-1

Which axioms are assumed to hold for any standard inductive proof of a formula using only nonnegative integers (e.g. of $1+2+\ldots+n=\frac{n(n+1)}{2}$, where $n\ge0$) ? For example, are Peano axioms assumed to hold for this proof by induction?

Astronut
  • 7
  • Why not $n \geq 0$? That aside, it depends greatly on your level when proving it. The principle of induction is one of the Peano axioms, so that could be used. But most people are on a much lower level when first encountering this proof, and not anywhere near thinking axiomatically about the naturals. – Arthur Dec 01 '16 at 18:08
  • I'm not sure what you mean, "in a standard inductive proof for only nonnegative integers". In general, you may need all of the Peano axioms. But for this particular statement, you may or may not need all of them, although you will certainly need some form of induction and you will probably need most of them. – Caleb Stanford Dec 01 '16 at 18:25
  • @Arthur ok, $n$ is nonnegative for the example. So are you saying that every such inductive proof inherently assumes Peano's axioms? – Astronut Dec 01 '16 at 18:26
  • @6005 Are you saying that you don't need any axioms for a simple inductive proof or not? – Astronut Dec 01 '16 at 18:29
  • @Astronut No, I specifically said you will probably need most of the axioms. – Caleb Stanford Dec 01 '16 at 18:30
  • @6005 Can you then specify which axioms are necessary? – Astronut Dec 01 '16 at 18:32
  • I certainly could. I don't have time to work it out now. But I didn't think that would answer your question since you only listed that induction proof as an example. – Caleb Stanford Dec 01 '16 at 18:33
  • @6005 I'm not trying to catch you out, but your remarks in their current form appear to me to be quite vague so far. If you believe you can provide an answer even for that specific inductive proof example, I'd be most grateful. – Astronut Dec 01 '16 at 18:39
  • 1
    Um actually, I am not sure how you would formalize this result in the language of peano arithmetic without going through a lot of hoops – mercio Dec 01 '16 at 18:39
  • 1
    @mercio Lol, good point. We don't have sigma notation. We can assume a function symbol f(n) such that f(n+1) = (n+1) + f(n), I suppose. Astronut, it's true what I said was vague. If I have a chance later and you clarify your question to be more precise I'll write an answer. – Caleb Stanford Dec 01 '16 at 18:41
  • @mercio Surely some set of axioms need to be assumed, don't they? – Astronut Dec 01 '16 at 18:41
  • no this is not even a question of axiom, but of language. "1+2+...+n" is not a well-formed term in that language (0, successor, +, and * only, iirc ?). – mercio Dec 01 '16 at 18:43
  • @6005 The question is now crystal clear and it only concerns that particular inductive proof. I look forward to your forthcoming answer. – Astronut Dec 01 '16 at 18:46
  • @mercio So are you saying that no axioms are necessary? – Astronut Dec 01 '16 at 18:48
  • What about the axiom "$1+ 2+ \dots +n = \frac{n(n+1)}{2}$"? This is enough to conclude that $1+ 2+ \dots +n = \frac{n(n+1)}{2}$ is true. – Crostul Dec 01 '16 at 18:50
  • @Astronut No, mercio is saying that even writing down the statement you're trying to prove in the language of PA is not something you can obviously do ("..." is not part of first-order logic). And it doesn't even make sense to ask whether a statement that can't be expressed in a given language can be proved in a theory in that language. Luckily, we can express such statements in PA with nontrivial effort; see my answer. – Noah Schweber Dec 01 '16 at 18:52
  • @Crostul I don't think that's correct. It wouldn't be an axiom if several proofs (see https://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-ldotsn/2028238#2028238) of it exist. – Astronut Dec 01 '16 at 18:55
  • @NoahSchweber I'm not asking whether a statement that can't be expressed Peano arithmetic can be proved in a theory in that uses Peano arithmetic. I'm asking whether any axioms are a necessary starting point for the inductive proof. – Astronut Dec 01 '16 at 19:41
  • @NoahSchweber The comments in the second answer about PA in its modern format not including addition and multiplication suggest that your answer is incomplete. The other remarks suggest that your answer may lack important details. – Astronut Dec 02 '16 at 20:44
  • 1
    @Astronut The other answer refers to second-order PA, which is a very different object; see here. Although second-order version came first, when mathematicians speak of "PA" they almost always mean the first-order version, partly because (as Dan's answer suggests) the second-order version is really just set theory in disguise. – Noah Schweber Dec 02 '16 at 20:49
  • 1
    In particular, second-order logic lacks a proof system - there's no good notion of what a "second-order proof" is. We can speak of a second-order sentence being a semantic consequence of some others if we fix an ambient set theory to do so, but then this reduces to the question of proving things in that set theory; and if that set theory isn't first-order, then we need to embed it in a larger set theory, and so forth. – Noah Schweber Dec 02 '16 at 20:50
  • 1
    The question of how much set theory is needed for second-order logic to work right is a more interesting one. While Quine essentially equated the two (I believe he is responsible for the line "second-order logic is set theory in sheep's clothing), others such as Shapiro (and to an extent, Vaanaanen) have argued that it is intermediate between first-order logic and background set theory. Still, since there is a second-order sentence which is valid iff the Continuum Hypothesis holds, second-order logic certainly subsumes some set theory, enough to make questions about axiomatizations break. – Noah Schweber Dec 02 '16 at 20:53

2 Answers2

2

First of all, an elaboration of Mercio's comment: in the language of $PA$, there is no direct way to talk about something like "$1+2+...+n$". That's simply not something $PA$ can express.

. . . without work. It turns out we can talk about finite sequencs in $PA$, and hence summations. But this takes real effort - Goedel was the first to do this in his proof of the incompleteness theorem, and it involved clever use of the Chinese Remainder Theorem.

So even expressing the theorems you want to prove in the language of PA is hard. However, once we've done so, it turns out that very few PA-axioms are actually needed.

PA consists of basic algebraic axioms (together called $P^-$), together with an induction scheme. By restricting attention to certain types of formulas in the induction scheme, we get a family of weak versions of PA - for instance, $I\Sigma_1$ is the theory gotten from $P^-$ by adding induction for $\Sigma_1$ formulas. $I\Sigma_1$ is enough to run essentially every "basic" induction argument, including the one you describe.

The study of exactly how much induction is needed to prove a given theorem is part of Reverse Mathematics.

Noah Schweber
  • 245,398
  • Note that you need some induction to even prove that something like "$1+2+...+n$" exists at all. – Noah Schweber Dec 01 '16 at 18:53
  • So, to give me a simple answer, are you saying that the inductive proof of the given summation formula depends (non)exclusively upon Peano axioms? – Astronut Dec 01 '16 at 19:20
  • Thanks for the answer. Would you possibly provide me with some links which detail $P^−$ and explain the meaning of $I\sum_1$. Am I right in understanding that the notation $\sum_1$ is used to express the standard summation formulae? – Astronut Dec 01 '16 at 19:25
  • @Astronut $P^-$ just says that $(\mathbb{N}, +,\times, <)$ is an ordered semiring. Meanwhile, $\Sigma_1$ refers to the arithmetical hierarchy. A formula $\varphi(x)$ in the language of arithmetic is $\Sigma_1$ if it has the form "$\exists k[stuff]$", where every quantifier in $[stuff]$ is bounded (= of the form "$\forall y<a$" or "$\exists y<a$"); $I\Sigma_1$ is then (roughly) $P^-$ together with, for each $\Sigma_1$ formula $\varphi$, the axiom "If $\varphi(0)$, and $\varphi(x)\implies \varphi(x+1)$, then $\forall x\varphi(x)$." – Noah Schweber Dec 01 '16 at 19:39
  • @Astronut I don't know what "depends (non)exclusively upon Peano axioms" means. Can you elaborate? – Noah Schweber Dec 01 '16 at 19:39
  • Firstly, are you saying that Peano axioms can be considered to be a necessary starting point for the inductive proof? Is it also possible to use ZFC arithmetic instead of PA as a starting point? – Astronut Dec 01 '16 at 19:49
  • @Astronut Of course we can use ZFC instead of PA (and I wouldn't call it "ZFC arithmetic" - it's set theory, not arithmetic) - ZFC is a much stronger theory than PA. Anything PA can prove, ZFC can prove as well. And there are lots of theories in between PA and ZFC - for instance, $Z, KP, Z_2, ATR_0$, just to name a few. – Noah Schweber Dec 01 '16 at 19:51
  • Ok, but just to confirm, are you saying that Peano axioms are sufficient assumptions on which an inductive proof can be based on? – Astronut Dec 01 '16 at 19:59
  • @Astronut Well, it depends on the inductive proof. More complicated proofs require more complicated axioms. But yes, for every proof by induction seen in (say) an undergrad math course, PA is enough. – Noah Schweber Dec 01 '16 at 20:09
  • Ok, PA is sufficient for an inductive proof using only $P^{-}$. – Astronut Dec 01 '16 at 20:20
0

It wouldn't be easy, but you should eventually be able to prove the above sum starting with only the usual five Peano Axioms (defining a successor function, but not addition or multiplication), and some set theory and logic.

To do so, you would first need to construct the addition and multiplication functions on $N$ as well as the finite summation operator. Then you should be able to formally prove your sum by induction. Note that the principle of mathematical induction is included in the Peano's Axioms.

To give you an idea of the magnitude of such a task, here is just my formal construction of the addition function on $N$ starting from Peano's Axioms: Construct Add on N (727 lines of formal proof). And that's probably only about a quarter of the job done toward your sum!

EDIT: To do even the simplest formal proofs by induction on the natural numbers, you will, of course, need to use the principle of mathematical induction from Peano's axioms which essentially define the set of natural numbers. If you want to prove things about addition, multiplication and summations (as in your example), you will have to define/construct each of these. Older versions of Peano's Axioms included definititions of addition and multiplication. If you are allowed the axioms of set theory, you do not need to include them in your definition of $N$; you can construct them, i.e. you can actually prove their existence.

Dan Christensen
  • 14,529