How could there possibly be any doubt about the existence of the
square root of two? This is a very reasonable question, but one that
few people dare to ask when their lecturers solemnly stand in front of
them and prove that the square root of two exists, either by defining
x to be the supremum of all rationals r such that r^{2}<2 and
proving that x^{2}=2, or by applying the intermediate value
theorem (which itself confuses people by seeming too obvious to need a
proof).

Imagine that you did not know any advanced mathematics (if you actually don't, then that is fine) and were confronted by somebody who denied the existence of the square root of two. What would you say? The conversation might go something like this.

What do you * mean * when you say that the square root
of two doesn't exist?

I mean the obvious thing: if you take any real number x and square it, the answer is never 2.

But what about 1.41421356237309... ?

What about it? You haven't told me how the sequence continues.

Well, all I'm doing is taking the decimal expansion of the square root of two.

That sounds pretty circular to me.

You're right. I'm sorry. But it isn't really as circular as it
sounds. What I mean is that I am calculating the decimal expansion
of the real number x with the property that x^{2}=2.

That still sounds circular. Aren't you still assuming that a number with this property exists?

No, because I can tell you how to calculate the sequence of digits, and that will be my proof that the number exists.

Go on then.

Well, 1^{2}=1<2 and 2^{2}=4>2 so I know that
the the number must be one point something. Then by trial and
error I discover that 1.4^{2}=1.96<2 and
1.5^{2}=2.25>2 so the decimal expansion must start with
1.4. I then just continue this process: if I have calculated the
first 38 digits, say, then I try all the possibilities for the
39th, picking the largest one that results in a number whose square
is less than 2.

All right, I see what it is you are doing, and that it leads to an unambiguously defined infinite decimal. But what makes you call that a real number and what makes you so sure that it squares to two?

I don't understand the first question. Surely a real number is something with a possibly infinite decimal expansion.

That sounds fishy to me. I notice that you say * something
with * a decimal expansion, rather than just a decimal expansion.
So what is the actual thing, of which you calculate the expansion?

Well, it's just a number ... you know, something like 1,2,3,.. or 25/38 or pi or the square root of 3.

I notice you didn't have the courage to say the square root of 2! I'm beginning to think that you don't really have any idea what a real number is. You've given me a few examples, but you haven't said what they have in common.

I think you are being quite unnecessarily pedantic. Just think of the number line. It's got all the numbers on it, in order (ignoring the complex numbers for now). We know that some numbers are irrational, but we can still describe them, by means of their decimal expansions.

Describe what?

Positions on the number line, lengths, whatever you want to call them.

That's no good at all. What * is * this number line that
you assume I am familiar with? What is a length? Note that for the
second question you can't fob me off with an answer about rulers
and so on, because they only work to a certain accuracy.

All right, I take the point, but I still don't think it is a
serious problem. If it makes you feel better, I shall simply
* define * a real number to be a decimal expansion.

So when you say "the real number x" what you really mean is "the decimal expansion x"?

Well, it's not always what I think of when I talk about real numbers, but if you insist on a precise definition, then I can fall back on this one.

Does that mean that 0.999999.... and 1 are different numbers?

Oh yes, I forgot about that. Different decimal expansions correspond to different real numbers except in cases like 2.439999999.... equalling 2.44. So I suppose my definition is that real numbers are finite or infinite decimals except that a finite decimal can also be written as, and is considered equal to, the "previous" finite decimal with an infinite string of nines on the end. Happy now?

We've hardly started, because you haven't told me how to do arithmetic with these real numbers of yours, and you certainly haven't convinced me that there is a real number that squares to give 2.

Are you going to ask me how to multiply two infinite decimals together?

Yes.

Well, you just do it in the obvious way, by a sort of infinite long multiplication.

It sounds to me as though long multiplication would be a pretty accurate description of whatever process you have in mind, but I notice that you are somewhat vague about it.

Do I really have to go into this? Surely you can see how it would work.

No I can't.

Well, let's take the example of the square root of two. If you
take the numbers 1, 1.4, 1.41, 1.414 and so on, then their squares,
1, 1.96, 1.9881, 1.999396 and so on, get closer and closer to 2.
On the other hand, if you take the numbers 2, 1.5, 1.42, 1.415
and so on, their squares, 4, 2.25, 2.0164, 2.002225 and so on,
also get closer and closer to 2. And they don't just get closer
and closer, but they get * as close to 2 as you like *,
as long as you take enough digits. It follows that x^{2}=2,
where x is once again the infinite decimal resulting from the
procedure I told you before.

I don't see that it follows at all.

Oh come on. It's obvious that x^{2} can't be * less
* than 2, because if it was, then it would have a decimal
expansion which at some point was definitely less than 1.99999....
Suppose for example that the decimal expansion of x^{2}
began 1.9999736... This leads to an easy contradiction, because
x is obviously greater than 1.41421 (which is given by the first
few digits of x) and 1.41421^{2}=1.9999899241, which is
definitely bigger than any number that starts 1.9999736... A
similar argument shows, for any number y less than 2, that
x^{2}>y, and for any number z greater than 2, that
x^{2} < z. Therefore x^{2}=2.

I have several objections to what you have just said. Perhaps
the most fundamental one is how you are so certain that x^{2}
exists at all. You seem to assume it in your argument. You don't
really * calculate * x^{2} at all. The only calculation
involved is with finite parts of x. Looking at the eventual argument
you came up with, which is quite ingenious I suppose, it appears to
rest on the following assumption: if a and b are positive real numbers
and a < b, then a^{2} < b^{2}. That is basically what you
were using. I accept that this rule is true for finite decimals. However,
at no point did you tell me how to put infinite decimals in order, and
you have yet to justify that this rule applies to them. You also used
the famous * law of trichotomy * when you said that
x^{2} must either be less than two, greater than two or
equal to two and then ruled out the first two possibilities.
Finally, your argument seemed to be saying that * if *
x^{2} is anything at all, then it is 2.

Well I hope it's pretty obvious how to put infinite decimals in order.

I must admit, it is.

The law of trichotomy is then pretty easy to justify.

I suppose it's not that hard. Anyway, I'm prepared to accept
it because it seems to me that the main problem is with the
assumption that x^{2} exists in the first place.

I'm tempted to say that of course it exists, and I've shown you how to calculate it, but you'll probably tell me that I assumed that it existed in the course of my calculation.

Exactly.

But surely you see how I did the calculation? I mean, that
is how I work out x^{2}.

Work out what? What are you working out, if you're not even sure it exists?

But I * am * sure it exists. The argument I gave is
what I * mean * by calculating x^{2}.

Go on.

Well, I suppose you could say that I * define *
x^{2} to be 2, simply because the finite truncations
of x square to numbers that get arbitrarily close to 2. In
fact, what my argument showed is that if I want to define
x^{2} in some way, consistent with the principle that
0 < a < b implies that a^{2} < b^{2}, then I have
no choice but to define it to be 2.

I see. So you are basically defining the square root of two into existence. Smart move.

Well, it's more than that. Although I haven't given the
details, the ideas I have discussed can be used to show that
the real numbers, defined as decimals, can be added, multiplied
and so on, in a natural way, and that adopting this natural
way as a definition, one finds that there is indeed a number
x that squares to 2. (See here
for more details.) In other words, I didn't * just *
define the square root of two. Rather, I defined an entire
number system and showed, by which I mean actually proved,
that the square root of 2 exists in that system.

Oh, well if that's what you mean by the existence of the square root of two, then I suppose I accept it.