Non-existence of irrational numbers?

I realize the title of my question will probably cause the raising of some eyebrows, so let me explain. Not sure whether to file this under "math" or "philosophy". This also might be able to be explained away due to some possible misconceptions I have about irrational/transcendental numbers, so please correct me where I'm wrong, since I'm quite certain I am...but this is nagging me, so here it is:

Ok, so imagine a non-repeating decimal with an infinite number of digits in it. The numbers are random, but for the sake of argument, let's say the number starts with "0.2368210236..." and continues from there.

So I look at the first decimal digit - 2 - and think to myself "in the rest of this decimal, there are an infinite number of times that '2' shows up."

So I look at the first TWO decimal digits - 23 - and again, think to myself "due to the randomness and infinity of this decimal, there is statistically a 100% certainty that the numbers "23" show up, in that order, somewhere in the rest of this decimal.

Continuing this line of thought, I reach the conclusion that if I choose ANY number of digits (variable: x) that start off this infinite decimal number, then a) there is statistically a 100% certainty that particular sequence shows up somewhere else in the decimal, and b) that it shows up an infinite number of times.

If such is the case, then what happens as x approaches infinity? Would that not mean that, given enough length in the decimal, there would eventually come a point where the decimal starts repeating itself, ad infinitum? Would that not mean that what was originally thought of as an irrational number is actually rational? Would this line of thinking not apply to ALL irrational numbers?

I guess what I'm asking is this: could it be that there ARE no irrational numbers, just rational numbers whose scope and vastness outstrip our ability to see their rationality?

Alright, mathphiles; have at me!

Solution 1:

No. If you choose the digits independently and randomly, the resulting number is transcendental almost surely. (This follows from the fact that there are only countably many algebraic numbers, let alone rational numbers, and a countable subset of the reals has measure zero.)

You are confused about several things. Mathematicians have a precise and completely formal way of defining numbers that, in particular, does not depend on decimal expansions. Using these definitions, there are also precise and completely formal proofs that the resulting numbers are irrational or even transcendental.

Solution 2:

This is the problem Pythagoras faced. We once believed all numbers could be expressed as a ratio of two integers, hence the term rational number. The diagonal of a unit square is $\sqrt 2$ which is irrational. This is easy to see. Take two unit squares and cut them along their diagonals. You now have four right triangles whose legs are each equal to $1$. Arrange them into a square by putting the four right angles together. Since this one square is made from two unit squares, the whole thing has area equal to $2$. Therefore, the sides of the square which correspond to the diagonals of our original unit squares have length of $\sqrt{2}$. Therefore, this number is real. You may like neither limits nor non-repeating decimals, but irrational numbers exist.

Solution 3:

Let me answer your question with another question. What rational number contains ever sequence in its decimal expansion? It turns out none. Either the decimal terminates or it has some point where it becomes periodic (a terminating decimal could be seen as 0 repeating).