Elements of Algebra

160 ELEMENTS SECT. III. 489. A relation continuing always the same, when we divide both its terms by the same number, in order to abridge it, we may greatly facilitate the above composition by comparing the antecedents and the consequents, for the purpose of making such reductions as we performed in the last chapter. For example, we find the compound relation of the following given relations thus: Relations given. $12 : 25, 28 : 33$, and $55 : 56$. Which becomes $(12 \cdot 28 \cdot 55) : (25 \cdot 33 \cdot 56) = 2 : 5$ by cancelling the common divisors. So that $2 : 5$ is the compound relation required. 490. The same operation is to be performed, when it is required to calculate generally by letters; and the most remarkable case is that in which each antecedent is equal to the consequent of the preceding relation. If the given relations are $a : b$ $b : c$ $c : d$ $d : e$ $e : a$ the compound relation is $1 : 1$. 491. The utility of these principles will be perceived when it is observed, that the relation between two square fields is compounded of the relations of the lengths and the breadths. Let the two fields, for example, be A and B; A having 500 feet in length by 60 feet in breadth; the length of B being 360 feet, and its breadth 100 feet; the relation of the lengths will be $500 : 360$, and that of the breadths $60 : 100$. So that we have $(500 \cdot 60) : (360 \cdot 100) = 5 : 6$. Wherefore the field A is to the field B, as 5 to 6. 492. Again, let the field A be 720 feet long, 88 feet broad; and let the field B be 660 feet long, and 90 feet broad; the relations will be compounded in the following manner: Relation of the lengths $720 : 660$ Relation of the breadths $88 : 90$ and by cancelling, the Relation of the fields A and B is $16 : 15$.