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Not sure if the following argument is circular in nature or breaks down in some other way.

We've defined the counting numbers $n \ge 1$ with the two familiar binary operations of addition and multiplication.

From this point all numbers will be greater than $1$.

We don't rush into things so we have no symbol for the number succeeding $1$, and also haven't defined the notion of divisibility or formulated Euclidean division.

Definition: A number greater than $1$ is said to be a prime number if it is not in the range of the multiplication operator, $(m,n) \mapsto m \times n$, when it is restricted to $\Bbb N^{\gt 1} \times \Bbb N^{\gt 1}$. All other numbers greater than $1$ are said to be composite numbers.

Proposition 1: Every composite number can be expressed as the product of primes.
Proof
Assume $n$ is a composite number that can't be expressed as a product of primes. Since it is a composite we can write $c = ab$. Now if both $a$ and $b$ can be written as a product of primes, then $c$ has such a representation. So, wlog, assume that $a$ can't be written as a product of primes. But then we have found a number $a \lt c$ that has no such representation.

By the method of infinite descent, we've reached an absurdity. $\quad \blacksquare$

Corollary 2: There exist both composiste and prime numbers.
Proof
The multiplication operator has a nonempty domain, so there exist at least one composite number $a$. Since $a$ can be written as a product of primes, there must also exist prime numbers. $\quad \blacksquare$

CopyPasteIt
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    Well, how do you start the induction? Surely you must begin by exhibiting a prime (such as $2$) no? And once you have done that then the existence of primes is obvious as is the existence of composites (e.g. $2^2$). – lulu Jul 21 '19 at 19:54
  • @lulu I put in a link to infinite descent. It is also on wikipedia. I believe this form of induction pertains here, – CopyPasteIt Jul 21 '19 at 20:10
  • Well, if you assume enough about the order maybe that works...but is it actually easier or clearer than just exhibiting $2$? – lulu Jul 21 '19 at 20:23
  • @lulu You would have to develop Capital Pi notation: $\quad {\displaystyle \prod _{i=1}^{k} a_i}\quad$ to express proposition 1. – CopyPasteIt Jul 21 '19 at 20:31
  • @lulu This is an exercise in logic - just for fun! – CopyPasteIt Jul 21 '19 at 20:38
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    or $\underbrace{a_1\cdot a_2 \cdots a_{k-1} \cdot a_k}_{\text {k times}}$ notations –  Jul 21 '19 at 23:49
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    Apparently your booting up included, along with the operations of addition and multiplication, the order relation $<$ and the method of infinite descent, but not the symbol $2$. Do you have the symbol $1$? If so, then having $+$ available, you can just use $1+1$ instead of $2$. – Andreas Blass Jul 22 '19 at 01:37
  • @AndreasBlass yes to all – CopyPasteIt Jul 22 '19 at 02:17

2 Answers2

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You need the well-ordering principle, which is equivalent to the induction principle: every non-empty set of natural numbers has a least element.

lhf
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  • I used the method of infinite descent, a form of induction using the well-ordering principle. So it is assumed to be part of the 'boot-up'.. Sorry - I should have made that more explicit. – CopyPasteIt Jul 22 '19 at 00:59
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The next number after 1, would need itself to be it's own square to avoid being prime by your own definition.

  • Square? Please carefully look over my definition. $\quad$

    $\Bbb N^{\gt 1} = {2,3,4,\dots}$ $\quad$ $2 \times 2 = 4, $ $2 \times 3 = 6, $ $2 \times 4 = 8, $ $3 \times 3 = 9, $ $3 \times 4 = 12, $ $3 \times 5 = 15$ $\quad$ The number $4$, $6$, $8$, $9$, $12$, $15$ are examples of composite numbers.

     – CopyPasteIt Jul 27 '19 at 14:35
  • yes the composite number produced when a natural number is multiplied by itself. –  Jul 27 '19 at 14:51
  • You need the multiplication of higher numbers to hit lower numbers, or you can't give 2 a least prime factor above it. So it's least prime factor is itself, and it can't have a higher one so it's greastest prime factor is itself, you then eitjer conclude it's its own square, have two factors greater than 1, or it's prime, which contradicts its assumed identity as composite. –  Jul 29 '19 at 00:36