# 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

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $3$