Proposition: 8.04: Construction of Sequence of Numbers with Given Ratios

(Proposition 4 from Book 8 of Euclid's “Elements”)

For any multitude whatsoever of given ratios, (expressed) in the least numbers, to find the least numbers in continued proportion in these given ratio.

Let the given ratio, (expressed) in the least numbers, be the (ratios) of $A$ to $B$, and of $C$ to $D$, and, further, of $E$ to $F$.
So it is required to find the least numbers in continued proportion in the ratio of $A$ to $B$, and of $C$ to $B$, and, further, of $E$ to $F$.

fig04e

Modern Formulation

Let the proportions $\frac AB,$ $\frac CD,$ and $\frac EF$ be given. It is to find least numbers $N,O,M,P$ with $\frac AB=\frac NO,$ $\frac CD=\frac OM,$ and $\frac EF=\frac MP.$

Euclidean Algorithm to construct such a sequence of numbers

To continue the continued proportion $$\cfrac{a}{\cfrac{\vdots}{b}}$$ with the ratio $\frac cd.$ * Multiply each of the numbers in the proportion by $\frac{\operatorname{lcm}(b,c)}{b}$ * Continue the proportion with the number $\frac{d\cdot \operatorname{lcm}(b,c)}{c}.$

Examples

To take $\cfrac{4}{5}$ and continue it by $\cfrac 23.$ * Factor to multiply all numbers with: $\frac{\operatorname{lcm}(5,2)}{5}=\frac{10}{5}=2.$ * It follows $\cfrac{4\cdot 2}{5\cdot 2}=\cfrac{8}{10}.$ * New number to continue with: $\frac{3\cdot \operatorname{lcm}(5,2)}{2}=\frac{3\cdot 10}{2}=15.$ * New continued proportion: $$\cfrac{8}{\cfrac{10}{15}}.$$

To take $\cfrac{8}{\cfrac{10}{15}}$ and continue it by $\cfrac 47.$ * Factor to multiply all numbers with: $\frac{\operatorname{lcm}(15,4)}{15}=\frac{60}{15}=4.$ * It follows $$\cfrac{8\cdot 4}{\cfrac{10\cdot 4}{15\cdot 4}}=\cfrac{32}{\cfrac{40}{60}}.$$ * New number to continue with: $\frac{7\cdot \operatorname{lcm}(15,4)}{4}=\frac{7\cdot 60}{4}=105.$ * New continued proportion: $$\cfrac{32}{\cfrac{40}{\cfrac{60}{105}}}.$$

Proofs: 1

Proofs: 1 2

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