If an infinite sequence has no last element, why do we say the $\sqrt2$ is irrational?

Question

The arguments I've seen for certain square roots being irrational (like $\sqrt2$) boil down to proving by contradiction that if we assume p/q to be fully reduced we'll find that both p and q must be even (in the case of $\sqrt2$) and :boom: contradiction.

But if I allow p to have an infinite number of digits (a valid member of the integers) then it has no last digit and cannot be classified as even or odd. Right? The proof by infinite descent, for example, seems to assume we can't descend infinitely but why not? I must be missing something important here about the integers.

Similarly, why not define pi as the fraction 314159.../100000...? I believe both are valid integers.

I've often heard people loosely describe rational numbers as the set of numbers whose decimal representations terminate or repeat, but I don't see what limits the construction of an arbitrarily large integer by an infinite process. Lets for example attempt to construct the numerator in the quotient above:

Step 1: Construct the number 3 (1+1+1 = 3)
Step 2: Construct 31 (3+1+1...+1 = 28)
Step 3: Construct 314 (you get the idea)

It seems I could continue the process above indefinitely and produce a valid integer at every step. Because it is an infinite process any one step has a finite integer, but the number this process describes is unlimited.

Is the number this process describes then not an integer?

---

Update As commenters and answerers of this question have pointed out, my question assumes that a number with an infinite number of digits could be called an integer. That underlying assumption has confused some readers of my question so I've edited the above in an attempt to make the meat of my question clearer.

To that end, I'd like to point out that the essence of my question is either very similar to or a duplicate of A "number" with an infinite number of digits is a natural number?.

Answer 1

Neither $314159\ldots$ nor $100000\ldots$ are valid integers. Integers are characterized by being finite sequences of digits. Integers cannot be infinite; that's exactly what we mean by the word "integer".

The integers are counting numbers (and their negatives); that is to say, they are numbers one might use to count. You can't count to $100000\ldots$ - for one thing, there's no "previous" number to count from. The key idea is that the integers are the numbers we can "see"; I can show you $43$ sheep, and in principle I could show you $43000000000$ sheep, but I can't show you a convincing $\pi$ sheep. Rational numbers are numbers we can build using comparisons between the numbers we can "see"; Jack can have $1.5$ times as many sheep as Jill, but he can't have $\sqrt{2}$ times as many sheep as Jennifer, no matter what number of sheep Jennifer has. Allowing integers to have infinitely many digits would defeat this "concrete" idea of what an integer is, so we define integers to be only finite.

Answer 2

There's no such thing as an integer with infinitely many digits.

With a bit of ingenuity it is possible to define arithmetic on infinite strings of digits (it's easier if they are infinite towards the left, leading to "10-adic integers", which despite the confusing name are not really integers) -- but the result of this will not be what we call integers.

One definition of what a (positive) integer is, is:

• $1$ is a positive integer.
• If $a$ is a positive integer, then $a+1$ is also a positive integer.
• Nothing else is a positive integer.

Now consider the set of numbers that can be written down such they have a first and last digit. This certainly includes $1$, and adding $1$ to something in this set will again lead to a member of the set.

According to the definition above, this means that everything outside that set falls under the "nothing else is a positive integer" clause -- or in other words, every positive integer has the property that it has a first and last digit (when written down in base ten).

---

In your edited question you describe a sequence of integers, but having a sequence of ever-larger integers is by far not the same as having constructed one integer with infinitely many digits. Mathematics is in general wary of notions of "completed infinite process" -- they are very useful as a visualization technique, so you will often see that kind of description in informal discussions, but they can also lead to false results, so they are not accepted as arguments for truth unless it is clear how they can be reduced to precise definitions without any infinitary handwaving.

In calculus and analysis, this reduction-to-definition usually goes via the concept of limits, but that doesn't work in your case because not every sequence has a limit, and yours in particular doesn't. So, since the usual assumption doesn't work here, the onus would now be on you to explain how your infinite-process imagery reduces to concrete timeless facts. And you can't do that without somehow leaving the realm of what we call integers.

(The closest thing I can think of here would be the somewhat esoteric area of "non-standard analysis via ultraproducts", in which your process does make some sense and produces a "non-standard rational number whose difference from $\pi$ is infinitesimal but nonzero". Of course, the non-standard rationals of non-standard analysis are not what we usually call rationals either).

Intuitively though, you can think about this: You seem to say that your infinite process does not preserve the property of "having a first and a last digit" -- each element in your sequence has this property, but the thing you claim results at the end doesn't. On the other hand, you're tacitly assuming that your process does preserve the property of "being an integer" -- why is that? Once we accept that the process does not preserve all properties, there has to be something in particular about "being an integer" that makes it be preserved, before we can just assume it is.

Answer 3

But if I allow $p$ to have an infinite number of digits (a valid member of the integers) then it has no last digit and cannot be classified as even or odd. Right?

One has to be careful about numbers and their representation as a sequence of symbols.

Let me try to point out some issues with extension to representations with infinite many digits:

For our usual decimal representation of a non-negative integer with $m\in \mathbb{N}$ digits from $D = \{ 0, \dotsc, 9 \}$ and semantic $$ (\cdot)_{10} : D^m \to \mathbb{N} \\ (d_{m-1} \dotsb d_0)_{10} = \sum_{k=0}^{m-1} d_k 10^k $$

we need a last digit $d_0$.

In fact the meaning of the first digit $d_{m-1}$ of that word, as a multiple of $10^{m-1}$, we only know because we start assigning the powers of $10$ from the right end of the sequence of digit symbols.

An infinite sequence of digits ($\omega$-word) given as $$ 3\dotsb $$ which we interpret as an infinite sequence of digits starting with a $3$ digit symbol, is not useful, as it gives us no clue what $3$-multiple of some power of the base $10$ it is: Three tens? Hundreds? Bazillions?

If we would grow the digit sequence to the other side $$ \dotsb 3 $$ or use a less ambigious notation to specify the $\omega$-word, e.g. as word of the accepted language of a Büchi automaton, the situation, is not much better: If the sequence honours the convention of not using leading $0$ digits there must show up non-zero digits repeatedly (otherwise we had a finite sequence) and the extended semantic $$ (.)_{10} : D^\omega \to \mathbb{N} \\ ((d_k))_{10} = \sum_{k=0}^{\infty} d_k 10^k $$ would be infinite. Alas all integer numbers are of finite value, and there is no largest integer. So this is not a useful representation for an integer number.