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I'm learning mathematical induction and during this process I became impressed and confused by the manner in which several mathematicians speak about the logical properties of mathematical induction (for clarity, some utilize phrases like “complete induction”.) For example, Richard E. Hodel’s intro to mathematical logic details the method as “establishing” or “proving” that for any n ≥ 2, n is a product of primes. (Or, relatedly, let S( ) assert that something(?) is a product of primes. Mathematical induction is used as a means of proving that S( ) is true for all natural numbers.) I can’t help but maintain that, minimally, something logically peculiar and potentially wrong occurs here if we consider the words “prove” or “demonstrate” in the way logicians and philosophers consider them. One can summarize the dialectic found in Hodel’s intro as follows:

P1: The natural number 2 is a product of primes. (P1 follows from the fact that every number is a product of itself, and 2 is prime.)

P2: Each number is either prime or not prime (i.e., composite). (P2 is a principle of mathematics)

Assumption: Assume that every natural number is a product of primes. That is, assume, for the induction hypothesis proving that n + 1 is a product of primes, that S(2),...S(n) is true for all natural numbers. Assume that n + 1 is not prime.

Intermediary Conclusion 1: n + 1 is composite. That is, n + 1 = a x b. (P2, Assumption, and disjunctive syllogism)

Intermediary Conclusion 2: a and b are products of primes. That is, a = p₁ x…, pₜ and b = q₁ x…qₛ. (Assumption and principle of identity)

Intermediary Conclusion 3: n + 1 = p₁ x…, pₜ x q₁ x…qₛ (IC 3, IC 1, and principle of Identity)

Therefore: n + 1 is a product of primes.

The above conclusion along with the obvious truth that 2 is prime sufficiently establishes that all natural numbers greater than or equal to 2 are products of primes. However, I’m concerned that the “proof” for the ultimate claim that all n greater than or equal to 2 are products of primes isn't exactly logically viable. A pessimist may look at these steps and maintain, somewhat plausibly, that one’s established only conditional acceptance of the conclusion. That is, we can accept the conclusion, given acceptance of the relevant assumption. Yet, the assumption is the major claim induction aims at proving, so this would render mathematical induction in this case at least a question begging argument. In other words, in this case, induction assumed the conclusion p as a means of deriving some claim q functioning as evidence for p. This is question begging, a logical mistake according to standard understandings of logic. Or at least this is what it appears like to me.

KuraiKojo
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    "Assume that every natural number is a product of primes." is not the inductive hypothesis in a well written induction proof. (I do not have "Richard E. Hodel’s intro to mathematical logic" in front of me to know if the issue is with the book or your quotation.) – Mark S. Dec 24 '24 at 01:04
  • The induction step shows that if $S(1)$ through $S(n)$ are true, then $S(n+1)$ is true. The base case is to observe that $S(1)$ is true. Therefore, by the reasoning in the inductive step, we can conclude that $S(2)$ is true. Therefore, again by the reasoning in the inductive step, we can conclude that $S(3)$ is true. And so on. For every natural number $n$, we can conclude that $S(n)$ is true. – Karl Dec 24 '24 at 01:11
  • @Karl I have read the posted question that you linked to, and I disagree that the current posted question is a duplicate. See my answer. – user2661923 Dec 24 '24 at 01:23
  • "Assume that every natural number is a product of primes. That is, assume, for the induction hypothesis proving that n + 1 is a product of primes, that S(2),...S(n) is true for all natural numbers. Assume that n + 1 is not prime." This is a mess. You don't assume it is true for EVERY natural number; you only assume it is true for natural number *equal or less than $n$ and you don't assume $S(2),... S(n)$ is true for all* natural numbers; you assume it is true for all $k: 2\le k \le n$. – fleablood Dec 24 '24 at 01:39
  • "Intermediary Conclusion 1: n + 1 is composite. " Um... what if $n+1$ is prime? Are really trying to prove that there are no primes larger than $2$? ... Your error is in my comment above. Your assumption is not for all natural numbers-- it's only for all natural numbers up to $n$. As $n+1 > n$ we can't assume we know anything about it. – fleablood Dec 24 '24 at 01:44
  • I wouldn't say this proof is begging. I'd just say this proof is wrong. The above link is not a duplicate but it does explain your error. When you assume something is true for some $n$th case you aren't assuming it is true for all natural numbers, just for some $n$. And at this point of the proof the only case $n$ that you know it is true for is the first case $n=2$ only. If you read the linked post and understand it, you'll see what is wrong with this proof. (Which can be fixed!) – fleablood Dec 24 '24 at 01:55
  • @fleablood I disagree. The original poster's error was in wrongly concluding that the given proof assumes that every number ($\geq 2$) is the product of primes. The linked MathSE posting does not address the original poster's error. See also the comment left by the original poster, following my answer. – user2661923 Dec 24 '24 at 02:05
  • Maybe some basic logic (e.g. how to make universal generalizations) should be a prerequisite for learning proofs by induction. – Dan Christensen Dec 24 '24 at 18:18
  • @DanChristensen Your responses lack of utility primarily results from the fact that the relevant induction principle invoked in my proof isn't related to it. The principle you're referencing permits inferring a general statement from a particular statement (meeting certain requirements, of course). By contrast, the principle invoked in the informal proof outlined here doesn't invoke a particular statement; rather, it erroneously assumed the very generalization induction aims at proving. Hence, from what I can see, point doesn't "get at the issue". – KuraiKojo Dec 28 '24 at 13:40
  • @KuraiKojo You wrote: "Induction assumed the conclusion p as a means of deriving some claim q functioning as evidence for p." It seems you do not understand how one goes about proving $\forall x: [P(x)\implies Q(x)]$. First, you make a provisional assumption (a premise) $P(y)$, then derive $Q(y)$. Then you can infer the required result thereby discharging your original assumption. – Dan Christensen Dec 28 '24 at 17:42
  • @DanChristensen No, I understand precisely how to prove an indicative conditional--that is, show that some condition B functions as the essential condition for the occurrence of another condition A. Actually, depending on the nature of the derivation of Q(y) from P(y), one may or may not appropriately infer the universal quantifying statement associated with the relevant conditional. Since knowing this ambiguity strongly suggests knowing the nature of the derivation here, I claim I do know how to articulate this derivation appropriately. – KuraiKojo Jan 01 '25 at 04:00
  • @KuraiKojo I may have over simplified a bit, but there is no "ambiguity" in how universal generalizations are made. It gets a little more complicated if, subsequent to introducing the initial assumption $P(y)$, additional assumptions or free variables are introduced, but this is easily handled. Before generalizing, those additional assumptions must have been discharged. Also, those new free variables must not occur in $Q(y)$. This is common practice in mathematical proofs, including proofs by induction. – Dan Christensen Jan 01 '25 at 14:58

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You have misinterpreted the assumption on which the proof is based.

It is not being assumed that every number is the product of primes.

Instead, a prime number is defined as any positive integer $~p > 1~$ whose only factors are $~1~$ and $~p.~$

Then, every positive integer $~n > 1~$ is either prime, or it is not prime.

The label composite is used to describe any positive integer $~n > 1~$ such that $~n~$ is not prime.

So, by the definition of a prime number, if a number $~n~$ is composite, then there must exist a positive integer $~a : ~1 < a < n,~$ such that $~a~$ is a factor of $~n.~$ This implies that $~b = \dfrac{n}{a}~$ is also a positive integer such that $~1 < b < n.$

This also implies that $~a \times b = n.$

So, to restate the analysis used to prove that every positive integer $ ~\geq 2~$ is a product of primes:

Let $~S~$ denote the complete set of all positive integers $~\geq 2~$ that are not the product of primes.

Clearly $~S~$ is bounded below, and so (being a subset of the positive integers) must have a minimum element.

Denote this minimum element as $~M.~$

Then, clearly, $~M~$ can not be prime, and so must be composite.

Therefore, there exists $~a,b \in \Bbb{Z^+}~$ such that $~1 < a,b < M~$ and $~a \times b = M.$

However, since $~1 < a,b < M,~$ and since $~M~$ is assumed to be the minimum value of the set $~S,~$ you must have that $~a,b~$ are each the product of primes.

This implies that $~M~$ is the product of primes, which implies that $~M~$ is not an element in $~S.~$

This yields a contradiction.

Therefore, the set $~S~$ has no minimum element.

Therefore, the set $~S~$ must be the empty set.

In effect, the above analysis represents an argument based on the idea of descent, that every non-empty subset of the positive integers has a minimum element.

The original assertion could also be proved by analyzing forward, rather than backward.

That is, if you assume that every positive integer $~n~$ such that $~2 \leq n \leq N~$ is a product of primes, then you can use analysis similar to the analysis that I have given to establish that this implies that the positive integer $~(N+1)~$ is also the product of primes.

user2661923
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  • Thank you for your reply! Your proof, as far as I can tell, clearly avoids the logical mistake attributed to the original proof I reviewed and shows my mistake. Thank you again! – KuraiKojo Dec 24 '24 at 02:02
  • I should add that my original interpretation isn't flawed because it misrepresents an instance of question begging per se. If one did utilize the series of premises and assumptions found in my original post as a basis for accepting the relevant claim, one would likely question beg. That said, my original interpretation and critique is flawed because it potentially misrepresents the more adequate and logically precise variant of the proof, the variant Hodel most likely intended. – KuraiKojo Dec 24 '24 at 02:09
  • Your understanding of mathematical logic and set theory is clearly pristine. Could you recommend any books to better understand the subject at your level? – KuraiKojo Dec 24 '24 at 02:22
  • @KuraiKojo You are asking the wrong question. You want the pertinent math textbook that is the right book for you. I suggest that you print out this entire webpage, show it to a (college) math professor (rather than a high school math teacher), discuss your math background with the math professor, and then have the math professor recommend the right math textbook for you. – user2661923 Dec 24 '24 at 02:24
  • @KuraiKojo You might find this helpful: https://matheducators.stackexchange.com/questions/10021/why-are-induction-proofs-so-challenging-for-students/10057#10057 – Ethan Bolker Dec 24 '24 at 02:47
  • @EthanBolker thank you!!!! – KuraiKojo Dec 28 '24 at 13:42