Why does a theorem in G.H. Hardy's Introduction to the Theory of Numbers looks like it implies that 1 is prime?

Question

In G.H. Hardy's 6th edition of Introduction to the Theory of Numbers, there is a theorem that looks like it implies that $1$ is prime:

"Theorem 1. Every positive integer, except $1$, is a product of primes. (p. 2)"

$17$ is a positive integer, but the factors of $17$ are not both prime.

Is this because of how Euclid defines a prime number in Book VII of Euclid's Elements? The translation I have says:

"A prime number is that which is measured by an unit alone".

If I were to take a couple of guesses, either I ignore $1$ as a factor or I bought a bad copy of G.H. Hardy's Introduction to the Theory of Numbers.

Edit: Not the Fundamental Theorem of Arithmetic, it is just Theorem 1.

score 3 · Accepted Answer · answered Nov 28 '17 at 04:11

3

Hardy never said more than one prime - $17$ is the product of the single number "$17$," which is prime.

answered Nov 28 '17 at 04:11

Noah Schweber

245,398

4

And, indeed, 1 is also a product of primes, the number of primes in the product being zero. – Gerry Myerson Nov 28 '17 at 04:13
@GerryMyerson But Hardy didn't say "iff"! :) – Noah Schweber Nov 28 '17 at 04:13
But Hardy apparently wrote, Every positive integer except 1 is a product of primes (emphasis added). To me, that says he didn't recognize empty products as products. – Gerry Myerson Nov 28 '17 at 04:15
Wow I didn't know a product could be a single value.
So a product results from multiplying at least one number?
– Chaz Biroan Nov 28 '17 at 04:16
@ChazBiroan Actually, as Gerry Myerson observed you can even take the product of the emptyset! See here. – Noah Schweber Nov 28 '17 at 04:18
Super glad I asked. I was being driven mad that I might have bought a badly printed copy. Thanks all! – Chaz Biroan Nov 28 '17 at 04:19

Why does a theorem in G.H. Hardy's Introduction to the Theory of Numbers looks like it implies that 1 is prime?

1 Answers1