Existence of Canonical Form of Rational Number

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 field of quotients of the set of integers.

From Divided by Positive Element of Field of Quotients:

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

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

Otherwise, let:

$\gcd \set {s, t} = d$

where $\gcd \set {s, t}$ 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:

\(\ds \frac s t\)

\(=\)

\(\ds \frac {p d} {q d}\)

\(\ds \)

\(=\)

\(\ds \frac p q \frac d d\)

\(\ds \)

\(=\)

\(\ds \frac p q 1\)

\(\ds \)

\(=\)

\(\ds \frac p q\)

Thus:

$r = \dfrac p q$

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

$\blacksquare$

Also see

• Canonical Form of Rational Number is Unique

Sources

• 1971: Wilfred Kaplan and Donald J. Lewis: Calculus and Linear Algebra ... (previous) ... (next): Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-1}$: The Real Numbers