Irrational numbers are uncountable Rational numbers are countable. They are also order dense. Intuitively shouldn't it make irrational numbers also countable. I have seen proofs explaining R is uncountable . This along with countability of rationals implies irrationals are uncountable. 
Can anyone provide me with good intuitive explanation why irrationals are uncountable? I am a beginner and any help would be greatly appreciated. Thanks.
 A: Let's push your intuition a little further.  Because there is a rational number between every two irrational numbers, perhaps your intuition tells you that maybe there should be an injective function $\mathbb{R} \backslash \mathbb{Q} \times \mathbb{R} \backslash \mathbb{Q} \to \mathbb{Q}$ which sends every pair of irrational numbers $(\alpha, \beta)$ to some rational number between them.  Certainly if such a function existed then $\mathbb{R} \backslash \mathbb{Q}$ would be countable.
But no such function can exist, and you can prove it by adapting Cantor's diagonal argument.  Suppose that a function $f \colon \mathbb{R} \backslash \mathbb{Q} \times \mathbb{R} \backslash \mathbb{Q} \to \mathbb{Q}$ of the sort described above exists, and let $A \subseteq \mathbb{Q}$ denote the range of $f$.  $f^{-1}(A)$ (which we know to be $\mathbb{R} \backslash \mathbb{Q} \times \mathbb{R} \backslash \mathbb{Q}$) is countable since $A$ is countable and $f$ is injective, so there is an enumeration $(\alpha_n, \beta_n)$ of $f^{-1}(A)$.  Arrive at a contradiction by constructing $(\alpha, \beta)$ such that the $n$th digit of $\alpha$ is different from the $n$th digit of $\alpha_n$ ($\beta$ can be anything).
To summarize the argument, the problem with your intuition that the order density of the rational numbers should imply the countability of the irrationals is that to "fill the gap" between every pair of irrational numbers with a rational number you have to reuse many rational numbers over and over.
A: I understand you already know of the Cantor diagonal argument? It is proof that real numbers are uncountable. With such a knowledge, it's simple to prove irrationals are uncountable. If irrationals are countable, then $\mathbb R = \mathbb Q \cup (\mathbb R\setminus \mathbb Q)$. This leads to a contradiction: because both sets here are countable, and the union of countable sets is countable, this means reals are countable.
This contradiction means one of the assumptions was wron. Since the only assumption is that irrationals are countable, this means they are uncountable.
A: You put your finger on the proof. As for the intuition, and this is not a proof substitute by any means, I think of the irrationals like this.
With a rational number, even if the decimal representation would require infinite digits to represent, there is a pattern. An example is $\frac13 = 0.33333....$  
With an irrational number however there is no pattern, you will have $0.333...4333...$ where the 4 is twenty billion digits in, and a 5 twenty billion digits in, and a 5 six trillion digits in... All of these are not only possible but have to exist in a small area around $0.33333...$
And look at the real number representation of a integer, you may have $1.0000$.... for that number to be an integer then all "infinite" digits must be $0$.
I hope this helps your intuition. 
