Quotients of 3 Unequal Numbers are Unequal

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Theorem

Let $x, y, z \in \R_{\ne 0}$ be non-zero real numbers which are not all equal.

Then $\dfrac x y, \dfrac y z, \dfrac z x$ are also not all equal.


Proof

Aiming for a contradiction, suppose $\dfrac x y = \dfrac y z = \dfrac z x$.

\(\displaystyle \) \(\) \(\displaystyle \dfrac x y = \dfrac y z = \dfrac z x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(\) \(\displaystyle x^2 z = y^2 x = z^2 y\) multiplying top and bottom by $x y z$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(\) \(\displaystyle x z = y^2, x y = z^2, y z = x^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(\) \(\displaystyle x y z = y^3, x y z = z^3, z y z = x^3\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(\) \(\displaystyle x^3 = y^3 = z^3\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(\) \(\displaystyle x = y = z\)


This contradicts the assertion that $x, y, z$ are all unequal.

Hence the result by Proof by Contradiction.

$\blacksquare$


Sources