# Positive Rational Numbers are Closed under Addition

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## Theorem

Let $\Q_{\ge 0}$ denote the set of positive rational numbers:

- $\Q_{\ge 0} := \set {x \in \Q: x \ge 0}$

where $\Q$ denotes the set of rational numbers.

Then the algebraic structure $\struct {\Q_{\ge 0}, +}$ is closed in the sense that:

- $\forall a, b \in \Q_{\ge 0}: a + b \in \Q_{\ge 0}$

where $+$ denotes rational addition.

## Proof

Let $a$ and $b$ be expressed in canonical form:

- $a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$

where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.

As $\forall a, b \in \Q_{\ge 0}$ it follows that:

- $p_1, p_2 \in \Z_{\ge 0}$

By definition of rational addition:

- $\dfrac {p_1} {q_1} + \dfrac {p_2} {q_2} = \dfrac {p_1 q_2 + p_2 q_1} {q_1 q_2}$

From Integers form Ordered Integral Domain, it follows that:

\(\ds p_1 q_2\) | \(\ge\) | \(\ds 0\) | ||||||||||||

\(\ds p_2 q_1\) | \(\ge\) | \(\ds 0\) | ||||||||||||

\(\ds q_1 q_2\) | \(>\) | \(\ds 0\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds p_1 q_2 + p_2 q_1\) | \(\ge\) | \(\ds 0\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \dfrac {p_1 q_2 + p_2 q_1} {q_1 q_2}\) | \(\ge\) | \(\ds 0\) |

$\blacksquare$