Ytdyklly

Although the proof by contradiction is correct, your feeling of unease is fine, because the direct proof by induction is so much clearer:

Take an integer $N$. If $N$ is prime, there is nothing to prove. Otherwise, we must have $N = mn$, where $1 < m, n < N$. By induction, since $m, n$ are smaller than $N$, they must each be a product of primes. Then so is $N = mn$. Done.

The proof is not circular, the key is in the second sentence: Let N be the smallest positive integer with this property.

We are allowed to say a least $N$ exists because of the well-ordering principle.

I feel like this proof kind of presupposes the lemma.

Because it does.

It says so right in the first two sentences, which can be rephrased as:

Let $N$ be the smallest positive integer that cannot be written as a product of primes.

So yes, the proof assumes that all positive integers smaller than $N$ can be written as a product of primes.

This is OK, though, because it is trivially true for the smallest integers: 1, 2. The proof builds on that to infer that no such an $N$ exists where the lemma is not true.

I can definitely understand how this can feel a little off.

1) The lemma (as stated in the question) says all nonzero integers. Primes are integers and, by definition, cannot be products of primes. So, I think the lemma probably is actually more along the lines of: "all positive non-prime integers can be written as a product of primes".

2) Also, the statement "since 𝑚,𝑛 are positive and smaller than 𝑁 they must each be a product of primes" doesn't really explain why they must be a product of primes. Since, 𝑁 is the smallest positive non-prime integer that cannot be written as a product of primes (by supposition of the lemma), then 𝑚,𝑛 are either prime themselves or a product of primes (as they are less than 𝑁 and 𝑁 is the smallest number that isn't a product of primes). Either way, they will provide the primes necessary to create 𝑁, making 𝑁 able to be constructed as a product of primes.

Hopefully this helps to see why the proof by contradiction works.

An integer $n$ is said to be a composite if it can be expressed as the product of two integers $a$ and $b$ with $a notin {-1,0,1}$ and $b notin {-1,0,1}$.

An integers $p notin {-1,0,1}$ that is not a composite is called a prime number.

Recall the method of infinite descent used in mathematical proofs.

Suppose $m notin {-1,0,1}$ and it can't be expressed as a product of primes. If $m lt 0$ then it is certainly true that the positive number $-m$ can't be factored into primes. So the existence of $m$ allows us to assert that there are positive integers greater than $1$ that can't be factored into a product of prime numbers.

So using infinite descent, we have a minimal $n > 1$ that can't be written as a product of primes. In particular, $n$ can't be a prime. But then it must be a composite, and we can write

$quad n = st text{ with } s,t gt 1$

Note: The composite factors $s$ and $t$ must both be positive or negative.

If they are both negative, replace $s$ with $-s$ and $t$ with $-t$.

But then $s lt n$ and so it can be written as a product of primes. Similarly, $t$ can be written as a product of primes. But then $n$ itself is a product of primes. But this is not possible by our choice of $n$. So the initial assumption of the existence of $m notin {-1,0,1}$ with no prime factorization leads to a contradiction.

So every $n notin {-1,0,1}$ has a prime factorization.

I'm new to number theory. This might be kind of a silly question, so I'm sorry if it is.

No apology is necessary since your question is by no means silly. It is not at all surprising that you are puzzled by the cited exposition since it is incredibly sloppy. Kudos to you for reading it very carefully and noticing these problems.

Edit: I'd like to add that this textbook states that if $p$ is a prime number, then so is $-p$. That's where my confusion stems from. The textbook is A Modern Introduction to Classical Number Theory by Ireland and Rosen.

Let's examine closely that initial section on primes and prime factorizations.

On page $1$ begins a section titled "Unique Factorization in $Bbb Z$" where they briefly review divisibility of "natural numbers $1,2,3ldots"$ This leads to the following "definition" of a prime:

Numbers that cannot be factored further are called primes. To be more precise, we say that a number $p$ is a prime if its only divisors are $1$ and $p.$

This is imprecise. Is $1$ a prime by this definition? In the next paragraph we find

The first prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,ldots$

So $1$ is not prime. That agrees with modern conventions.

On the next page they segue into factorization in the ring of integers $Bbb Z$ where they write

If $p$ is a positive prime, $-p$ will also be a prime. We shall not consider $1$ or $-1$ as primes even though they fit the definition.

This poses a few problems. They now claim that $1$ does fit the prior definition of a prime, but they didn't list it above (or explain why it was excluded). Further it implies that $ p = -2$ is a prime but it doesn't fit the above definition (it has divisors $,pm1, pm 2,,$ not only $1$ and $p$). They don't give any definition of a prime integer (vs. natural).

Readers familiar with basic ring theory and factorization in integral domains will likely have no problem inferring what is intended (the notion of an irreducible or indecomposable element), but any careful reader lacking such background will likely be quite puzzled by these inconsistencies and gaps.

As such, it comes as no surprise that the following proof employing these fuzzy notions may well prove troublesome for readers unfamiliar with the intended notions.

Lemma $1.$ Every nonzero integer can be written as a product of primes.

PROOF $ $ Assume that there is an integer that cannot be written as a product of
primes. Let $N$ be the smallest positive integer with this property. Since $N$
cannot itself be prime we must have $,N = mn,,$ where $1 < m,, n < N.,$ However, since $m$ and $n$ are positive and smaller than $N$ they must each be a
product of primes. But then so is $N = mn.$ This is a contradiction.

The proof has many problems. It doesn't properly handle the (implied) prime factorization of $pm1$ and they forgot to handle the possibility that the counterexample is negative (w.l.o.g. reducing to a positive counterexample).

Considering all of the above problems, it is no wonder that you found this proof confusing.

The proof can be given in a more positive way by using mathematical
induction. It is enough to prove the result for all positive integers. $2$ is a
prime. Suppose that $2 < N$ and that we have proved the result for all
numbers $m$ such that $2 < m < N.$ We wish to show that $N$ is a product of
primes. If $N$ is a prime, there is nothing to do. If $N$ is not a prime, then
$N = mn,$ where $2 < m,, n < N.$ By induction both $m$ and $n$ are products of
primes and thus so is $N.$

Here they've reformulated the induction from negative form - an (infinite) descent on counterexamples (or a "minimal criminal") - into a positive ascent, i.e. into a complete (or strong) induction, and they give some hint about the reduction to the positive case, but still there is no handling of $pm1$. What is actually intended can be inferred from the next theorem they present.

Theorem $1.$ For every nonzero integer $n$ there is a prime factorization

$$ n, = (-1)^{e(n)} prod_p p^{a(p)}$$

with the exponents uniquely determined by $n$. Here $e(n) = 0$ or $1$ depending on whether $n$ is positive or negative and where the product is over all positive primes. The exponents $a(p)$ are nonnegative integers and, of course, $a(p) = 0$ for all but finitely many primes.

That explains how they handle the prime factorization of $pm1$ and the reduction to positive primes. With that in mind you should be able to fix the proof of the lemma.

As above, often when there is puzzling exposition in textbooks it can be clarified by reading a bit further to help infer what was intended. But - of course - that is no excuse for sloppy exposition.

A proof by induction has base case(s), Let m and n be said base cases. if it's true for m and true for n (not necessarily distinct), then because it's a product, it follows for mn. All the proof supposes, is N=mn for some base case ( primes or prime powers to start, in these cases) with m and n proved. Then it follows for N, which by saying N which
originally was consider the least element of a set of counterexamples, has one, it eliminates all possible least elements for the set we originally supposed existed.

There is a property of the natural numbers called well-order. A set is well-ordered if every non-empty subset has a least element. So given any property $P$:

1. The set of numbers for which $P(n)$ is false is either empty or has a least element.




2. Suppose there is some number $n_0$ such that $P(n_0)$ is false. If $n_0$ is the least such number, then obviously $P(n_0-1)$ is true ^[1] (otherwise $n_0-1$ would be a number for which $P$ is false that is smaller than $n_0$, and so $n_0$ wouldn't be the smallest such number).




3. Thus, if we can prove that there is no number $n_0$ such that $P(n_0-1)$ is true and $P(n_0)$ is false (i.e. "$neg exists n_0: (P(n_0-1) land neg P(n_0))$", then we have shown that the set of numbers for which $P$ is false has no least element.




4. "$neg exists n_0: (P(n_0-1) land neg P(n_0))$" is equivalent to "$forall n_0: (neg P(n_0-1) lor P(n_0))$", which is in turn equivalent to "$forall n_0: (P(n_0-1) rightarrow P(n_0))$".




5. Thus, if we can prove $forall n_0: (P(n_0-1) rightarrow P(n_0))$, then it follows that the set of numbers for which $P(n)$ is false does not have a least element. Since all non-empty sets of natural numbers have a least element, this set must be empty. That is, there are no numbers for which $P(n)$ is false, i.e. $P(n)$ is true for all $n$.

[1] There is also the possibility that $n_0-1$ isn't a natural number, which happens when $n_0=0$. Dealing with this possibility requires proving that $P(0)$ is true separately, which is why induction proofs require a base case.

So that's the concept behind induction proofs: if the proposition isn't true for all numbers, then there is a non-empty set of numbers for which it is false, which has to have a least element, which means that we have to go from "true" to "false" at some point. Inductive proofs thus look a bit like circular reasoning: you start assuming that the proposition is true, and use that to prove that the proposition is true. But what makes it non-fallacious is that you prove that the proposition is true for a later number by assuming that it's true for an earlier number.

The proof that you cite is using the same basic principle as induction, namely the well-order of the natural numbers, but it is skipping the one-by-one sort of process that induction proofs usually use. Instead of saying "If $P(n_0)$ is false, then $P(n_0-1)$ being true leads to a contradiction", it's saying "If $P(n_0)$ is false, then $P(n)$ being true for $n<n_0$ leads to a contradiction". Like a standard induction proof, it superficially looks like circular reasoning, but isn't, because it's proving that the proposition is true for $N$ using the fact that it's true for smaller numbers.