University of Toronto – MAT246H1-S – LEC0201/9201 – Winter 2021
Concepts in Abstract Mathematics
3 - Prime numbers
Jean-Baptiste Campesato
Informally, prime numbers are the integers greater than 1 which can’t be factorized further. More pre- cisely they are the natural numbers admitting exactly two positive divisors. Otherwise stated, a natural number𝑛 ≥ 2is a prime number if and only if its only positive divisors are1and𝑛itself.
They play a crucial role in number theory since every natural number admit a unique expression as a product of prime numbers. They will also appear quite often later when we will study modular arithmetic.
All the results presented below were already known inEuclid’s Elements(circa 300BC). Nonetheless, there are still many conjectures involving prime numbers which are easy to state but still open (some of them despite several centuries of attempts). For instance:
• Goldbach conjecture (1742): any even natural number greater than2may be written as a sum of two prime numbers (e.g.4 = 2 + 2,6 = 3 + 3,8 = 5 + 3,10 = 5 + 5 = 7 + 3…).
• The twin prime conjecture (1849):there are infinitely many prime numbers𝑝such that𝑝+2is also prime (e.g.(3, 5),(5, 7),(11, 13)…).
• Legendre conjecture (1912):given𝑛 ∈ ℕ ⧵ {0}, we may always find a prime between𝑛2and(𝑛 + 1)2.
1 Prime numbers
Definition 1. We say that a natural number𝑝is aprime numberif it has exactly two distinct positive divisors.
A positive natural number with more than2positive divisors is said to be acomposite number.
Remark 2.
• 0is not a prime number since any natural number is a divisor of0.
• 1is not a prime number because it has only one positive divisor.
Hence a natural number𝑝is prime if and only if𝑝 ≥ 2and the only positive divisors of𝑝are1and𝑝.
Example 3. The first prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97…
We face two natural questions:
1. How to check whether a natural number is a prime number?
2. How many prime numbers are there?
Proposition 4. Let𝑛 ∈ ℕ. Then𝑛is composite if and only if there exist𝑎, 𝑏 ∈ ℕ ⧵ {0, 1}such that𝑛 = 𝑎𝑏.
Proof. Let𝑛 ∈ ℕ.
⇒assume that𝑛is a composite number, then it admits a divisor𝑘 ∈ ℕsuch that𝑘 ≠ 1and𝑘 ≠ 𝑛.
So𝑛 = 𝑘𝑚for some𝑚 ∈ ℕ. Note that𝑘, 𝑚 ≠ 0since otherwise𝑛 = 0. Note that𝑚 ≠ 1since otherwise𝑘 = 𝑛.
⇐Assume that𝑛 = 𝑎𝑏for some𝑎, 𝑏 ∈ ℕ ⧵ {0, 1}.
Note that𝑎 ≠ 𝑛, since otherwise𝑏 = 1and that𝑛 ≠ 1since otherwise𝑎|1, i.e. 𝑎 = 1.
Therefore1, 𝑎, 𝑛are three distinct positive divisors of𝑛, so that𝑛is a composite number. ■ Proposition 5. A composite number𝑎admits a positive divisor𝑏such that1 < 𝑏2 ≤ 𝑎.
Proof. Write𝑎 = 𝑏1𝑏2for some𝑏1, 𝑏2∈ ℕ ⧵ {0, 1}. Then𝑏21, 𝑏22> 1.
Assume by contradiction that both𝑏21 > 𝑎and𝑏22 > 𝑎. Then𝑎2 = (𝑏1𝑏2)2= 𝑏21𝑏22> 𝑎2. Hence a contradiction.
■ Example 6. We want to prove that97is a prime number.
Since102 = 100 > 97, it is enough to check that none of2,3,4,5,6,7,8and9are divisors of97.
We will see later criteria to check divisibility.
2 Prime numbers
Lemma 7. A natural number𝑛 ≥ 2has at least one prime divisor.
Proof. We are going to prove with a strong induction that every natural number𝑛 ≥ 2has a prime divisor.
Base case at𝑛 = 2:2admits a prime divisor (itself).
Induction step:assume that all the natural numbers2, … , 𝑛admit a prime divisor for some𝑛 ≥ 2.
• First case: 𝑛 + 1is a prime number, then it has a prime divisor (itself).
• Second case: 𝑛 + 1is a composite, then𝑛 + 1 = 𝑎𝑏where𝑎, 𝑏 ∈ ℕ ⧵ {0, 1}.
Note that𝑎 ≠ 𝑛 + 1since otherwise𝑏 = 1.
Since2 ≤ 𝑎 ≤ 𝑛,𝑎admits a prime divisor𝑝by the induction hypothesis, i.e. 𝑎 = 𝑝𝑘for some𝑘 ∈ ℕ.
Then𝑛 + 1 = 𝑎𝑏 = 𝑝𝑘𝑏. Thus the prime number𝑝is a divisor of𝑛 + 1.
Which proves the induction step. ■
Theorem 8. There are infinitely many prime numbers.
Proof. Assume by contradiction that there exist only finitely many prime numbers𝑝1, 𝑝2, … , 𝑝𝑛.
We set𝑞 = 𝑝1𝑝2⋯ 𝑝𝑛+ 1. By Lemma7,𝑞has a prime divisor. Thus there exists𝑖 ∈ {1, 2, … , 𝑛}such that𝑝𝑖|𝑞.
Then, since𝑝𝑖|𝑝1𝑝2… 𝑝𝑛and𝑝𝑖|𝑞, we have that𝑝𝑖|(𝑞 − 𝑝1𝑝2… 𝑝𝑛), i.e.𝑝𝑖|1.
Therefore𝑝𝑖= 1, which is a contradiction because1is not a prime number. ■
2 The fundamental theorem of arithmetic
Lemma 9(Euclid’s lemma). Let𝑎, 𝑏 ∈ ℤand𝑝be a prime number. If𝑝|𝑎𝑏then𝑝|𝑎or𝑝|𝑏(or both).
Proof. Let𝑎, 𝑏 ∈ ℤand𝑝be a prime number such that𝑝|𝑎𝑏.
Assume that𝑝 ∤ 𝑎then gcd(𝑎, 𝑝) = 1since the only positive divisors of𝑝are1and itself.
Hence, by Gauss’ lemma,𝑝|𝑏. ■
Theorem 10(The fundamental theorem of arithmetic). Any integer greater than 1 can be written as a product of primes, moreover this expression as a product of primes is unique up to the order of the prime factors.
Remark 11. The above theorem states two things: theexistenceof a prime factorization, and itsuniqueness.
Proof.
• Existence.We are going to prove with a strong induction that𝑛 ≥ 2admits a prime factorization.
Base case for𝑛 = 2: 2is a prime number, so there is nothing to do.
Induction step:assume that all the integers2, 3, … , 𝑛have a prime factorization for some𝑛 ≥ 2.
We want to prove that𝑛 + 1admits a prime factorization.
By Lemma7,𝑛 + 1admits a prime factor, so𝑛 + 1 = 𝑝𝑘where𝑝is a prime number and𝑘 ∈ ℕ ⧵ {0}.
If𝑘 = 1then there is nothing to do. So we may assume that𝑘 ≥ 2.
Since1 < 𝑝, we have that𝑘 < 𝑝𝑘 = 𝑛 + 1.
Since2 ≤ 𝑘 ≤ 𝑛, by the induction hypothesis,𝑘admits a prime factorization𝑘 = 𝑝1𝑝2… 𝑝𝑙. Finally𝑛 + 1 = 𝑝𝑝1𝑝2… 𝑝𝑙, which proves the induction step.
• Uniqueness (up to order).
Assume by contradiction that there exists an integer greater than 1 with (at least) two distinct prime factorizations. Denote by𝑛the least such integer (which exists by the well-ordering principle).
Let𝑛 = 𝑝1𝑝2… 𝑝𝑟and𝑛 = 𝑞1𝑞2… 𝑞𝑠be two distinct prime factorizations of𝑛.
Then𝑝1𝑝2… 𝑝𝑟 = 𝑞1𝑞2… 𝑞𝑠.
By Euclid’s lemma𝑝1divides one of the𝑞𝑗.
Up to reordering the indices, we may assume that𝑝1|𝑞1. Since𝑞1is a prime number, either𝑝1= 1or𝑝1= 𝑞1.
And thus𝑝1= 𝑞1since𝑝1is also a prime number (and1is not).
Therefore, by cancellation,𝑚 = 𝑝2… 𝑝𝑟= 𝑞2… 𝑞𝑠is a number with two distinct prime factorizations.
Note that𝑚 > 1since otherwise𝑛 = 𝑝1 = 𝑞1is not two distinct prime factorizations.
And, since1 < 𝑝1we get that𝑚 = 𝑝2… 𝑝𝑟 < 𝑝1𝑝2… 𝑝𝑟= 𝑛.
Which contradicts the fact that𝑛is the least integer greater than1with two prime factorizations. ■
MAT246H1-S – LEC0201/9201 – J.-B. Campesato 3
Corollary 12. Any natural number𝑛 ∈ ℕ ⧵ {0}admits a unique expression𝑛 = ∏
𝑝prime
𝑝𝛼𝑝where𝛼𝑝 ∈ ℕ (i.e. the𝛼𝑝are uniquely determined).
Remarks 13.
• The above product is finite since all but finitely many exponents are equal to0.
• 1is the special case when𝛼𝑝= 0for all prime numbers𝑝.
Example 14. 60798375 = 32× 53× 11 × 173 Corollary 15. Write𝑎 = ∏
𝑝prime
𝑝𝛼𝑝 and𝑏 = ∏
𝑝prime
𝑝𝛽𝑝with𝛼𝑝, 𝛽𝑝∈ ℕall but finitely many equal to0. Then
• 𝑎|𝑏if and only if for every prime number𝑝,𝛼𝑝≤ 𝛽𝑝.
• gcd(𝑎, 𝑏) =
𝑝∏prime
𝑝min(𝛼𝑝,𝛽𝑝).
Example 16. gcd(32× 53× 11 × 173, 3 × 55× 172× 23) = 3 × 53× 172 Corollary 17. Write𝑛 = ∏
𝑝prime
𝑝𝛼𝑝 with𝛼𝑝 ∈ ℕall but finitely many equal to0. Then the positive divisors of𝑛are exactly the numbers of the form𝑛 =
𝑝∏prime
𝑝𝛾𝑝with0 ≤ 𝛾𝑝≤ 𝛼𝑝for all prime numbers𝑝.
Particularly,𝑛has ∏
𝑝prime
(𝛼𝑝+ 1)positive divisors.
