Multiplication of Positive Number by Real Number Greater than One

Theorem

Let $x$ and $y$ be real numbers.

Let $x > 1$.

Let $y > 0$.

Then $\dfrac y x < y$.

Proof

\(\ds x\)

\(<\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds \frac 1 x\)

\(<\)

\(\ds 1\)

Ordering of Reciprocals

\(\ds \leadsto \ \ \)

\(\ds \frac y x\)

\(<\)

\(\ds y\)

Real Number Ordering is Compatible with Multiplication

$\blacksquare$