OF STAGE. Q5 



triangles. In tbc same manner it may be shown that GKzrGB ; 

 tlierefore the whole CDIG is equal to the sum of CH and GK. 



119. Problem. To find a common measure of any 

 two quantities. 



Subtract the less continually from the greater, the remainder from 

 the less, the next remainder from the preceding one, as often as pos- 

 sible, and proceed till there bo no further remainder ; then the last 

 remainder will be the common measure required. For since it mea- 

 sures the preceding remainder, it will measure the preceding quan- 

 tities in which that remainder was contained, together with itself, 

 and which, increased at each step by these remainders, makes up the 

 original quantities. 



For example, if the numbers 54 and 21 be proposed, 64 — ^21 — 21 

 -Zl2, 21—12=9, 12— 9z=3, 9—3—3—3=0, tlierefore 3 is the com- 

 mon measure, for it measures 9, nnd 9+3 or 12, and 12+9 or 21, and 

 2x21+12 or 54. 



Scholium. Hence it is obvious, that there can be no greater 

 common measure of the two quantities than the quantity thus found ; 

 for it should measure the difference of the tw^o quantities, and all the 

 successive remainders down to the last, therefore it cannot be greater 

 than this last. It must also be remarked, that in some cases no ac- 

 curate common measure can be found, but the error, or the last re- 

 mainder, in tliis process, may always be less than any quantity that 

 can be assigned, since the process may be continued without limit. 

 That there are incommensurable quantities, may_be thus shown : 

 every number is either a prime number, that is, a number not capable 

 of being composed by multiplication of other numbers, or it is com- 

 posed by the multiplication of factors, which are primes. Let the 

 number a be composed of the prime numbers led, or azzbcd, then 

 aazz.hcd.hcdzzhh.ee. dd and eacli prime factor of aa occurs twice ; so 

 that every square number must be composed of factors in pairs; and 

 a square number multiplied by a number which is not composed of 

 factors in pairs cannot be a square number : for instance, 2aa or 3«a 

 cannot be a square number, since the factors of 2 are only 1.2, and of 

 3, 1.3, and not in pairs : therefore the square root of 2 or 3 cannot be 

 expressed by any fraction, for the square of its numerator would be 

 twice or thrice the square of its denominator. But the ratio of tlie 

 hypotenuse of a triangle to its side may be that of s/Z or v^3 to 1 ; so 



