594 
NUMBERS. 
[795 
Ordinary Theory, First Part. 
1. We are concerned with the integer numbers 0, +1, ±2, + 3, &c., or in the 
first place with the positive integer numbers 1, 2, 3, 4, 5, 6, &c. Some of these, 
1, 2, 3, 5, 7, fee., are prime, others, 4, = 2 2 , 6, = 2 .3, «fee., are composite; and we have 
the fundamental theorem that a composite number is expressible, and that in one way- 
only, as a product of prime factors, JT = a a 6^(+... (a, b, c, ... primes other than 1; 
a, ft, y,... positive integers). 
Gauss makes the proof to depend on the following steps: (i) the product of two 
numbers each smaller than a given prime number is not divisible by this number; 
(ii) if neither of two numbers is divisible by a given prime number the product is 
not so divisible; (iii) the like as regards three or more numbers; (iv) a composite 
number cannot be resolved into factors in more than one way. 
2. Proofs will in general be only indicated or be altogether omitted, but, as a 
specimen of the reasoning in regard to whole numbers, the proofs of these funda 
mental propositions are given at length, (i) Let p be the prime number, a a number 
less than p, and if possible let there be a number b less than p, and such that ab 
is divisible by p; it is further assumed that b is the only number, or, if there is 
more than one, then that b is the least number having the property in question; 
b is greater than 1, for a being less than p is not divisible by p. Now p as a 
prime number is not divisible by b, but must lie between two consecutive multiples 
mb and (m + 1)5 of b. Hence, ab being divisible by p, mab is also divisible by p; 
moreover, ap is divisible by p, and hence the difference of these numbers, = a (p — mb), 
must also be divisible by p, or, writing p — mb = b', we have ab' divisible by p, where 
b' is less than b; so that b is not the least number for which ab is divisible by p. 
(ii) If a and b are neither of them divisible by p, then a divided by p leaves a 
remainder a which is less than p, say we have a = mp + a; and similarly b divided 
by p leaves a remainder ft which is less than p, say we have b = np + /3; then 
ab = (mp + a) (np + /3), = (mnp + na + m/3) p + a/3, 
and aft is not divisible by p, therefore ab is not divisible by p. (iii) The like proof 
applies to the product of three or more factors a, b, c, ... (iv) Suppose that the 
number N, = a a №c y ... (a, b, c, ... prime numbers other than 1), is decomposable in 
some other way into prime factors; we can have no prime factor p, other than 
a, b, c,..., for no such number can divide a a Wc* ...; and we must have each of the 
numbers a, b, c,..., for if any one of them, suppose a, were wanting, the number N 
would not be divisible by a. Hence the new decomposition if it exists must be a 
decomposition N = a a '№cy'...; and here, if any two corresponding indices, say a, a!, are 
different from each other, then one of them, suppose a, is the greater, and we have 
N-t-p a = №c y ... = a a '~ a №'c y '... That is, we have the number 4-p a expressed in two 
different ways as a product, the number a being a factor in the one case, but not a 
factor in the other case. Thus the two exponents cannot be unequal, that is, we 
must have a = a!, and similarly we have /3 = /3', y = y',...; that is, there is only the 
original decomposition N — a a №cy ...