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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.

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    $\begingroup$ The "sufficiently large" depends on $a$. $\endgroup$ Sep 14, 2020 at 10:25
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    $\begingroup$ 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 $p$ $\endgroup$
    – marwalix
    Sep 14, 2020 at 10:26
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    $\begingroup$ In this case: Any prime bigger than $a$, for example. For any specific $a$ you might be able to find a better bound. $\endgroup$
    – lulu
    Sep 14, 2020 at 10:26
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    $\begingroup$ en.wikipedia.org/wiki/Eventually_(mathematics) $\endgroup$
    – user76284
    Sep 14, 2020 at 20:59

5 Answers 5

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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 $a$.

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  • $\begingroup$ 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 $p$ $\endgroup$ Sep 15, 2020 at 10:00
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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 $p>37$.

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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.

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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

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

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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 $p_N \le a$.

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