The purpose of this section is to prove the fundamental theorem of arithmetic. We start with a
very basic notion of divisibility. Let n be a positive integer. Let a be any integer. We said n divides a if
there exists an integer 𝑞1 such that n * 𝑞1 = a. In this case, n is called a divisor of a. We further define a
prime to be an integer p greater than 1 that has the following divisors: ±1 and ±p. Some example of
prime are 2, 3, 5, 7, 11. . . Given two integers a and b, we define the common divisors of a and b to be
an integer n that divides both a and b, and the greatest common divisor to be the largest among the
common divisors. For example, if a = 8, b=12, then the divisors of 8 are 1,2,4,8, and the divisors of 12 are
1,2,3,4,6. So, the common divisor of 8 and 12 are 1,2,4, and 4 is the greatest common divisor.
Once again, let n be a positive integer, and a be any integer. Observe that either n divides a, (in
this case, there exists an integer 𝑞1 such that n * 𝑞1 = a), or a is between n * 𝑞1 and n * (𝑞1 +1). So, we
can write a = n * 𝑞1 + r, where r belong to {0,1,2,3, |n|-1} . This observation is called the division
algorithm. For example, let a = 101 and n= 6. Then 101=16*6+5.
Given two positive integers a and b, a≥b, we can find their greatest common divisor very
efficiently by the euclidean algorithm. To do this, first we divide a by b, and write a as a = b* 𝑞1 + 𝑟1 ,
where 0 ≤ 𝑟1 ≤b. Then we divide b by 𝑟1 , and write b as b = 𝑟1 ∗ 𝑞2 +𝑟2 , 0≤ 𝑟2 ≤ 𝑟1 . We continue the
process with 𝑟1 and 𝑟2 . Note that the remainder form a decreasing sequence, so, eventually, we arrive
at a 𝑟𝑛+1 =0. So, we have the following:
a = b* 𝑞1 + 𝑟1
b = 𝑟1 ∗ 𝑞2 +𝑟2
𝑟1 = 𝑟2 *𝑞3 + 𝑟3
.
.
.
𝑟𝑛−2 =𝑟𝑛−1 *𝑞𝑛 +𝑟𝑛
𝑟𝑛−1 =𝑟𝑛 *𝑞𝑛+1.
For example, let a = 2320 and b = 1708, we can use the euclidean algorithm on these two
numbers and get the following:
2320=1708*1+612
1708=612*2+484
612=484*1+128
484=128*3+100
128=100*1+28
100=28*3+16
28=16*1+12
16=12*1+4
12=4*3.
Note that in the euclidean algorithm, the second to the last line contains the last remainder.
This last remainder is the greatest common divisor of a and b. To see this, we have to show that
𝑟𝑛 divides a, 𝑟𝑛 divides b, and 𝑟𝑛 is the largest number among the common divisors.
First, we see that 𝑟𝑛 divides 𝑟𝑛−1 from the last equation. Then, combine the last two equation,
we see that 𝑟𝑛−2 =𝑟𝑛 *𝑞𝑛+1*𝑞𝑛 +𝑟𝑛 , so 𝑟𝑛 divides 𝑟𝑛−2 . Also, working backward, we see that because 𝑟𝑛
divides 𝑟𝑛−1 and 𝑟𝑛−2 , it divides 𝑟𝑛−3 , because 𝑟𝑛−3= 𝑟𝑛−2 *𝑞𝑛−1 +𝑟𝑛−1. Eventually, we see that 𝑟𝑛 divides
all 𝑟𝑖 ’s, and so it divides b and a. Secondly, if we can show that every other common divisor of a and b
must divides 𝑟𝑛 , then it automatically satisfies that 𝑟𝑛 is the largest number among the common divisors.
To this end, suppose that c is a common divisor of a and b. Then c divids a and c divids b implies c
divides 𝑟1 , by the first line of the algorithm. Similarly, c divides 𝑟1 and c divides b implies c divides 𝑟2 , by
the second line of the algorithm. Working forward, we see that c divides 𝑟𝑛−2 , 𝑟𝑛−1 , and finally 𝑟𝑛 .
Thus, 4 is the greatest common divisor of 2320 and 1708.
Finally, we note that from the algorithm, we see that there exist integer x, y such that
ax+by=gcd(a,b)= 𝑟𝑛 , where gdc(a,b) denotes the greatest common divisors of a and b. (Work backward
from the second to the last equation in the algorithm). We define two integers a and b to be relatively
prime when gcd(a,b)=1. We then have the following observations:
Let p be a prime, a and b be integers. If p divides a*b, then either p divides a, or p divides b.
Suppose p does not divides a, then since p has only 1 and p as divisors, it follow that p and a are
relatively prime. So, there exists integers x and y such that xa+yp=1. Multiply b on both sides, we get
bxa +byp = b. Observe that p divides bxa because p divides ab, and p also divides byp. So, p divides b.
The fundamental theorem of arithmetic state that any integer n can be written as a product of
prime, and this product is unique up to ±1. To show this, let n be an arbitrary integer. Suppose it’s
prime, we’re done. If it’s not prime, it has divisors other than 1 and n. Called one of the divisor a. So, n
= a*b . with a, b < n. If a is prime, we’re done. If not, we further decompose a into smaller part. Since n
is finite, we must get a prime factor. We then repeat the step for other divisors to get other prime
factors. Thus, eventually, n can be written as n=𝑝1 𝑝2 𝑝3….𝑝𝑘 , where each 𝑝𝑖 is a prime. To show
uniqueness, suppose n = 𝑝1 𝑝2 𝑝3….𝑝𝑘 = 𝑞1 𝑞2 𝑞3 …𝑞𝑙 , since each 𝑝𝑖 ’s divides the product of q’s, it must
divides at least one 𝑞𝑗 . Hence 𝑝𝑖 ’s must equal 𝑞𝑗 . We cancel them, and proceed with the next p.
eventually we have 1=1. Hence, we have uniqueness.
Reference:
Ore, Oystein: Number Theory and Its History, Dover Publication Inc., New York, 1976