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 Axioms: $\R \text O 2$: 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:

$a, b > 0 \implies a \times b > 0$

$\blacksquare$


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