MATH 57091 - Algebra for High School Teachers
Fundamental Theorem of Arithmetic
Professor Donald L. White
Department of Mathematical Sciences
Kent State University
D.L. White (Kent State University)
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Statement of Theorem
Fundamental Theorem of Arithmetic
Every integer n > 1 is either prime or a product of primes.
The expression of n as a product of primes is unique,
except for the order of the factors.
If we specify the order of the prime factors, we get a unique factorization:
Corollary
Every integer n > 1 can be written uniquely in the form
n = p1a1 p2a2 · · · prar ,
where p1 < p2 < · · · < pr are primes and ai > 1 is an integer for each i.
This unique expression is the Canonical Prime Factorization of n.
Example: The canonical prime factorization of 720 is
720 = 24 · 32 · 51 = 24 · 32 · 5.
2
4
It is not 3 · 2 · 5 or 24 · 32 · 5 · 70 .
D.L. White (Kent State University)
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Prime Factorization for Comparison
In the canonical prime factorization, we do not allow exponents to be 0.
Otherwise we could include p 0 as a factor for any prime p.
It will be useful for discussing divisibility in terms of prime factorizations
to allow exponents to be 0 in certain situations, however.
We have the following slight modification of the canonical factorization:
Prime Factorization for Comparison
Let a and b be positive integers.
Let p1 < p2 < · · · < pr be all primes that divide at least one of a or b.
The integers a, b can be expressed uniquely in the form
a = p1a1 p2a2 · · · prar
b = p1b1 p2b2 · · · prbr
where ai > 0 and bi > 0 are integers for all i.
Example: If a = 22 · 35 · 73 · 11 · 135 and b = 25 · 3 · 52 · 113 , then
a = 22 · 35 · 50 · 73 · 11 · 135
b = 25 · 3 · 52 · 70 · 113 · 130 .
D.L. White (Kent State University)
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Sketch of Proof
We now sketch a proof of the Fundamental Theorem of Arithmetic.
Existence:
Suppose there exists an integer greater than 1
that is neither a prime number nor a product of primes.
By the Well-ordering Principle, there is a smallest such integer; call it n.
Since n is not prime, there are integers a and b such that
n = ab, 1 < a < n, and 1 < b < n.
Thus by the minimality of n, a is prime or a product of primes
and b is prime or a product of primes.
Hence n = ab is a product of primes in any case, a contradiction.
Therefore, no such integer n exists,
and so every integer greater than 1 is a prime or a product of primes.
D.L. White (Kent State University)
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Sketch of Proof
Uniqueness:
Suppose n = p1 p2 · · · pr and n = q1 q2 · · · qs , where each pi and qj is prime.
We have p1 p2 · · · pr = q1 q2 · · · qs .
Dividing both sides by each prime that appears in both factorizations,
we end with either
1 = 1, so r = s and (after possible rearrangement) pi = qi for each i,
or
0 , where p 0 6= q 0 for all i and j.
p10 p20 · · · p`0 = q10 q20 · · · qm
i
j
In the first case, the prime factorization of n is unique (except for order).
0 ,
In the second case, p10 | q10 q20 · · · qm
and so by Euclid’s Lemma for Primes, p10 | qj0 for some j.
But qj0 is prime, so the only positive factors of qj0 are 1 and qj0 .
Since p10 6= 1, we must have p10 = qj0 , a contradiction.
Therefore, the second case does not occur.
D.L. White (Kent State University)
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Another Return to Square Roots
Theorem
If p is a prime number, then
√
p is an irrational number.
√
Proof: Suppose to the contrary that p is rational.
√
We can then write p = ba , where a and b are integers.
√
We then have a = b p, and so a2 = pb 2 .
Observe that if an integer n has canonical prime factorization
n = p1a1 p2a2 · · · prar , then
n2 = (p1a1 p2a2 · · · prar )2 = p12a1 p22a2 · · · pr2ar .
Hence the number of prime factors in the factorization of n2 is
2a1 + 2a2 + · · · + 2an = 2(a1 + a2 + · · · + an ),
and so is even.
Therefore, the number of prime factors of a2 is even,
while the number of prime factors of pb 2 is odd.
Since a2 = pb 2 , this contradicts the uniqueness of prime factorization.
√
Hence no such fraction ba exists, and so p is irrational.
D.L. White (Kent State University)
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