Strictly Positive Rational Numbers are Closed under Addition

Theorem

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

$\Q_{>0} := \set {x \in \Q: x > 0}$

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

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

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

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 $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.

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:

\(\ds p_1 q_2\)

\(>\)

\(\ds 0\)

\(\ds p_2 q_1\)

\(>\)

\(\ds 0\)

\(\ds q_1 q_2\)

\(>\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds p_1 q_2 + p_2 q_1\)

\(>\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {p_1 q_2 + p_2 q_1} {q_1 q_2}\)

\(>\)

\(\ds 0\)

$\blacksquare$

Sources

• 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Introduction