Where Do Irrational Numbers Come From?

I was recently involved in a philosophical exchange elsewhere about some of Wittgenstein’s remarks about whether in a space containing rational numbers, the existence of ‘irrational points’ is ‘prejudged’. I also recently helped a friend to do some of her Analysis in problems, and had a lot of fun. I realised that I’ve missed doing proof-based math in the last year or so, and also that I’m rather rusty in my proof skills. So I’m going to go through my battered copy of Fitzpatrick’s Advanced Calculus (1st Ed.) again and do some of the problems, and maybe explain some of the important axioms and theorems.

Today we’ll start off with the axioms that let us show the existence of irrational numbers. We build calculus by first defining the natural numbers inductively, then by defining the integers and rational numbers ‘from’ the field axioms. (If we ’start’ with only natural numbers the field axioms give us all rational numbers.) In order to do calculus, we need a continuum, and hence we need irrational numbers. (This is not a logical deduction but a quick explanation for why mathematicians wanted to ‘get’ irrational numbers in their system.) It would obviously be tedious and probably impossible to define the irrational numbers one by one until we have all of them. But we can get plenty of [maybe all?] irrational numbers into the system by introducing the Completeness Axiom, thus giving us all the real numbers.

The Completeness Axiom says that every nonempty set of real numbers that is bounded above has a least upper bound. An upper bound, as its name suggests, is a number that is larger than all the members of the set. A least upper bound is the smallest number amongst all the upper bounds of a set. It is thus a sort of ‘lowest ceiling’ of the set. It may or may not be itself a member of the set.

I won’t go through the whole proof leading from the Completeness Axiom to the existence of irrational numbers (specifically, of irrational square roots). It’s rather involved and not particularly thrilling. But I can give you an intuitive feel for why the Axiom should have that consequence. It’s because we can conceive of sets like < , where . With the Axiom, this set must have a least upper bound. The proof involves showing that if the least upper bound of the set is b, then b²=c. And since c can be any number more than or equal to 0, this means that any positive real number has a square root. Since we know (from other independent proofs) that there some square roots are not rational, there exist irrational numbers.

The reason why b²=c can also be explained intuitively. For suppose that b² is larger than c. Then it would seem that there is ’space’ between b and the ‘ceiling’ of the set; b can no longer be the least upper bound. Suppose b² is less than c. Then b isn’t even an upper bound. So it must be the case that b²=c. (Clearly, I have left a lot out of the explanation, but I have sketched the main moves that one would have to make.)

So, just by assuming that all bounded sets have to have a definite ‘ceiling’, we get a whole bunch of irrational numbers.