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
\(\ds x\) | \(<\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 x\) | \(<\) | \(\ds 1\) | Ordering of Reciprocals | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac y x\) | \(<\) | \(\ds y\) | Real Number Ordering is Compatible with Multiplication |
$\blacksquare$