I've been studying the construction of the natural numbers, and I can't solve my own question, namely

Why is it necessary to use mathematical induction?

Let me clarify this. For example, we know that, for all $n,m \in \mathbb{N}$, then $n \cdot m = m \cdot n$. In order to prove this, we use mathematical induction, but when we think about $n,m \in \mathbb{R}$ (real numbers), in order to prove $n \cdot m= m \cdot n$, we don't need mathematical induction.

Why sometimes in a proof it's enough to take $x \in \mathbb{R}$, arbitrary number, but for natural numbers mathematical induction is necessary?

Anyone can help me, please?

I hope somebody can give me a hint in order to understand this question.

  • 12
    $\begingroup$ How do you define multiplication in the real numbers? It is usually defined in terms of the multiplication of rational numbers. How do you define multiplication of rational numbers? It is defined in terms of multiplication of the integers. How do you define multiplication of the integers? It is defined in terms of multiplication of the natural numbers. If you try to prove that multiplication of natural numbers is commutative by invoking that multiplication of real numbers is commutative, you are engaging in a circular argument. $\endgroup$ Nov 3, 2020 at 21:59

2 Answers 2


There are two ways to introduce the real numbers.

First way: a set of layers, starting from the natural numbers, then the integers, then the rational numbers, then the real numbers.

Second way: a set of axioms for the real numbers is taken and a structure satisfying them is assumed to exist.

In the first way, we need to prove commutativity of multiplication in the integers, then in the rational numbers, and then in the real numbers. This chain of proofs is based on the proof of commutativity of multiplication in the natural numbers.

In the second way, commutativity of multiplication is taken as an axiom. But we need to embed the natural numbers in the real numbers and this again requires induction and commutativity of multiplication in the natural numbers.


First of all, induction only really works on well-ordered, countable sets: things like $\mathbb{Z}$ and $\mathbb{N}$, where you can always define a "next" number. You can well-order $\mathbb{Q}$, but not in any way that makes induction useful: the most common well-ordering of $\mathbb{Q}$ starts $0,1,-1,2,\frac12,-\frac12,-2,3,\frac13,-\frac13,-3,4,\frac32,\frac23,\frac14,-\frac14,-\frac23,-\frac32,-4$ and continues on like that. (See if you can spot the pattern!) For any rational number, it really doesn't make sense to talk about the "next" rational number, and so induction isn't really useful.

$\mathbb{R}$ has all the issues of $\mathbb{Q}$, but it's not even countable. An induction on $\mathbb{R}$ that considers each real number separately requires all sorts of weirdness like the Axiom of Choice and maybe even some form of Transfinite induction in order to make it work, and it'll be a mess. Of course, there are ways around this, like slicing up \mathbb{R} into countably many pieces, but that's usually just standard induction with intervals.

Incidentally, though, we do need to establish that multiplication on $\mathbb{R}$ is commutative, and the way we do it depends on the way we're constructing $\mathbb{R}$. Sometimes, we poof $\mathbb{R}$ into existence using the magic spell, "the unique Dedekind-complete ordered field up to isomorphism" and then we don't need to do much work; the field axioms establish that the operation we call "multiplication" must be associative. Careful, though; if you're doing this for the first time, you may want to establish that this approach actually does give a unique, well-defined set, and that this set is indeed the "real numbers" that you've grown to know.

The other way to define real numbers is by "building them up" in some way from the rationals (which, incidentally, are defined by "building up" from the integers, which are in turn defined by "building up" from the naturals). There are a few ways of doing this, and some of them make showing commutativity easier than others do, but either way we do have to prove it. It just turns out that induction is usually totally useless, because by the time you've shown that this new set behaves the way you want it to behave (i.e. being Dedekind-complete, closed under certain operations, etc) you've already probably shown that the multiplication operator is commutative. If you're curious about how that's done, here's a paper that goes into depth on two of the most common ways to construct the reals, Dedekind cuts and Cauchy sequences.

The naturals, on the other hand, are far more fundamental; there's really not much that you can "build up" from in a rigorous and meaningful way, and there's really only one commonly accepted definition of the naturals as far as I know. In that definition, commutativity is a theorem, not an axiom. I suppose if you really tried, you could find some contrived definition of the naturals where commutativity of multiplication has to be an axiom, but it's so much simpler to just use Peano and induct.

  • $\begingroup$ Well, there’s also ordinal induction; induction works well with any well-ordered set, though the inductive step may take a slightly different form which is equivalent to the usual “next” induction of $\mathbb{N}$. $\endgroup$ Nov 3, 2020 at 23:39

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