Product of Positive Element and Element Greater than One

Theorem
Let $\left({R,+,\cdot,\le}\right)$ be an ordered ring with unity.

Let $x,y \in R$.

Suppose that $x > 0$ and $y > 1$.

Then $x \cdot y > x$.

Proof
By the definition of an ordering compatible with a ring, $\ge$ is compatible with $+$.

By the definition of a relation compatible with an operation, the premise that $y > 1$ implies that $y + (-1) > 1+(-1)$.

Thus $y - 1 > 0$.

Since $x>0$, Properties of Ordered Ring (property 6) implies that $x \cdot (y - 1) > x \cdot 0$.

By Ring Product with Zero, $x \cdot 0 = 0$.

Thus $x \cdot (y-1) > 0$.

By the definition of a ring, $\cdot$ is distributive over $+$, so $x \cdot y - x \cdot 1 > 0$.

By the definition of the unity of a ring, $x \cdot 1 = x$, so

$x \cdot y - x > 0$.

Then by the definition of an ordering compatible with a ring, $(x \cdot y - x) + x > 0 + x$.

Since $+$ is associative, $(-x)+x = 0$, and $0 + x = x$,
 * $x \cdot y > x$.