

From Divided by Positive Element of Quotient Field:

$\exists s \in \Z, t \in \Z_{>0}: r = \dfrac s t$

Now if $s \perp t$, our task is complete.

Otherwise, let:

$\gcd \left\{{s, t}\right\} = d$

where $\gcd \left\{{s, t}\right\}$ denotes the greatest common divisor of $s$ and $t$.


Let $s = p d, t = q d$.

We have that $t, d \in \Z_{>0}$

Therefore $q \in \Z_{>0}$ also.

From [[Integers Divided by GCD are Coprime]:

$p \perp q$


Also:

\(\displaystyle \frac s t\) \(=\) \(\displaystyle \frac {p d} {q d}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac p q \frac d d\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac p q 1\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac p q\) $\quad$ $\quad$

Thus:

$r = \dfrac p q$

where $p \perp q$ and $q \in \Z_{>0}$.

$\blacksquare$


Also see