Elements of Algebra

150 ELEMENTS SECT. III. So that $24$ is the greatest common divisor; and consequently the relation $504:312$ is reduced to the form $21:13$. 454. Let the relation $625:529$ be given, and the greatest common divisor of these two numbers be required. $529) 625 (1$ $529$ --- $96) 529 (5$ $480$ --- $49) 96 (1$ $49$ --- $47) 49 (1$ $47$ --- $2) 47 (23$ $46$ --- $1) 2 (2$ $2$ --- $0.$ Wherefore $1$ is, in this case, the greatest common divisor, and consequently we cannot express the relation $625:529$ by less numbers, nor reduce it to simpler terms. 455. It may be necessary, in this place, to give a demonstration of the foregoing Rule. In order to this, let $a$ be the greater, and $b$ the less of the given numbers; and let $d$ be one of their common divisors; it is evident that $a$ and $b$ being divisible by $d$, we may also divide the quantities, $a-b$, $a-2b$, $a-3b$, and, in general, $a-nb$ by $d$. 456. The converse is no less true: that is, if the numbers $b$ and $a-nb$ are divisible by $d$, the number $a$ will also be divisible by $d$; for $nb$ being divisible by $d$, we could not divide $a-nb$ by $d$, if $a$ were not also divisible by $d$. 457. We observe farther, that if $d$ be the greatest common divisor of two numbers, $b$, and $a-nb$, it will also be the greatest common divisor of the two numbers $a$ and $b$; for if a greater common divisor than $d$ could be found for these numbers $a$ and $b$, that number would also be a common divisor of $b$ and $a-nb$; and consequently $d$ would not be the greatest common divisor of these two numbers; but we have supposed $d$ to be the greatest divisor common to $b$ and