Strictly Positive Real Numbers are Closed under Multiplication/Proof 2

Proof
Let $b > 0$.

From :
 * $a > c \implies a \times b > c \times b$

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