If four magnitudes are proportional then they will also be proportional alternately. * Let $A$, $B$, $C$ and $D$ be four proportional magnitudes, (such that) as $A$ (is) to $B$, so $C$ (is) to $D$. * I say that they will also be [proportional] alternately, (so that) as $A$ (is) to $C$, so $B$ (is) to $D$.
In modern notation, this proposition reads that if \[\frac\alpha\beta=\frac\gamma\delta,\] then \[\frac\alpha\gamma=\frac\beta\delta,\]
for all positive real numbers \(\alpha,\beta,\gamma,\delta\).
This is a special case of the fact that the multiplication of real numbers is cancellative. Since for $\frac\beta\gamma\neq 0$:
$$\frac\alpha\beta=\frac\gamma\delta\Leftrightarrow \frac \beta\gamma \frac\alpha\beta=\frac \beta\gamma \frac\gamma\delta.$$
Proofs: 1
Proofs: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
Sections: 27
