Strictly Positive Real Numbers are Closed under Multiplication
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Theorem
The set $\R_{>0}$ of strictly positive real numbers is closed under multiplication:
- $\forall a, b \in \R_{> 0}: a \times b \in \R_{> 0}$
Proof 1
Let $a, b \in \R_{> 0}$
We have that the Real Numbers form Ordered Integral Domain.
It follows from Positive Elements of Ordered Ring that:
- $a \times b \in \R_{> 0}$.
$\blacksquare$
Proof 2
Let $b > 0$.
From Real Number Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication:
- $a > c \implies a \times b > c \times b$
Thus setting $c = 0$:
- $a > 0 \implies a \times b > 0 \times b$
But from Real Zero is Zero Element:
- $0 \times b = 0$
Hence the result:
- $a, b > 0 \implies a \times b > 0$
$\blacksquare$
Also see
Sources
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(b)}$