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