# Meaning of "for sufficiently large" in infinite products

The book I'm reading states that any positive integer $$a$$ greater than 1 can be expressed as a product of primes,

$$a=\prod_p{p^{\alpha{(p)}}}$$ where $$\alpha{(p)}$$ is a non-negative integer. And that it is understood for sufficiently large primes $$p$$, $$\alpha{(p)}=0$$.

My question is: what is considered to be a large prime? And how can the statement $$\alpha{(p)}=0$$ for large primes $$p$$ be true? Does that mean large primes can never be factors of any integers? If so I find this very unintuitive.

• The "sufficiently large" depends on $a$. Sep 14, 2020 at 10:25
• This is just a pedantic way to say that the product is made of a finite number of terms because $p^0=1$ beyond a certain $p$ Sep 14, 2020 at 10:26
• In this case: Any prime bigger than $a$, for example. For any specific $a$ you might be able to find a better bound.
– lulu
Sep 14, 2020 at 10:26
• en.wikipedia.org/wiki/Eventually_(mathematics) Sep 14, 2020 at 20:59

This is just another way of saying that for all but finitely many primes $$p$$, we have $$\alpha(p) = 0$$. In particular, the "largeness of the prime" depends upon the $$a$$ you are given to start with.

Specifically how this relates to your case. You are given $$a \geq 1$$, then, for the most unrefined bound, $$p > a$$ can never be factors of $$a$$.

• I think the thing the OP may have been missing, given that it's omitted from the notation, is that $\alpha(p)$ depends on $a$ as well as on $p$ Sep 15, 2020 at 10:00

What that means is that, for each $$n>1$$, you can write $$n$$ as$$2^{\alpha(2)}\times3^{\alpha(3)}\times5^{\alpha(5)}\times7^{\alpha(7)}\times\cdots,$$where $$\alpha(p)=0$$ if $$p$$ is sufficiently large; in other words, you only have $$\alpha(p)\ne0$$ for finitely many primes.

Note that here “sufficiently large” depends on $$n$$. For $$n=10$$, “sufficiently large” means $$p>5$$, whereas for $$n=74$$, it means $$p>37$$.

Does that mean large primes can never be factors of any integers?

Of course not: each prime is a factor of itself.

What's sufficiently large depends on $$a$$. it means for each $$a$$, there is some $$N$$ for which any $$p>N$$ is sufficiently large. We can take $$N$$ to be $$a$$'s greatest prime factor.

Here we formally tweak the books's statement of the theorem to add some insight.

Let $$P$$ denote the set of all prime numbers. Clearly $$P$$ is a well-ordered set.

If $$a$$ is an integer and $$a \ge 2$$ then any prime factor (and at least one exists) of $$a$$ must be less than or equal to $$a$$. Said another way, we can associate to $$a$$ a prime factor $${p^{max}_a}$$ satisfying

$$\tag 1 [\; {p^{max}_a} \mid a \;] \land [(p \in P) \land (p \mid a) \implies p \le {p^{max}_a}]$$

You can easily prove $$\text{(1)}$$ without the FTA.

Here is the fundamental theorem of arithmetic:

For every integer $$a \ge 2$$ there exist there exist one and only function

$$\quad \alpha: P \setminus \{p \in P \mid p \gt {p^{max}_a} \} \to \Bbb N$$

satisfying the following property

$$\tag 2 \displaystyle a=\prod_{p \in \text{domain}(\alpha)} {p^{\alpha{(p)}}}$$

This is equivalent to say that, for any fixed $$a$$, the product is for a finite number of terms

$$a=\prod_{p=p_1}^{p_N}{p^{\alpha{(p)}}}$$

with $$p_N \le a$$.