PROOF 13

Prove.

1 is not prime

On the naturals ℕ.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

Definition D1

b | a means that a natural number k exists satisfying a = bk.

Proof.

This renders divisor as a single multiplication.convention

∎

Definition D2

p is prime means that p > 1 and the only divisors of p are 1 and p.

Proof.

This is the definition in use today. It carries two conditions.convention

∎

Definition D3

Let D3 be D2 with the condition p > 1 removed: a natural number whose only divisors are 1 and itself is called prime.

Proof.

D3 does not exclude 1.convention

∎

Lemma L1

Under D3, 1 is prime.

Proof.

(1)Since 1 = 1 × 1, 1 | 1.A2, D1

(2)If a | 1, there is a k with 1 = ak.D1

(3)Since 1 ≤ k, a = a × 1 ≤ ak = 1.(2), A1, A2, A4

(4)Since 1 ≤ a, a = 1.(3), A1

(5)Therefore the only divisor of 1 is 1.(1), (4)

(6)The divisors of 1 are 1 and itself alone, so under D3, 1 is prime.(5), D3

∎

Theorem T1

Under D3, there are infinitely many ways to write 12 as a product of primes.

Proof.

(1)12 = 2 × 2 × 3.A3

(2)1 is prime.L1

(3)Since 1 × 12 = 12, 12 = 1 × 2 × 2 × 3.(1), A2, A3, E

(4)In the same way, 12 = 1 × 1 × 2 × 2 × 3.(3), A2

(5)Attaching any number of 1s leaves the value 12.A2

(6)By (2), these expressions are all products of primes.(2), (5)

(7)The counts of 1s differ, so these are not reorderings of one another.(3), (4)

(8)Therefore such expressions are infinitely many.(5), (6), (7)

∎

Corollary C

1 is not prime.

Proof.

(1)Prime factorization is unique.A5

(2)Under D3, 12 has infinitely many expressions.T1

(3)Therefore D3 and A5 cannot stand together.(1), (2)

(4)D2 requires p > 1, and so excludes 1 from the primes.D2

(5)Under D2, the only expression of 12 is 2 × 2 × 3.(4), A5

(6)Excluding 1 is not the result of a computation, but a choice of definition made to preserve A5.(3), (4)

∎

Remark

Can A5 be removed from this proof?

A5 is a theorem proved in the previous walk. Here it serves as a requirement to be preserved. Remove it and see what remains.