Peano Axioms and Arithmetic: How we define how we count

Caution

It's another expository post (*I mean it's better than just ranting, but Samyak please stop playing an expert*). The same cautions ahead(I'm no expert; I may be wrong). With that out of the way, let's begin:

Foreplay Introduction

So! How do you define natural numbers? As is the case with other topics in intro analysis, you'll think (I definitely thought) that this is so boringly obvious. I mean, it's just how we count right? But many things about $\mathbb{N}$ aren't so obvious. For example, how would you prove that $a+b=b+a$ for $a,b\in \mathbb{N} $? You'd say it's blindingly obvious, but commutativity is not obvious for many other things(try, for example, rotating a Rubik's cube and you'll get it). Note that you can't justify with examples (you'd have to give infinite examples to prove this property)

Now, this was a problem that presented itself in the mid-to-late $19^{th}$ century. People always assumed that numbers, in general, had some external source of existence. But, as you may have heard from other debates around this time(*cough*Hilbert, Godel*cough*), we wanted to derive everything in math from (almost) scratch(everything in math comes from axioms, everything!). After all, physical objects ain't gonna cut it when you work in advanced number theory.

 And so, Hermann Grassmann showed that many facts of Number theory could be derived just from two $``$things": The successor function(++) and the principle of mathematical induction(don't worry, I am gonna touch both today). Richard Dedekind(of Dedekind cut fame) then came up with a set of axioms(ah! exactly what we needed) and as far as I can guess, Peano acted like I am acting right now, and simplified everything and published a nice book with a long Latin name.

The Meat of it

(Trigger Warning to Indian Students: Please don't cancel me when I declare $0$ to be a Natural Number)

We at least need a few building blocks, and in this case, they're gonna be $5$ axioms, with $2$ $``$things" at the beginning to work with: $0$ and an incrementing function($++$)

Axiom 1: $0$ belongs to the set of Natural Numbers, $\mathbb{N}$

$$i.e. 0 \in \mathbb{N}$$

Now, that was a little bit obvious and necessary step. You need to have some starting point right? Up till now, our set of Natural Numbers looks like this $$\mathbb{N}=\{0\}$$

To populate the set, we make use of the incrementing function, which I've been talking about all the time. However, we'll define this function in a slick way, using the second axiom:

Axiom 2: If a number  $n$ belongs to the set of Natural numbers, then

$ f(n)=n++$ belongs to Naturals too

$$i.e. if\ n\in \mathbb{N}, \Rightarrow f(n)=(n++)\in \mathbb{N}$$

Now, by this logic, our set becomes

$$\mathbb{N}=\{0,0++,(0++)++,.......\}$$

which we, for our convinience define as:

$$0++=1; (0++)++=1++=2;....$$

$$\Rightarrow \mathbb{N}=\{0,1,2,3,...\}$$

Note that, we are defining $`1',`2',...$ as  the numbers $`0++',`(0++)++',...$

It seems as if our task is done and this is of course what we think of as Natural Numbers. But there are still a few gaps to close.

Consider for example, that we define $2++=0$($=3$ by definition) and so $4$ becomes $1$, $5$ becomes $2$ on and on and on...

We don't even need to look too far in Math jargon to even look for this scenario. Take your nearest computer and start writing a C program to iterate infinitely, and wait until you get an *overflow*$(2^{31}=0)$(this teeny-tiny problem ended my non-existent career in competitive programming)

So, to fix things up, and as a condolence to my CP chapter, I introduce the third axiom

Axiom 3: $0$ is not a successor of any Natural Number

i.e.

$$n++\neq 0 \forall n \in \mathbb{N}$$

We can now confidently assure you that $3$ is not $0$(nor is any other non-zero natural number)

However, even with this axiom in place, we can still have weird and wacky versions of natural numbers. For eg. consider that we have a $``$ceiling" at $5$ i.e. $5++=5$($=6$, by definition). Thus, $7=5,8=5...$

DUH!!!SAMYAK!! STOP TROLLING!!!!!!

  (My feelings when I was reading this)

Try as you might, you can't disprove that the set I described above isn't $\mathbb{N}$. So, in the spirit of the last axiom, we axiomatize this problem away:

Axiom 4:Different natural numbers must have different successors

i.e.

$$If\ n,m \in \mathbb{N} \ and\ n\neq m, \Rightarrow n++\neq m++$$

which can also be stated as:

$$If\ n++=m++, \Rightarrow n=m$$

Phew, now we're done... or are we?[Play vSauce music]

It turns out, there can still exist, in the words of Terence Tao, $``$rogue" elements(basically just extra elements to the numbers which make sense to us here) in our oh so precious $\mathbb{N}$. For eg., consider this set:

$$\mathbb{N}=\{0,0.5,1,1.5,2,....\}$$

Note that, we can once again prove that this doesn't violate any of the above axioms.

To remove this exoticism, we define an $``$axiom schema", which is basically just saying:$``$Hey, I have this cool new way of doing things, and you can use this method to make infinite axioms about stuff" and this is called the Principle of Mathematical Induction$^{TM}:$

Axiom 5: Let $P(n)$ be any property pertaining to a natural number n. Suppose that $P(0)$ is true,

and suppose that whenever $P(n)$ is true, $P(n++)$ is also true. Then

$P(n)$ is true for every natural number $n$

Now, since we have a way to select the true naturals from the mess above, we can ignore those $``$rogue" elements.

And now, all problems go away, finally.

Assumption: There exists a set $\mathbb{N}$ whose elements we call the Natural numbers which satisfy Axioms $1-5$

Why should you care?

Why should you go through mental gymnastics to prove something so obvious?

Well, aside from historical reasons, the fact that we can base the entirety of the mathematical subject called analysis and huge swathes of Number Theory, can be based on a few simple axioms about $0$ and a successor function. Isn't that amazing?

(Fun fact: you can now boss around any math teacher you encountered until $4^{th}$ grade)