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I'm trying to understand why the fundamental theorem of arithmetic is phrased like this:

Every positive integer, except 1, is a product of primes.

A prime number is an integer but it is not a product of primes. It seems to me that "every positive integer" generalization is not justified. Something like, "every non-prime integer is a product of primes" should have been a better definition. This phrasing would except 1 as well.

What am I missing?

According to the definition of prime numbers, a prime is not defined as "a product of primes":

A number p is said to be prime if (i) p > 1, (ii) p has no positive divisors except 1 and p.

I took these definitions from Introduction to the theory of numbers, by Hardy and Wright, p.2.

In his proof to Theorem 1, "Every positive integer, except 1, is a product of primes" Hardy states, "Either $n$ is prime, when there is nothing to prove, or $n$ has divisors between 1 and $n$."

But this doesn't make sense to me. The prime $n$ is a positive integer and according to the theorem it must be a product of primes but prime $n$ is not made of products of primes. As stated, every $n$ which is prime will contradict the theorem.

And if in the case of prime $n$, there is nothing to prove, then, Hardy is implicitly assuming the phrasing of the theorem as "every non-prime integer is a product of primes."

I assume there must be a reason why the prime number theorem is stated this way. Can you please explain?

Note:

As I was posting my question SE suggested this post Adequately defining the fundamental theorem of arithmetic. where Gerry Myerson comments that "mathematicians are happy with [...] products of one prime..." So is this the answer? Do we consider primes made of product of one prime, i.e., itself?

user3840170
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zeynel
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    IMO the trick is formula (1.2.2) page 2: $n = p_1^{a_1} \ldots p_k^{a_k}$ for $n$ prime $k=1$ and $p_1=n$ – Mauro ALLEGRANZA Mar 22 '24 at 08:56
  • Even though a prime is not defined to be a product of primes, doesn't mean it isn't. - Which is basically what Gerry Myerson said. – Henrik supports the community Mar 22 '24 at 09:02
  • And it's not the phrasing, it the phrasing used in the book you're reading. In the first book where I saw it (Judson: "Abstract Algebra", not to be recommended),it's phrased differently (but of course says the same - that's not one of the problems with that book). – Henrik supports the community Mar 22 '24 at 09:10
  • @MauroALLEGRANZA So, do we write something like this for e.g., $n=5$? $5=5^1\cdot 7^0 \cdot 11^0 \cdot 13^0 \ldots$ – zeynel Mar 22 '24 at 09:20
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    No, Mauro meant $a_i\ge1$, but $k=a_1=1$ is allowed. What you are missing is that in a "product" $\prod_{i=1}^rp_i$, $r$ can be equal to $1$. Actually, with the convention $\prod_{i\in\varnothing}p_i=1$, $r$ could even be allowed to be equal to $0$, and then "except $1$" could be removed from the "phrasing". – Anne Bauval Mar 22 '24 at 09:22

2 Answers2

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This is not the fundamental theorem of arithmetic, and Hardy & Wright don’t call it that – in fact, the next theorem on the next page is referred to as the fundamental theorem of arithmetic; it states that the prime factorization is unique.

As regards your criticism of the treatment of special cases, my criticism would be the opposite: Not only is a prime a product of $1$ prime, but $1$ is a product of $0$ primes, the empty product, so excluding $1$ is unnecessary. The Wikipedia article on the empty product cites two quotes from Edsger Dijkstra where he specifically criticizes this definition of Hardy & Wright and a similar one by Harold Stark. So you’re not the first to take issue with it, but consider doing away with unnecessary special cases rather than adding more.

To see why it makes sense to define products to include zero or one factors, consider, for example, the map from the positive integers to the exponents in their representation as a product of primes (i.e. by a function from $\mathbb Z^+$ to $\mathbb Z_0^+$ that maps $n$ to the exponent of $p_n$). This map tells us a lot about the structure of multiplication. Multiplication of numbers corresponds to addition of the corresponding functions. Taking $\gcd$ and $\operatorname{lcm}$ corresponds to taking $\min$ and $\max$, respectively. It would be a pity if we couldn’t use this map in full generality because we made exceptions and refrained from mapping $1$ to the zero function and the primes $p_n$ to the functions with a single non-zero value of $1$ at $n$, from which all other functions can be built by addition.

joriki
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Every positive integer, except 1, can be expressed as a product of prime(s)

There, fixed it.

Jokes aside, the theorem can be stated more rigorously as: For any given integer $n>1$, there exists a prime sequence $p_i$ such that $$n=\prod p_i$$

This explained why $1$ is treated differently, cause $1$ is an empty product

ioveri
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    To the contrary, it explains why $1$ should have been included, because there is a prime sequence (the empty prime sequence) whose product is $1$. – joriki Mar 22 '24 at 10:14