Multiplication of Positive Number by Real Number Greater than One

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Theorem

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

Let $x > 1$.

Let $y > 0$.


Then $\dfrac y x < y$.


Proof

\(\displaystyle x\) \(<\) \(\displaystyle 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 1 x\) \(<\) \(\displaystyle 1\) Ordering of Reciprocals
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac y x\) \(<\) \(\displaystyle y\) Real Number Ordering is Compatible with Multiplication

$\blacksquare$