# 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:

\(\displaystyle \frac s t\) | \(=\) | \(\displaystyle \frac {p d} {q d}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac p q \frac d d\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac p q 1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac p q\) |

Thus:

- $r = \dfrac p q$

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

$\blacksquare$