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

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.


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.