In class, we talked a little bit about Peano’s axioms. Specifically, we talked about how the induction axiom allows us to, well, do induction. But what are Peano’s axioms, exactly, and what are their importance?
Essentially, Peano’s axioms are the axioms we use to form the Natural numbers. They are as follows:
1. There is a natural number 0.
2. For every natural number x, x=x.
3. For all natural numbers x and y, if x=y, then y=x.
4. For all natural numbers x, y and z, if x=y and y=z, then x=z.
5. For all a and b, if a is a natural number and a=b, then b is a natural number.
6. For every natural number n, S(n) is a natural number. S(n) is the successor function (1 is defined as the successor of 0, 2 the successor of 1, and so on). We would know it as n+1, but we haven’t defined addition yet so we don’t “know” that’s the case yet.
7. For every natural number n, S(n)=0 is false. In other words, 0 is not the successor of any natural number n.
8. For all natural numbers n and m, if s(n)=s(m), then n=m.
And finally, we have the induction axiom:
9. If K is a set such that 1 is in K and for every natural number n, if n is in K then so is S(n), then K contains every natural number.
Along with these axioms, one can define the addition operation as follows:
a. a + 0 = a
b. a + S(b) = S(a+b)
Let us calculate 5+3 to see how this works. According to the formula, this equals 5+S(2), which equals S(5+2), which equals S(5+S(1)), which equals S(S(5+1)), which equals S(S(5+S(0))), which equals S(S(S(5))), which equals 8.
Multiplication is defined similarly, as such:
Lastly, let us prove that induction is possible. Specifically, we’d like to prove that for some set of statements P (P(m), for example, might be the statement “m is odd or even”), if P(0) is true and if P(n) implies P(n+1), then P(n) is true for all natural numbers.
To the surprise of perhaps nobody, we use the induction axiom to prove this. Let K be the set of numbers n for which P(n) is true. Since 0 is in K and for every n in K, n+1 is also in K, K must have all the natural numbers, so P(n) must be true for every natural number.
Now, at first glance these axioms seem a bit excessive. I mean, they’re so obvious, right? But, as it turns out, they’re all extremely necessary. To show this, I will prove for you something that should sound absurdly trivial: that m+n=n+m. Note that from the definitions of addition, this isn’t apparent at all.
We’ll start with another necessary result: That (k+m)+n=k+(m+n) for all k,m,n. We can write this as P(n), where k and m are assumed to be any integers and n is a specific one we choose. First, let us prove it for the case P(0): (k+m) = k + (m). So far so good.
Now to prove that P(n) implies P(n+1), or that (k+m)+n=k+(m+n) implies (k+m)+S(n)=k+(m+S(n)). Well, (k+m)+S(n) = S((k+m)+n)=S(k+(m+n)) on the left side. On the right side, we have k+(m+S(n))=k+S(m+n)=S(k+(m+n)), so we are done.
Now, we must prove 3 more propositions: Q(m), which is that m+0=0+m for all m, T(m), which is that m+1=1+m, and R(n), which is for all m and n, m+n=n+m.
To start, we prove that 0+0=0+0. Even this is non-trivial: We have to use property 1 of the addition definition to show that 0+0=0, and then use axiom 2 to deduce 0=0.
Now, we prove that if m+0 = 0+m, then (m+1)+0=0+(m+1). The left hand side is simply m+1, which is s(m). For the right hand side we use P to rewrite it as (0+m)+1, we use our induction hypothesis to rewrite this as (m+0)+1, then we use the definition of addition to write this as m+1=s(m), thus showing Q(m) holds for all m.
Now, we must show T(m). Again, we start with our base case: 0+1=1+0. By Q(1), this is true.
Now, we must show that if m+1=1+m, then (m+1)+1=1+(m+1). On the left hand side this simply becomes S(m+1). On the left hand side, we use P to rewrite this as (1+m)+1, and use our induction hypothesis to rewrite this as (m+1)+1, which becomes S(m+1). So we are done.
Finally, we must prove that if R(n), then R(n+1), that is if m+n=n+m, then m+(n+1)=(n+1)+m. Using P, rewrite the left side as (m+n)+1=S(m+n). Using P, we can write the right hand side as n+(1+m). Using T, we can write this as n+(m+1)=n+S(m)=S(n+m), which we then use the induction step to write as S(m+n). And now we are finally done.
Notice how long it took to prove even such a trivial looking proposition. In a sense, that’s the trade-off: Peano’s axioms are incredibly general and assume very little, but in order to prove even basic results with them a lot of work is required. In the end though, it’s quite worth the trouble, because you can form (at least with natural numbers alone) the foundations required to do anything you would be doing anyways with just these simple axioms.
It’s important to note that it’s impossible just with Peano’s axioms to develop anything like the real numbers. In fact, all of Peano’s axioms (besides the induction axiom) are part of what is called First-Order logic, which is not enough to describe mathematics.
The development of Peano’s axioms was extremely important. In fact, they are still used today, nearly unchanged from when Peano developed them, and they are used in the research of very fundamental questions about mathematics, such as asking about the consistency and completeness of number theory itself. The fact that such a large amount of mathematics can be built off of such a seemingly small foundation is, to me at least, quite interesting.