Relatively Prime Integers and the Fundamental Theorem of Arithmetic
I. Introduction.
At the moment, you probably have no idea what the title of this handout means,
but that is only because it is using unfamiliar words for ideas that you have already encountered. In the first place, the Fundamental Theorem of Arithmetic
is the standard name given to what we have been calling the “unique factorization property” of the integers—namely, that if two people express a given
integer as a product of primes, they will wind up using the same primes, each
one the same number of times. This fact will be proved in this handout.
The other new term in the title is relatively prime. Two positive integers are
said to be relatively prime when gcd(a, b) = 1. Note carefully that the terms
relatively prime and prime have very little to do with each other. For example,
the integers 18 and 35 are relatively prime, even though 18 and 35 are both
nonprime1 numbers. It turns out that pairs of relatively prime integers have
many special properties.
II. When a and b are Relatively Prime· · ·
· · ·many things happen. This handout will introduce only two of these (Theorems RP#1 and RP#2); you will see some others later in the semester. Theorem
RP#1 is a basic fact that is used to prove many other facts about relatively
prime integers, including RP#2.
Theorem RP# 1 Let a and b be positive integers. Then
gcd(a, b) = 1
⇐⇒
1 = ax0 + by0 for some x0 and y0 .
Proof (=⇒): We proved in the handout on linear combinations that in all cases,
gcd(a, b) = d0 = ax0 + by0 for some x0 and y0 ;
in this case, since we’re given that gcd(a, b) = 1, we get
gcd(a, b) = 1 = ax0 + by0 for some x0 and y0 .
Proof (⇐=): We’re given that 1 = ax0 + by0 . I will prove that gcd(a, b) =
1 by
a and
showing
that
any
common
divisor
of
a
and
b
is
also
a
divisor
of
1.
Let
d
d b. We proved in the linear combinations handout that d (ax + by) for every
x and every y. The special case x = x0 and y = y0 gives
d (ax0 + by0 ) = 1.
Proof is complete.
I will introduce the next property with some questions, which we will discuss in
class.
1. If 15 † n, can 15 ever divide 20n?
2. If 15 † n, can 15 ever divide 18n?
3. If 15 † n, can 15 ever divide 4n?
1 The
word for nonprime is “composite.”
1
By trying different values forn, one
finds that the answer to the first two
 15 60 = 20 × 3 
questions is yes—for example,
—but the answer to the third
 
15 90 = 18 × 5
question seems2 to be no. The answer to the third question does indeed turn
out to be no;3 this fact is a consequence of Theorem RP#2.
Theorem RP#
2 Let a and b be relatively prime positive integers. For any
integer n: if a bn, then a n.
Proof . Let n be any integer such that a bn, so that for some integer k,
ak = bn.
(1)
Since a and b are relatively prime, Theorem RP#1 tells us that
1 = ax0 + by0 for some x0 and y0 ;
(2)
and multiplying (2) through by n gives
n = n(ax0 + by0 ) = (na)x0 + (nb)y0 .
(3)
Now, equation (1) allows us to replace nb with ak in (3), which gives
n = (na)x0 + (ak)y0 .
(4)
Finally, factoring a out of the right side of (4) gives
n = a(nx0 + ky0 ),
(5)
which shows that n is a multiple of a. Proof is complete.
III. Factorization into Primes.
Whether an integer p > 1 is prime4 obviously depends upon what numbers
divide p. It turns out that proving the Fundamental Theorem of Arithmetic
requires a different property of primes, one that concerns what numbers p divides. I will introduce this property with some questions, which we will discuss
in class.
1. If 15 † a and 15 † b, can 15 divide ab?
2. If 9 † a and 9 † b, can 9 divide ab?
3. If 11 † a and 11 † b, can 11 divide ab?
2 Note the word seems: the fact that we did not find such an n does not prove that there
is no such n.
3 in other words: if 15 † n then 15 † 4n. The contrapositive of this statement is one instance
of Theorem RP#2.
4 Recall that a number p > 1 is prime if the only divisors d of p are d = 1 and d = p.
2
By trying different values for a and
 b, one finds that
 the answer to the first
 15 60 = 20 × 3 
two questions is yes—for example,
—but the answer to the
 
9 18 = 6 × 3
third question seems to be no. In fact, the answer to the third question IS no,
because it turns out that the following claim is true:
Claim 1 If a prime p divides a product, then p will divide at least one of the
factors.
(∗)
Theorem RP#3 proves this claim.5
Theorem
RP# 3 Let p be prime, let c and d be positive integers, and suppose
that p cd. Then
• either p c
• or p d (or both).
Proof . I will prove the equivalent statement: If p † c, then p d. Suppose
that p † c. Since
gcd(p, c) is a divisor
of the prime p, there are only two


 either gcd(p, c) = p 
possibilities:
. Moreover, since p † c, we know that gcd(p, c) 6= p;


or gcd(p, c) = 1
therefore,
gcd(p, c) = 1.


 p in for a 
We can now apply Theorem RP#2, by putting
c in for b ;


d in for n
RP#2 then says that either p c or p d (or both). Proof is complete.
We can now prove the Fundamental Theorem of Arithmetic.
Theorem RP# 4 (Fundamental Theorem of Arithmetic) Let n ≥ 2. If
n = p1 p2 p 3 p 4 · · · pk
and
n = q1 q2 q3 q4 · · ·
(where the p’s and q’s are all prime), then (after rearranging the q’s), we have
p1
p2
p3
p4
= q1 ,
= q2 ,
= q3 ,
= q4 ,
..
.
5 RP#3 proves (∗) for a two-factor product, but by using RP#3 over and over, you get (∗)
for longer products.
3
Proof . We’re given that
p1 p2 p 3 p 4 · · · pk = n = q 1 q 2 q 3 q 4 · · · .
(6)
Now, obviously p1 p1 (p2 p3 p4 · · · pk ), and by (6), p1 (p2 p3 p4 · · · pk ) = q1 q2 q3 q4 · · ·,
so
p1 q 1 q 2 q 3 q 4 · · · .
(7)
Applying (∗) to (7) then tells us that p1 divides one of the q’s, say q ∗ . Rearrange
the q’s so that q ∗ becomes q1 ; now we have
p1 q 1 .
(8)
Next: since q1 is prime, the only divisors of q1 are 1 and q1 , and since p1 6= 1,
(8) implies that
p1 = q 1 .
This means that we can cancel p1 = q1 from both sides of (6) to get
p2 p 3 p 4 · · · pk = q 2 q 3 q 4 · · · .
(9)
We now start all over again, applying the same reasoning to (9) that we did
to (6); we will find (after rearranging the remaining q’s) that
p2 = q 2 ,
and cancelling them from both sides of (9) will give us
p 3 · · · pk = q 3 · · · .
(10)
We now start over a third time, applying the same reasoning to (10) that we
did to (9); we wind up cancelling p3 and one of the remaining q’s from (10). We
keep on doing this over and over until we are finally left with our last p, namely
pk ; the equation at this point will be
pk = qk qk+1 · · · .
(11)
Now, the number we have on the left side of (11) is the prime pk , which cannot
be factored, so there can’t be more than one prime factor on the right side
of (11). In other words, there is no qk+1 in (11); equation (11) is actually
pk = q k .
Proof is complete.
4