February4 ,2021 PRIMENUMBERS ConceptsinAbstractMathematics

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MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

PRIME NUMBERS

February 4^th, 2021

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Introduction

Prime numbers play a crucial role in number theory: they are the integers greater than 1 which can’t be factorized further while any other positive integer admits a unique expression as a product of prime numbers.

While negative integers are relatively new, prime numbers have been studied for quite a long time (maybe even before human beings developped writing systems).

All the results from today’s lecture were already known inEuclid’s Elements(circa 300BC).

Nonetheless, there are still many conjectures involving prime numbers which are easy to state but are still open (some 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 + 2 is also prime (e.g. (3, 5),(5, 7),(11, 13)…).

• Legendre conjecture (1912):given𝑛 ∈ ℕ ⧵ {0}, we may always find a prime number between

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Definition

Definition: prime numbers

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.

Remarks

• 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.

Hence𝑝 ∈ ℕis a prime number if and only if𝑝 ≥ 2and the only positive divisors of𝑝are1and𝑝.

Examples

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…

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Definition

Definition: prime numbers

Remarks

Examples

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Definition

Definition: prime numbers

Remarks

Examples

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How to check whether a natural number is prime?

Proposition

A composite number𝑎admits a positive divisor𝑏such that1 < 𝑏²≤ 𝑎.

Proof.Write𝑎 = 𝑏₁𝑏₂for two natural numbers𝑏₁and𝑏₂which are not equal to1or𝑎. Assume by contradiction that both𝑏²₁> 𝑎and𝑏²₂> 𝑎. Then

𝑎²= (𝑏₁𝑏₂)²= 𝑏²₁𝑏²₂> 𝑎²

Hence a contradiction. ■

Example: 97 is a prime number

Since10²= 100 > 97, it is enough to check that none of2,3,4,5,6,7,8and9are divisors of97.

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How to check whether a natural number is prime?

Proposition

Proof.Write𝑎 = 𝑏₁𝑏₂for two natural numbers𝑏₁and𝑏₂which are not equal to1or𝑎.

Assume by contradiction that both𝑏²₁> 𝑎and𝑏²₂> 𝑎. Then 𝑎²= (𝑏₁𝑏₂)²= 𝑏²₁𝑏²₂> 𝑎²

Example: 97 is a prime number

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How to check whether a natural number is prime?

Proposition

Proof.Write𝑎 = 𝑏₁𝑏₂for two natural numbers𝑏₁and𝑏₂which are not equal to1or𝑎.

Assume by contradiction that both𝑏²₁> 𝑎and𝑏²₂> 𝑎. Then 𝑎²= (𝑏₁𝑏₂)²= 𝑏²₁𝑏²₂> 𝑎²

Example: 97 is a prime number

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Towards the fundamental theorem of arithmetic

Lemma

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 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 ∈ ℕsatisfy1 < 𝑎, 𝑏 < 𝑛 + 1. Since2 ≤ 𝑎 ≤ 𝑛,𝑎admits a prime divisor𝑝by the induction hypothesis,

i.e. 𝑎 = 𝑝𝑘for some𝑘 ∈ ℕ. Then𝑛 + 1 = 𝑎𝑏 = 𝑝𝑘𝑏.

Thus𝑝is a prime divisor of𝑛 + 1.

Which proves the induction step. ■

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Towards the fundamental theorem of arithmetic

Lemma

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 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 ∈ ℕsatisfy1 < 𝑎, 𝑏 < 𝑛 + 1.

Since2 ≤ 𝑎 ≤ 𝑛,𝑎admits a prime divisor𝑝by the induction hypothesis, i.e. 𝑎 = 𝑝𝑘for some𝑘 ∈ ℕ.

Then𝑛 + 1 = 𝑎𝑏 = 𝑝𝑘𝑏.

Thus𝑝is a prime divisor of𝑛 + 1.

Which proves the induction step. ■

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How many prime numbers are there? – A lot!

Euclid’s theorem

There are infinitely many prime numbers.

Proof.Assume by contradiction that there exist only finitely many prime numbers, namely: 𝑝₁, 𝑝₂, … , 𝑝_𝑛

Set𝑞 = 𝑝₁𝑝₂⋯ 𝑝_𝑛+ 1.

Since𝑞has a prime divisor, there exists𝑖 ∈ {1, 2, … , 𝑛}such that𝑝_𝑖|𝑞. Then, since𝑝_𝑖|𝑝₁𝑝₂… 𝑝_𝑛and𝑝_𝑖|𝑞, we have that

𝑝_𝑖|(𝑞 − 𝑝₁𝑝₂… 𝑝_𝑛) = 1

Therefore𝑝_𝑖= 1, which is a contradiction because1is not a prime number. ■

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How many prime numbers are there? – A lot!

Euclid’s theorem

There are infinitely many prime numbers.

Proof.Assume by contradiction that there exist only finitely many prime numbers, namely:

𝑝₁, 𝑝₂, … , 𝑝_𝑛 Set𝑞 = 𝑝₁𝑝₂⋯ 𝑝_𝑛+ 1.

Since𝑞has a prime divisor, there exists𝑖 ∈ {1, 2, … , 𝑛}such that𝑝_𝑖|𝑞.

Then, since𝑝_𝑖|𝑝₁𝑝₂… 𝑝_𝑛and𝑝_𝑖|𝑞, we have that

𝑝_𝑖|(𝑞 − 𝑝₁𝑝₂… 𝑝_𝑛) = 1

Therefore𝑝_𝑖= 1, which is a contradiction because1is not a prime number. ■

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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 ∤ 𝑎thengcd(𝑎, 𝑝) = 1since the only positive divisors of𝑝are1and itself.

Hence, by Gauss’ lemma,𝑝|𝑏. ■

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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 ∤ 𝑎thengcd(𝑎, 𝑝) = 1since the only positive divisors of𝑝are1and itself.

Hence, by Gauss’ lemma,𝑝|𝑏. ■

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The fundamental theorem of arithmetic – 1

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.

Proof.Existence.

We are going to prove with a strong induction that𝑛 ≥ 2admits a prime factorization. Base case for𝑛 = 2:2is a prime number.

Induction step: assume that all the integers2, 3, … , 𝑛have a prime factorization for some𝑛 ≥ 2. We want to prove that𝑛 + 1admits a prime factorization.

Since𝑛 + 1admits a prime factor, there exist𝑝prime and𝑘 ∈ ℕ ⧵ {0}such that𝑛 + 1 = 𝑝𝑘. If𝑘 = 1, then𝑛 + 1 = 𝑝. So we may assume that𝑘 ≥ 2.

Since1 < 𝑝, we also have that𝑘 < 𝑝𝑘 = 𝑛 + 1, i.e. 2 ≤ 𝑘 ≤ 𝑛.

Thus, by the induction hypothesis,𝑘admits a prime factorization𝑘 = 𝑝₁𝑝₂… 𝑝_𝑙. Finally𝑛 + 1 = 𝑝𝑝₁𝑝₂… 𝑝_𝑙.

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The fundamental theorem of arithmetic – 1

The fundamental theorem of arithmetic

Proof.Existence.

We are going to prove with a strong induction that𝑛 ≥ 2admits a prime factorization.

Base case for𝑛 = 2:2is a prime number.

Induction step: assume that all the integers2, 3, … , 𝑛have a prime factorization for some𝑛 ≥ 2.

We want to prove that𝑛 + 1admits a prime factorization.

Since𝑛 + 1admits a prime factor, there exist𝑝prime and𝑘 ∈ ℕ ⧵ {0}such that𝑛 + 1 = 𝑝𝑘.

If𝑘 = 1, then𝑛 + 1 = 𝑝. So we may assume that𝑘 ≥ 2.

Since1 < 𝑝, we also have that𝑘 < 𝑝𝑘 = 𝑛 + 1, i.e. 2 ≤ 𝑘 ≤ 𝑛.

Thus, by the induction hypothesis,𝑘admits a prime factorization𝑘 = 𝑝₁𝑝₂… 𝑝_𝑙. Finally𝑛 + 1 = 𝑝𝑝₁𝑝₂… 𝑝_𝑙.

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The fundamental theorem of arithmetic – 2

The fundamental theorem of arithmetic

Proof.Uniqueness.

Assume by contradiction there exists an integer greater than 1 with two prime factorizations. Denote by𝑛the least such integer (which exists by the well-ordering principle).

Let𝑛 = 𝑝₁𝑝₂… 𝑝_𝑟= 𝑞₁𝑞₂… 𝑞_𝑠be two distinct prime factorizations of𝑛. By Euclid’s lemma𝑝₁divides one of the𝑞_𝑗, WLOG assume that𝑝₁|𝑞₁. Since𝑞₁is a prime number, either𝑝₁= 1or𝑝₁= 𝑞₁.

Thus𝑝₁= 𝑞₁since𝑝₁is a prime number.

Therefore, by cancellation,𝑚 = 𝑝₂… 𝑝_𝑟= 𝑞₂… 𝑞_𝑠has two distinct prime factorizations. Note that𝑚 > 1since otherwise𝑛 = 𝑝₁= 𝑞₁.

But since1 < 𝑝₁we get𝑚 = 𝑝₂… 𝑝_𝑟< 𝑝₁𝑝₂… 𝑝_𝑟= 𝑛.

Which is impossible since𝑛is the least integer greater than1with two prime factorizations. ■

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The fundamental theorem of arithmetic – 2

The fundamental theorem of arithmetic

Proof.Uniqueness.

Assume by contradiction there exists an integer greater than 1 with two prime factorizations.

Denote by𝑛the least such integer (which exists by the well-ordering principle).

Let𝑛 = 𝑝₁𝑝₂… 𝑝_𝑟= 𝑞₁𝑞₂… 𝑞_𝑠be two distinct prime factorizations of𝑛.

By Euclid’s lemma𝑝₁divides one of the𝑞_𝑗, WLOG assume that𝑝₁|𝑞₁. Since𝑞₁is a prime number, either𝑝₁= 1or𝑝₁= 𝑞₁.

Thus𝑝₁= 𝑞₁since𝑝₁is a prime number.

Therefore, by cancellation,𝑚 = 𝑝₂… 𝑝_𝑟= 𝑞₂… 𝑞_𝑠has two distinct prime factorizations.

Note that𝑚 > 1since otherwise𝑛 = 𝑝₁= 𝑞₁.

But since1 < 𝑝₁we get𝑚 = 𝑝₂… 𝑝_𝑟< 𝑝₁𝑝₂… 𝑝_𝑟= 𝑛.

Which is impossible since𝑛is the least integer greater than1with two prime factorizations. ■

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The fundamental theorem of arithmetic – 3

Corollary

Any natural number𝑛 ∈ ℕ ⧵ {0}admits a unique expression

𝑛 = ∏

𝑝prime

𝑝^𝛼^𝑝

where𝛼_𝑝∈ ℕ(i.e. the𝛼_𝑝 are uniquely determined).

Remarks

• The above product is finite since all but finitely many exponents are equal to0.

• 1is the special case when𝛼_𝑝= 0for all prime numbers𝑝.

Example

60798375 = 3²× 5³× 11 × 17³

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The fundamental theorem of arithmetic – 3

Corollary

𝑛 = ∏

𝑝prime

𝑝^𝛼^𝑝

Remarks

Example

60798375 = 3²× 5³× 11 × 17³

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The fundamental theorem of arithmetic – 3

Corollary

𝑛 = ∏

𝑝prime

𝑝^𝛼^𝑝

Remarks

Example

60798375 = 3²× 5³× 11 × 17³

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The fundamental theorem of arithmetic – 4

Corollary

Write𝑎 = ∏

𝑝prime

𝑝^𝛼^𝑝 and𝑏 = ∏

𝑝prime

𝑝^𝛽^𝑝 with𝛼_𝑝, 𝛽_𝑝∈ ℕall but finitely many equal to0. Then

• 𝑎|𝑏if and only if for every prime number𝑝,𝛼_𝑝≤ 𝛽_𝑝.

• gcd(𝑎, 𝑏) = ∏

𝑝prime

𝑝^min(𝛼^𝑝^,𝛽^𝑝⁾.

Example

gcd(3²× 5³× 11 × 17³, 3 × 5⁵× 17²× 23) = 3 × 5³× 17²

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The fundamental theorem of arithmetic – 5

Corollary

Write𝑛 = ∏

𝑝prime

𝑝^𝛼^𝑝 with𝛼_𝑝∈ ℕall but finitely many equal to0.

Then the positive divisors of𝑛are exactly the numbers of the form𝑛 = ∏

𝑝prime

𝑝^𝛾^𝑝 with0 ≤ 𝛾_𝑝 ≤ 𝛼_𝑝. Particularly,𝑛has ∏

𝑝prime

(𝛼_𝑝+ 1)positive divisors.