prime numbers and expressing non-prime numbers

Question

My textbook says if $b$ is a non-prime number then it can be expressed as a product of prime numbers. But if $1$ isn't prime how it can be expressed as a product of prime numbers?

Answer 1

What is the sum of no numbers at all? Zero, of course, since it is the additive identity: $x + 0 = 0$, where $x \neq 0$, or even if it is.

Now, what is the product of no numbers at all? It can't be zero, since, maintaining the stipulation that $x \neq 0$, we have $x \times 0 = 0$, and we said $x \neq 0$. The multiplicative identity is $1$, since $x \times 1 = 1$.

Hence, the product of no primes at all is $1$. The fundamental theorem of arithmetic is a subtlety that's unnecessary for answering your question.

Answer 2

This is mainly just an extended comment on Peter Foreman's answer. The (relatively difficult) uniqueness aspect of the Fundamental Theorem of Arithmetic is not needed for the OP's question, just the (easier) existence aspect.

What's missing from the OP's textbook is the qualifier in the correct assertion that every non-prime number greater than $1$ can be expressed as a product of primes. This is the existence aspect of FTA, and it can be proved by strong induction: If $n\gt1$ is not a prime, then $n=ab$ for some pair of integers with $1\lt a,b$. Both $a$ and $b$ must be less than $n$ (otherwise their product would be more than $n$), so we can assume, by strong induction, that each of them can be written as a product of primes, hence so can their product, which is $n$.

Remark: "Strong" induction means that you don't just assume an assertion is true for $n-1$ and then prove it for $n$, you assume it's true for all positive integers $k\lt n$. In this case the assertion is "if $k\gt1$ and $k$ is non-prime, then $k$ can be written as a product of primes." Note that the base case, $k=1$, is vacuously true, because $1$ is not greater than $1$.

Answer 3

An empty product is still a product.

Lulu's comment is sufficient to answer what you asked. Although you didn't state any constraints on $b$, I understand your question to be specifically about the case $b = 1$. And that's neatly addressed just by the concept of the empty product alone.

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But then why is everyone talking about the fundamental theorem of arithmetic? Its overuse is part of the reason.

But also because it's kind of implied by the idea of $b$ being the product of primes. For, after all, if a number domain does not have unique factorization, then it probably has numbers that are divisible by irreducible non-primes but not by any primes.

However, even if we're only talking about $\mathbb Z$, I think it's still more productive (pun fully intended) to say numbers in a unique factorization domain are divisible by units and primes: the units may vary in infinitely many ways, but the primes can only vary in regards to order.

And so we're not the least bit baffled about the number 1, since it's a product of units in several different ways (e.g., $(-1)^2 =$ $ (-1)^4 \times 1 = $ $\ldots$) and an empty product of primes.

Nor are we baffled by negative numbers, even if we stubbornly refuse the sensible idea that negative numbers can be prime. For example, $-14 = (-1) \times 2 \times 7$.

And $-1$ is the product of the unit $-1$ and an empty product of primes, which is, of course, 1.

This only leaves out the very special case of 0.