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!
 A: 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.  
A: 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).
A: 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. 
A: Just because a string will repeat (even infinitely often), doesn't mean that the entire string must be a repeating sequence.
You can create such a sequence as follows:
Start with 1.
Concateate the sequence with itself and add a 0. This gives us 110.
Concatenate the sequence with itself and add a 0. This gives us 1101100.
Concatenate the sequence with itself and add a 0. This gives us 110110011011000.
Repeat to infinity.
Due to the concatenation, you can see why given any sequence, we can find it again later on. In fact, we can find it infinitely many times.
You can also see why this will never repeat itself.
A: I am answering this question 4 years late simply 
because someone showed me this link, and I think 
the existing answers ignore aspects of your
question. 

It seems you are actually talking about randomly 
generating a real number $R$ in the interval  $[0,1]$ by writing 
$$R=0.A_1A_2A_3… = \sum_{n=1}^{\infty} A_n 10^{-n}$$
and by independently selecting the $A_n$ 
digits uniformly over $\{0, 1, …, 9\}$. That is fine. 
So let us first agree that your conclusions should be
restricted to such randomly generated numbers in the interval $[0,1]$ 
(not all numbers in the interval $[0,1]$). 
You correctly infer that (with probability 1) the decimal expansion of our number $R$ has the property that any particular sequence of a given 
length $x$ (where $x$ is a positive integer) occurs an infinite number of times. This property 
can be formally proven about such randomly generated 
numbers. Clearly this property does not hold for all real numbers 
since the real number 0 = 0.0000000 does not even have 
any “1” digits. 
It indeed follows that the decimal expansion of this random number $R$ will (with probability 1), contain all possible finite length sequences, and that each of these finite length sequences will be repeated (at randomly placed locations) an infinite number of times. 
However, it is a mistake to conclude this means the decimal expansion will eventually be periodic.  As Paul's answer notes, it means it will not be eventually periodic.  An eventually periodic sequence has some finite transient and some finite period, of the form: 
$$ 0.\underbrace{b_1 b_2 ... b_m}_{\mbox{transient}} c_1 c_2 ... c_p c_1 c_2 ... c_p c_1 c_2 ... c_p... $$
where $m$ is the length of the transient and $p$ is the length of the period. A number is rational if and only if its decimal expansion has this eventually periodic structure.  It can be shown that this eventually periodic structure prevents the expansion from containing every possible finite length sequence:  Indeed, if $c_1 \neq 0$, then the above does not contain the all-zero sequence $000..000$ of length $(m+p+1)$.  This is because $m+p+1>m$ and $m+p+1>p$.  So if you take any $(m+p+1)$ consecutive numbers in the above sequence, at least one of them must be $c_1$ (which is nonzero). On the other hand, if $c_1=0$, then the above sequence does not contain the all-1 sequence $111...111$ of length $(m+p+1)$.      Thus, with probability 1, our randomly generated number $R$ will not have an eventually periodic decimal expansion, and so it will be irrational.
A: None of the other answers answer this question by figuring out where the author probably made the mistake and pointing out that mistake. Some of them prove that there are irrational numbers and that might enable the author to catch their own mistake but I prefer to explain the mistake.
Even if all random number are rational, that doesn't necessarily mean all real numbers are rational but I'm guessing you were probably thinking that once you prove that all random numbers are rational, it's easy to show that all real numbers are rational. I think you made a mistake in thinking that all random numbers are rational by using the following argument:
"For any natural number n, there exists a natural number m such that for all x from 1 to n, the m + x$^{th}$ digit is equal to the x$^{th}$ digit. Therefore there exists a natural number m such that for all natural numbers n, for any x from 1 to n, the m + x$^{th}$ digit is equal to the x$^{th}$ digit. From that, it's easy to show that for some natural number n, the first n digits keep repeating over and over."
The problem with that argument is that the second sentence doesn't follow from the first.
