Fundamental theorem of arithmetic for 2

Question

Is says every positive whole number can be written as a unique product of primes. Since $1$ is not a prime what is the product for $2$? Is it just $2$? But then it wouldn't be a product.

Answer 1

I am so old that I was actually taught as a child that $1$ is a prime number. It's not, and though I had a subconscious feeling I had been taught incorrectly, decades passed before I actually understood.

As you already know, $6 = 2 \times 3$ is the product of two distinct primes. By the way $3 \times 2$ is not a distinct factorization. We agree that reordering the prime factors does not change the factorization, even though we agree to write the primes in ascending order.

Now, look at the powers of $2$. You also know that $4 = 2^2$, also the product of two primes, but they're not distinct primes, they're both $2$. And also, $2^3 = 8$, $2^4 = 16$, $2^5 = 32$, etc.

Going in the other direction we have $2^1 = 2$. No surprise there. And $2^0 = 1$. And, as I hinted in a comment to another answer, $$2^{-1} = {1 \over 2}.$$

Clearly $${32 \over 2} = 16, {16 \over 2} = 8, {8 \over 2} = 4, {4 \over 2} = 2, {2 \over 2} = 1, {1 \over 2} = {1 \over 2}, {{1 \over 2} \over 2} = {1 \over 4}, {{1 \over 4} \over 2} = {1 \over 8}, {{1 \over 8} \over 2} = {1 \over 16}, \ldots$$

So, if $p^\alpha$ is the product of multiplying a prime $p$ by itself $\alpha - 1$ times, then $1$ is the product of multiplying a prime by itself $0$ times... wait, I seem to have painted myself into a corner: since $p$ can be any prime, or a composite number for that matter, that would mean that the factorization of $1$ is not actually distinct... But the point is that $1$ is nevertheless a product nonetheless.

Let me try to jump over the wet paint. You also know that $(-2)^2 = 4$. But why doesn't this count as a unique factorization? Because $-1$ is a unit, a number that does not multiplicatively change the norm of a number, which we agree in this domain to be the absolute value function.

Thus $(-2)^2$ is not a distinct factorization of $4$, and neither is $-1 \times 2^2$, $(-1)^3 \times 2^2$, $(-1)^5 \times 2^2$, etc. Neither is $1^\alpha 2^2$, whatever $\alpha$ happens to be, as long as it be a number. Whether you consider $-2$ to be a prime number or not, that's another can of worms. If you don't, you can choose to view it as the prime number $2$ multiplied by the unit $-1$.

Which means that neither order nor multiplication by units makes for a distinct factorization. You can view $1$ as the product of one unit times no primes. Or as the square of the unit $-1$. You can throw as many units in as you want, as long as it lands the product on the desired side of the number line, or the desired quadrant of the complex plane.

These concepts will be important if you go on to study algebraic number theory. For example, $(10 - 7 \sqrt 2)^2 (10 + 7 \sqrt 2)^2$ and $(\sqrt 2)^4$ are not distinct factorizations of $4$ in a certain domain of algebraic numbers that includes $\sqrt 2$ as a prime number and $2$ as correspondingly composite. But in that domain, $1$ and $-1$ are also units, and unique factorization also holds.

Answer 2

What it means is that any number can be factored into a product of primes. That is, every integer has a prime factorization. Of course, with prime numbers, their prime factorization product is just themselves.

Answer 3

1 is the product of no primes at all, it is a unit. And in that way, it is the most unique among the positive integers, among all numbers, actually. All other positive integers are each the product of at least one prime number (itself, in the case of a prime number).

It might be helpful to review the concept of "empty sum" and "empty product". Both of these are identities, too.

The empty sum is 0, right? You add 0 to a number $x$ and you still have $x$. You added nothing, you added emptiness. $x + 0 = 0$.

But 0 is not the empty product. Unless $x = 0$, we have $x \times 0 \neq x$. The empty product has to be something that does not change $x$. It has to be something that is the same as not performing multiplication, that is, empty multiplication. So $x \times 1 = x$. So 1 is the empty product.

Viewed in this way, then:

• 1 is the empty product of no primes at all, zero primes.
• A prime number $p$ is its own product, the product of a single prime.
• A semiprime is the product of two prime numbers, not necessarily distinct.
• Sphenic numbers are the product of three distinct primes. Cubes of primes also have three prime factors, though not distinct. Also in this bullet point, numbers of the form $p^2 q$, where $p$ and $q$ are distinct primes.
• Products of four primes, etc.

To really bend your mind, ask: what is the product of $-1$ primes?