The document defines prime numbers as integers greater than or equal to 2 that have no positive divisors other than 1 and itself. It then presents three lemmas: 1) An integer n ≥ 2 is composite if and only if it has factors a and b such that 1 < a < n and 1 < b < n; 2) If n > 1 then there is a prime p such that p | n; and 3) Euclid's theorem that there are infinitely many primes, proved by considering the number N formed by multiplying all primes together and adding 1, which must have a prime divisor not in the original list.

1. PRIME NUMBERS
2.1 Definition :
An integer p ≥ 2 is prime if it has no
positive divisors other than 1 and itself. An
integer greater than or equal to 2 that is not
prime is composite.

2. PRIME NUMBERS
2.2 LEMMA
An integer n ≥ 2 is composite if and only if
it has factors a and b such that 1 < a < n and 1 <
b < n.

3. PRIME NUMBERS
Proof:
Let n ≥ 2. The ‘if’ direction is obvious. For ‘only if’, assume that n
is
composite. Then it has a positive integer factor a such that a≠1, a≠n.
This means that there is a b with n = ab. Since n and a are positive, so is
b. Hence 1 ≤ a and 1 ≤ b. By Theorem 1.2, a ≤ n and b ≤ n. Since a≠1
and a≠n we have 1 < a < n. If b = 1 then a = n, which is not possible, so
b ≠ 1. If b = n then a = 1, which is also not possible. So 1 < b < n,
finishing this half of the argument.

4. PRIME NUMBERS
2.3 LEMMA
If n > 1 then there is a prime p such that p | n.

5. PRIME NUMBERS
Proof:
Let S denote the set of all integers greater than 1 that have no
prime divisor. We must show that S is empty. If S is not empty then by
the Well-Ordering Property it has a smallest member; call it m. Now m
> 1 and has no prime divisor. Then m cannot be prime (as every
number is a divisor of itself). Hence m is composite. Therefore by
Lemma 2.2, m = ab where 1 < a < m and 1 < b < m. Since 1 < a < m, the
factor a is not a member of S. So a must have a prime divisor p. Then p
| a and a | m, so by Theorem 1.2, p | m. This contradicts the
assumption that m has no prime divisor. So the set S must be empty.

6. PRIME NUMBERS
2.4 Theorem (Euclid’s Theorem) There are infinitely many primes.
Proof. Assume, to get a contradiction, that there are only a finitely
many primes p1 = 2, p2 = 3, . . . , pn. Consider the number N = p1 p2 · · ·
pn + 1.

7. PRIME NUMBERS
Since p1 ≥ 2, clearly N ≥ 2. So by Lemma 2.3, N has a prime divisor p.
That prime must be one of p1, . . . , pn since that list was assumed to be
exhaustive. However, observe that the equation
N = pi (p1 p2 · · · pi −1 pi+1 · · · pn ) + 1
along with 0 ≤ 1 < pi shows by Lemma 3.2 that n is not divisible by pi .
This is a contradiction; it follows that the assumption that there are
only finitely many primes is not true.

8. PRIME NUMBERS
2.5 REMARK Eucild’s Theorem, and its proof, is often cited as an
example of the beauty of Mathematics.

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