# Is induction something we take on faith?

I understand that in mathematics and logic we can continue to reduce things to simpler axioms, principles, and so on, and we have to "stop" at some point otherwise we're just going on forever. We eventually say that some axiom or principle is good enough, so that we accept it to be valid, true, useful, sensible, and so on.

That being said, my question is whether mathematical induction is one of these "fundamental" concepts we just accept, or if it follows from some even deeper or simpler concept.

Sometimes I see people say that it works because of the well-ordering principle of the natural numbers, but this doesn't satisfy me. In Tao's Analysis Vol I, we say $$m \leq n$$ (for natural numbers $$m$$ and $$n$$) iff $$m = n + a$$ for some natural number $$a$$. But then if I wanted to prove that any arbitrary set of natural numbers has a "least element" (the definition of the well-ordering principle), I'd be resorting to induction, the very thing I'm trying to "prove."

Does this mean the concept of induction is just something we all accept as one of those sufficiently simple, intuitive things that require no further proof, that comes from no simpler means?

• Another viewpoint is that induction is one of the defining properties of the set $\mathbb{N}$ - so "$\mathbb{N}$ satisfies induction" is a by-definition result. The place where axioms come into play is when we try to argue that $\mathbb{N}$ exists in the first place. It's also worth noting that we do often care about weakening the induction principle, such as in reverse mathematics, where induction and "comprehension" (= set existence principles) are analyzed to see exactly how much of each is necessary for various bits of mathematics. – Noah Schweber Nov 3 '18 at 19:02
• It depends on exactly how we construe the foundations of mathematics: is $\mathbb{N}$ a "given object" about which we have special axioms, or is $\mathbb{N}$ a "constructed object" in some broader mathematical universe (e.g. as in set theory) whose existence and properties are theorems of our more fundamental axioms? – Noah Schweber Nov 3 '18 at 19:04
• You're right that the principle of induction is more-or-less equivalent to the least-element principle in $\mathbb{N}$, and in an axiom system meant to capture the theorems of $\mathbb{N}$ it is necessary to include one or more "induction axioms" or "least-element axioms" (or some more powerful axiom from which it could be derived). – realdonaldtrump Nov 3 '18 at 19:48
• @user Mathematics is not philosophy so the meaning of "true" is not deep. If I want "the sky is purple" to be an axiom, then it's an axiom. Is it true in the real world? No. But it's a valid axiom. – Matt Samuel Nov 3 '18 at 20:53
• @DanChristensen Oh they are, but the OP's skepticism of induction makes me think they're not going to view ZF as non-controversial. – Noah Schweber Nov 6 '18 at 18:09

My viewpoint is the same as in Noah's first comment: for me, induction is part of the essence of what I mean when I talk about the natural numbers, so the thing I take on faith is not that induction is true but that the natural numbers "exist," whatever that means. Axioms don't tell you what to take on faith: they're a way for two people to agree that they're talking about the same thing.

Some people called ultrafinitists would in some sense deny that the natural numbers exist.

Here is a blog post which describes in detail the sense in which the Peano axioms are a way for two people to agree that they mean the same thing by "the natural numbers."

• @user525966: for example, when I write down the axioms of group theory I'm not telling you that any of those axioms are "true": I'm telling you what the word "group" means to me, so that if we're having a conversation about groups we can make sure that we're talking about the same thing. Similarly, when I write down the Peano axioms I'm telling you what the words "natural numbers" mean to me. Then I can say "also I believe that there is a thing that satisfies these axioms," which is a different statement from the statement of the axioms themselves. – Qiaochu Yuan Nov 3 '18 at 20:23
• Yes, that's one way to describe the modern understanding of what axioms are doing. A set of axioms is a definition of a certain class of mathematical objects, and once you've written down that definition you can ask whether anything satisfies it. I'm not sure how to answer your second comment any clearer than I've already answered it. If I say "definition: a brown cow is a cow that is brown," that is a different statement from "also I believe that there exists a brown cow." – Qiaochu Yuan Nov 3 '18 at 20:29
• Isn't this sort of saying the same thing though? We accept induction because it happens to be part of the natural number (Peano) axioms and those are what we take on faith? How is this (in effect) not equivalent to saying that we're also taking induction on faith? – user525966 Nov 3 '18 at 20:40
• It really isn't. What would you even mean by saying "I take induction on faith"? Induction is some sort of statement about the natural numbers. What do you mean by "the natural numbers"? One way to say it rigorously is by giving the Peano axioms, which include induction. So this is circular! – Qiaochu Yuan Nov 3 '18 at 21:05
• @user525966: but what's a chain? – Qiaochu Yuan Nov 3 '18 at 21:38

The question rather supposes that it is a straight choice, "give a proof from further axioms" vs "take on faith". But not so. We can offer reasoning which shows why induction is compelling, why it is bound up with the very conception of the natural series (so we are doing more than making a leap of faith in accepting the induction principle) -- but where this reasoning isn't a matter of deducing the induction principle from some more fundamental axiom.

Go back to basics. Suppose we want to show that all natural numbers have some property $$P.$$ We obviously can't give separate proofs, one for each $$n$$, that $$n$$ has $$P$$, because that would be a never-ending task. So how can we proceed?

Suppose we can show that (i) $$0$$ has some property $$P$$, and also that (ii) if any given number has the property $$P$$ then so does the next: then we can infer that all numbers have property $$P$$. Using $$\varphi$$ for an expression attributing some property to numbers, we can put the principle like this:

Given (i) $$\varphi(0)$$, and (ii) $$\forall n(\varphi(n) \to \varphi(n + 1))$$, we can infer $$\forall n\varphi(n)$$.

And the headline question seems to be: Why are arguments which appeal to this sort of principle good arguments? Is that just a matter of faith?

Well, suppose we establish both the base case (i) and the induction step (ii). By (i) we have $$\varphi(0)$$. By (ii), $$\varphi(0) \to \varphi(1)$$. Hence we can infer $$\varphi(1)$$. By (ii) again, $$\varphi(1) \to \varphi(2)$$. Hence we can now infer $$\varphi(2)$$. Likewise, we can use another instance of (ii) to infer $$\varphi(3)$$. And so on and so forth, running as far as we like through the successors of $$0$$ (i.e. through the numbers that can be reached by starting from zero and repeatedly adding one). But the successors of $$0$$ are the only natural numbers. So for every natural number $$n$$, $$\varphi(n)$$.

The arithmetical induction principle is underwritten, then, by the basic structure of the number sequence, and in particular by the absence of 'stray' numbers that you can't get to step-by-step from zero by applying and reapplying the successor function.

Now, the absence of 'stray' numbers (non-inductive numbers, if you like) isn't a matter of guesswork or faith. It isn't that we have a clear conception of the natural numbers which leaves it an open question whether there are natural numbers which aren't successors of zero -- and we have to take a leap into the dark and hope for the best in judging that there aren't any such! Rather, in elucidating what we mean by the natural number series (and distinguishing it e.g. from longer series of ordinals) we explain, precisely, that what we are after are the numbers that you can get to step-by-step from zero by applying and reapplying the successor function. And then, as explained, the induction principle can be seen to be just an elaboration of that understanding of what counts as the natural numbers.

• I am honestly really frustrated. I've read this answer like seven times and I just don't get it. Everyone seems to say no, it's not on faith, but then the actual explanations seem to suggest, "but it actually kind of is" – user525966 Nov 4 '18 at 23:52
• @user525966: The short answer: Yes it is on faith. You either need to assume induction 'inside the system' or you need to assume induction 'outside the system' (namely in the meta-system) in order to conclude that induction 'inside' is safe/sound. No induction at all gets you nothing. =) – user21820 Nov 5 '18 at 7:43

Peter Smith's answer is correct, but it seems that you do not grasp logic well enough to understand it. So I'm going to explain it differently:

Suppose you believe that there is such a thing as the counting number $$1$$ (whether as a concrete encoding in some physical medium such as a computer, or whether as an abstract notion), and that you can repeatedly add $$1$$ to itself to get more counting numbers, denoted using "$$+$$" for addition. So you can for example get the counting numbers $$1$$ and $$1+1$$ and $$1+1+1$$ and so on. Suppose you also consider as counting numbers the entities that you can obtain by this process. This means that for someone to convince you that they have a counting number $$n$$, they have to literally show you that it is of that form. For convenience we also include "$$0$$" as a counting number such that $$0+1 = 1$$.

Assuming those beliefs, you can analyze what induction says (when interpreted to be an assertion about counting numbers). Given any property $$P$$ of counting numbers that you can describe, induction asserts that if you can prove that $$0$$ satisfies $$P$$, and you can prove that ( for any counting number $$n$$ satisfies $$P$$, it also holds that $$n+1$$ satisfies $$P$$ ), then you can conclude that every counting number satisfies $$P$$.

Why should we accept this? Suppose you do not assume induction. Then all you can prove is: $$\def\nn{\mathbb{N}} \def\pp{\mathbb{P}} \def\imp{\Rightarrow}$$

$$P(0)$$.

$$\forall n \in \nn\ ( P(n) \imp P(n+1) )$$.

$$P(0) \imp P(0+1)$$.   [$$\forall$$-elim]

$$P(0+1)$$.   [$$\imp$$-elim]

$$P(1)$$.

$$P(1) \imp P(1+1)$$.   [$$\forall$$-elim]

$$P(1+1)$$.   [$$\imp$$-elim]

$$P(1+1) \imp P(1+1+1)$$.   [$$\forall$$-elim]

$$P(1+1+1)$$.   [$$\imp$$-elim]

...

You can obviously see that for any expression $$E$$ that represents a counting number, you can prove $$P(E)$$. So it is safe (based on your beliefs) to conclude "$$\forall n \in \nn\ ( P(n) )$$", but you cannot make this conclusion without induction! You can check for yourself that none of the rules in any deductive system permit you to 'go outside' the system and observe what you can prove and then 'go back inside' with some external conclusions.

That is what induction gives; the ability to transfer a certain particular kind of meta-logical reasoning into the system. Which kind? Exactly the above kind. That is why we have one induction axiom for each property $$P$$. Also, for the above reasoning to be meaningful, we must be able to explicitly write down $$P$$. For more detail, see this post.

Now that I have explained why induction actually makes sense and is not some ad-hoc assumption, your question of whether it is fundamental still remains. At this level, it suffices to say that induction is fundamental and cannot be non-circularly justified. See this post on the 'circularities' in mathematics, which mentions repetition as one of the main 'circularities'. Repetition can be said to be the core of the notion of induction. If you don't understand repetition, no amount of repeated explanation can help you to grasp it. =)

But if you want, there is a concrete justification of why induction cannot be non-circularly justified. The axioms of PA$$^-$$ are satisfied by not just the counting numbers but also the collection $$\pp$$ comprising the zero polynomial plus integer polynomials whose highest-degree term has positive coefficient. You can check that $$\nn$$ satisfies "every element is either even or odd", but $$\pp$$ does not. Specifically:

"$$\forall n \in \nn\ \exists k \in \nn\ ( n=k·2 \lor n=k·2+1 )$$" is true. (PA proves this, using induction.)

"$$\forall n \in \pp\ \exists k \in \pp\ ( n=k·2 \lor n=k·2+1 )$$" is false. (Exercise: Find a counter-example.)

Furthermore, and crucially, PA can prove this statement (for $$\nn$$). So we have an explicit example of a basic fact that cannot be proven by PA without using induction.

Of course, a follow-up question would be whether we only need a finite number of induction axioms. In the case of PA, the answer is no, but this fact cannot be explained simply. The gist of one explanation is to show that PA can prove the consistency of every finite fragment of PA, and hence by Godel's incompleteness theorem no finite fragment of PA is as powerful as the whole.

You can ask similar questions about what are the key philosophical assumptions that underpin stronger and stronger foundational systems. For a brief sketch from PA to ACA, see this post and the linked article by Peter Smith in the comments. Some logicians get uncomfortable somewhere between ACA and full second-order arithmetic Z2, due to the apparent circularity in impredicative definitions of subsets of $$\nn$$. But even if one is comfortable with iterating powersets, there are still further assumption you need to take on faith (without any non-circular justification as of today) before you can get to ZFC, such as full replacement. Alternative foundational systems will have their own share of (and potentially different) philosophical assumptions.

Is induction something we take on faith?

If you will grant the existence of an infinite set $$M$$ (perhaps not quite so large a leap of "faith"), then you can construct, i.e. prove the existence of the set of natural numbers $$N$$ (as a subset of $$M$$) on which all of Peano's Axioms, including Induction, will hold.

Using the Dedekind's definition of an infinite set, with $$M$$ being an infinite set, we have a function $$S: M \to M$$ and $$n_0 \in M$$ such that:

$$\forall a,b \in M:[S(a)=S(b) \to a=b]$$

$$\forall a\in M: S(a)\neq n_0$$

Using the axioms of set theory, we can then construct $$N\subset M$$ such that:

$$N=\{ n_0, \space S(n_0), \space S(S(n_0)), \space S(S(S(n_0))) \cdots \}$$

More formally:

$$\forall a: [a\in N \iff a\in M \land \forall b\subset M: [n_0\in b \land \forall c\in b: [S(c)\in b] \implies a\in b]]$$

(You can substitute $$0$$ for $$n_0$$ if you are comfortable using constants in that way. It shouldn't cause any problems.)

Using only the rules of FOL, we can also then derive each of the Peano Axioms (see my formal proof, commentary in blue font):

PA1: $$n_0 \in N$$

PA2: $$S: N\to N$$

PA3: $$\forall a,b\in N:[S(a)=S(b) \to a=b]$$

PA4: $$\forall a\in N:S(a)\neq n_0$$

PA5: $$\forall P\subset N: [n_0\in P \space \land\space \forall a\in P:[S(a)\in P]\space \to \space P=N]$$

• Why not modify and un-delete your previous answer? – Noah Schweber Nov 6 '18 at 15:54
• @NoahSchweber This is completely different. The previous one was also deservedly down-voted. – Dan Christensen Nov 6 '18 at 17:04
• I down voted, since a non-cheating statement would be: Using only the rules of FOL, DC proof sets and a formalization of Dedekind's definition of an infinite set, we can also then derive – j4n bur53 Nov 9 '18 at 19:51
• @j4nbur53f If you have set of natural numbers then you also have a Dedekind-infinite set. Does it really need a proof???? – Dan Christensen Nov 9 '18 at 23:51
• Well Frege did the same, and Russel found a bug. If you stick to FOL (and ZFC), such bugs cannot happen. If you leave FOL, as you do in DC proof, you need to be very very careful in a formal system. I showed you already a bug of such systems, its from 1922 Über die Antinomien der Prinzipien der Mathematik , eudml.org/doc/183579 – j4n bur53 Nov 22 '18 at 16:51

A couple of interesting things that may help the OP understand how induction 'fit in'.

According to wikipedia,

The earliest uses of the method of infinite descent appear in Euclid's Elements. A typical example is Proposition 31 of Book 7, in which Euclid proves that every composite integer is divided (in Euclid's terminology "measured") by some prime number.

But if you look at

Who introduced the Principle of Mathematical Induction for the first time?

it appears that only in the past 500 or so years has the induction technique been recognized as a way of 'doing business'.

Yet the two concepts are equivalent to those studying mainstream mathematics.

Now Euclid did not have the terminology of today to view the natural numbers as an infinite set, but he knew how to define a prime number.

Did Euclid's Elements have an axiom for using the method of infinite descent? I doubt it. It was so 'obvious' that the technique must have been considered 'logic'.

I may be biased, but once a 'rigorous and formal' math system start grappling with the infinite in a serious way, the induction technique comes along as a 'free ride'.

Does this mean the concept of induction is just something we all accept as one of those sufficiently simple, intuitive things that require no further proof, that comes from no simpler means?

The Principle of Mathematical Induction (PMI), despite its lofty sound, is indeed based on "one of those sufficiently simple, intuitive things that require no further proof."

Here are some other examples of simple, intuitive things that we have noticed about the natural (counting) numbers over the centuries:

• $$0$$ is a number.
• Every number has a unique successor (the "next" number).
• Different numbers have different successors.
• $$0$$ is the "first" number. It is not the successor of any number.
• You can reach any number by a process of repeated succession if you start from $$0$$.

The last item might be more usefully restated in terms of sets and subsets as follows:

Let $$P$$ be a subset of the set of natural numbers $$N$$. Suppose that $$0\in P$$. Suppose further that if $$x\in P$$ then we must also have the successor of $$x$$ in $$P$$. Then every natural number must be an element of $$P$$.

Symbolically, we can write this as:

$$\forall P\subset N:[ 0\in P \land \forall x\in P: [S(x)\in P] \implies N\subset P]$$

where $$S(x)$$ is just the successor of $$x$$.

This, or some slight variation, is how The Principle of Mathematical Induction is usually stated in more advanced textbooks.

It can be used to prove all sorts of cool things that are true of every natural number without having to test every number individually--an impossible task. You have only to prove that something is true about $$0$$, and that if it is true about any number $$x$$, then it must also be true of $$x+1$$.

Note that PMI, as formally stated above, can hold even on finite sets, e.g. when $$N=\{0, 1\}$$ and $$S(0)=1$$ and $$S(1)=0$$. Assuming $$0\neq 1$$.

Note, too, that PMI cannot hold if $$N=\{0, 1\}$$ and $$S(0)=0$$ and $$S(1)=1$$ since $$1$$ would not be "accessible" from $$0$$.

• Peano didn't have the P subset N assumption. But I guess he had ∀a∈N(a in P => S(a)∈P) and not what you wrote. – j4n bur53 Nov 15 '18 at 13:58

The peculiar thing about induction is the way in which it makes claims that we could never, even in principle, verify directly. For example, the claim that no natural number is its successor. It's impossible for finite beings to check infinitely many individual numbers. I think that this is the sort of claim you are asking about. We could assume that induction is valid (on faith, let us say), but how would that be different from assuming (on faith) that no natural number is its own successor, as well as infinitely many other claims about the natural numbers?

There are even more basic principles that can be used to derive induction, but you must keep in mind that induction doesn't work on all collections. Essentially, induction is a rigged game. We only play it where we know that we will win. The justification for induction (no, it's not faith) is in the selection/construction of a collection for which induction works.

I'm going to obscure, at first, part of an argument for induction, in order to emphasize what I consider its cornerstone. It seems to me that we should believe in induction not so much because of our concept of the natural numbers (which I won't even mention for a bit), but more because of how we construct our collection upon which we apply induction. Yes, what we construct matches our concept, that is certainly no accident. But we don't assume our construction matches our concept.

Suppose that we have property of collections called 'special' and a specific collection 'MM' such that no proper sub-collection pp of MM is special: $$\forall pp: pp \subsetneq MM \to \lnot special(pp)$$ Next, suppose that we determine some particular subcollection BB of MM has the special property. (Any subcollction, either proper ($$\subsetneq$$) or MM itself ($$=$$).) $$BB \subseteq MM \land special(BB)$$ The only sub-collection of MM that BB can be is MM itself. This is, in my opinion, the essence of induction. The actual application of induction involves successfully determining that a property B(x) defining a sub-collection BB of MM defines a special collection BB. If BB is special, BB is MM, and everything in MM for which B(x) is true is just everything in MM. $$\forall x:(x \in BB \leftrightarrow B(x) \land x \in MM )$$ $$special(BB)\to\forall x:( x\in MM \to B(x))$$ This result, $$\forall x:( x\in MM \to B(x))$$, is what I imagine some might imagine to be a product of "faith". Note, though, what all we haven't talked about yet, pretty much anything to do with natural numbers.

Now I "reveal" that the 'special' property is the 'inductive' property. Define $$special(pp) \leftrightarrow 0\in pp \land \forall x:( x \in pp \to Sx \in pp)$$ Remember, MM is a collection such that its only special sub-collection is MM itself.

Consider the collection BB, where BB is defined as all the objects in MM which are not their own successor. $$0$$ is in BB, because $$\lnot(Sx = 0)$$. And, if x is in BB, then Sx is in BB, because $$\lnot(x = y)\to \lnot(Sx = Sy)$$. BB has the special property. BB can only be MM. But, by definition, every object in BB is not its own successor. Every one of infinitely many objects in MM is not its own successor. And it is not by faith that we know this, as you can see.

Ah HAH! one may well exclaim. But, do we not take on faith the cornerstone property? $$\forall pp: pp \subsetneq MM \to \lnot special(pp)$$ No, we do not take it on faith. We can construct MM as the intersection of all special collections. $$\forall x:( x \in MM \leftrightarrow \forall pp:( special(pp)\to x \in pp) )$$ Suppose that CC is a special collection which is a proper sub-collection of MM. There must be some element c of MM which is not in CC. But every element of MM is in every special collection -- by definition. c must be in CC. Contradiction. Therefore, a proper subcollection CC of MM cannot be special. Note that showing this doesn't require 'special' to be 'inductive', nor to have anything else to do with natural numbers.

Of course, we must be taking something on faith. I would give quantifying, allowing a variable to range over some domain, the Most Valuable Player award here. Quantifying over the sub-collections of MM, quantifying over the objects in MM, quantifying over some domain of collections that includes all the special collections.

And in conclusion. Induction is not something we need to take on faith. If we sometimes seem dogmatic or "faithful" with regard to its claims, I would say that's because we have trimmed our inductive claims back far enough, trimmed our inductive collection back far enough, that -- despite working with infinitely many numbers -- we know whereof we speak.

The principle of mathematical induction essentially rules out the possibility that there exists any "isolated" subset $$P$$ of $$N$$ that does not contain $$0$$. By "isolated," I mean that there exists no element of $$P$$ that is accessible by means of the successor function from an element of $$N$$ that is outside of $$P$$.

(Edit: Note that being isolated in this sense does not rule out access in the reverse direction, i.e. an element is mapped by the successor function from an element of $$P$$ to an element of $$N$$ that is outside of $$P$$.)

I will attempt to formalize these notions here.

Suppose $$X$$ is a set (possibly finite), $$f: X \to X$$ and $$x_0\in X$$. Then induction is said to hold on $$(X,f,x_0)$$ iff

$$\forall P\subset X: [x_0 \in P \land \forall k\in P:[f(k) \in P] \implies P=X]$$

And accessibility (my term) is said to hold on $$(X,f,x_0)$$ iff

$$\forall P \subset X:[P\neq \emptyset \land x_0\notin P \implies \exists k\in X:[k\notin P \land f(k)\in P]]$$

The negation of "accessability" holding $$(X,f,x_0)$$ is

$$\exists P\subset X:[P\neq\emptyset \land x_0\notin P \land \forall k\in X:[k\notin P \implies f(k)\notin P]]$$

In this case, we can think of $$P$$ as being "isolated" from the elements of $$X$$ outside of $$P$$.

It can be formally proven that induction holds on a structure iff accessiblity holds there.

Example 1

Let $$N=\{ 0, 1 \}$$, $$S(0)=1$$ and $$S(1)=0$$. Then both induction and accessibility hold on $$(N,S,0)$$.

Example 2

Let $$N=\{ 0, 1 \}$$, $$S(0)=0$$ and $$S(1)=1$$. Then neither induction nor accessibility hold on $$(N,S,0)$$.

In the first several books of Aristotle you can find a discussion of different forms of "priority". Among them is a statement concerning priority of existence. In this account, Aristotle remarks that there must exist one individual if, in fact, two individuals are the subject of discussion. It is this account of relation which seems to ground the directionality of succession in natural numbers.

But, the situation is far more complicated. In the same discussion, Aristotle speaks of genera being prior to species. This is a motivation for arriving at the identity of indiscernibles -- a principle denied by first-order logic. Yet, it remains relevant in the guise of the principle of indiscernibility of non-existents from negative free logic. As Dana Scott has observed, this principle grounds Frege's definition of the number zero. Since Frege's definition is essentially that found in Zermelo-Fraenkel set theory, it serves as a condition of well-fomedness for the iterative hierarchy.

A good example in the literature can be found in Moschovakis' work on formal recursion theory where all meaningless syntactic forms are set equal to one another and denoted as the empty partial function.

The reason for mentioning the two directions (from below with succession and from above with "least" classes) is that this accounts for the difference between first-order and second-order notions of induction. The first-order account depends upon a well-formedness condition to assert the individuality of its base element. Note, however, that such a claim is opaque to formalist accounts of foundations based on formal systems and interpretations.

At a much deeper level, there is the issue of discernibility within a logical domain. Hilbert acknowledges that well formed identity relations are a presupposition of formalist mathematics. So, it is difficult to find relevant discussion. One place to find such considerations is in Russell's "Principles of Mathematics". He distinguishes between being and existence. Then he makes the claim that all things with being can be counted. This is motivated by the logical problem of negative existentials. But it is also a well ordering principle.

In other words, Russell is suggesting that ordinal numbers may be viewed as generic names for a consistent global labeling for any logical domain. This is implicit in his distinction between a logical class and a mathematical collection. And, it is very nearly explicit in the quote from Bolzano he uses to explain his account of numerical conjunction as the ground of logical domains. Bolzano is clear that each object be denoted uniquely.

Today one speaks of expanding a language with constants so that each object of the domain has a denotation. The procedure is attributed to Abraham Robinson. Such expansions are usually represented using ordinal subscripting.

If you wish to get a sense of how induction arises naturally in mathematical discourse, look at the first four chapters of Moschovakis' book, "Elementary Induction Over Abstract Structures".

• While there are many interesting points here, I think the relation to the actual question above is rather oblique (and in particular Moschovakis' EIOAS is definitely not about induction's natural role in mathematical discourse, but rather its technical logical aspects, and in my opinion while extremely interesting and well-written isn't relevant at all to the OP). – Noah Schweber Nov 18 '18 at 17:29
• Interesting answer but not quite relevant to my exact question – user525966 Nov 18 '18 at 19:00