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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.

Sriram Natarajan
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  • You named the correct explanation: because $\mathbb R$ is uncountable. It is often confusing to know that between two rationals there exists an irrational and between two irrationals there exists a rational and yet their cardinalities differ. – Hagen von Eitzen Jan 31 '14 at 14:54
  • Look up Cantor's diagonal argument. :) I agree; it is counter-intuitive! It's a strange world we live in. – Eric Auld Jan 31 '14 at 14:55
  • After looking up Cantor's diagonal argument, you can modify it slightly to directly produce an irrational number as the "diagonal example". In this way, you can prove the uncountability of the irrationals directly, without using the countability of the rationals. – Andreas Blass Jan 31 '14 at 14:56
  • More precisely, Cantor's argument will show that $\mathbb{R}$ is uncountable, and then since $\mathbb{Q}$ is countable, it follows that $\mathbb{R}\setminus \mathbb{Q}$ is uncountable. – Eric Auld Jan 31 '14 at 14:56
  • @AndreasBlass can you give me more detail on that? Thank you – Sriram Natarajan Jan 31 '14 at 14:59
  • For example, you can use every second digit for diagonalizing (i.e., make the $2n$-th digit of your diagonal number different from the $2n$-th digit of the $n$-th number in a given list) and then choose the odd-numbered digits of your diagonal number in some non-periodic way, thereby ensuring that the number is irrational. – Andreas Blass Jan 31 '14 at 15:01
  • There are literally an infinite number of questions on this topic on this very site. Did you try searching first? (There is a Related list on the side, and you can find two threads which might answer your question right there. And there are more. Many more.) – Asaf Karagila Jan 31 '14 at 15:02
  • The way the question is asked you may want to add the intuition tag. – kleineg Jan 31 '14 at 15:07
  • Here are some discussions on the topic: This one and that one and also this one. There is one more, which I am going to suggest as a duplicate, as it is centered about an intuitive explanation, which seems to be what this question is all about. – Asaf Karagila Jan 31 '14 at 15:14
  • Thank you. I am new to this site. I couldn't find it in my initial search. The answers in those thread makes sense. – Sriram Natarajan Jan 31 '14 at 15:48

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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.

Paul Siegel
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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.

5xum
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  • Mathematically it makes sense. But intuitively it is very hard to digest. – Sriram Natarajan Jan 31 '14 at 15:07
  • That's true. But really, is it any harder to digest than the uncountability of $\mathbb R$? From my perspective, once you accept there are "a lot more" real numbers than there are rational numbers, it sort of makes sense that in order to get all the reals, you have to add more than just double the number of rationals to fill your bag. You have "a lot"of rationals, if you add more, you still have "a lot". You need to add "a whole lot more" to get to the reals. – 5xum Jan 31 '14 at 15:42
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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.

kleineg
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  • Thank you. But with order denseness of rationals implies between any two irrationals there exists a rational between them. Shouldn't that make them of same cardinalities? – Sriram Natarajan Jan 31 '14 at 15:11
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    One of the problems is that in order to generate a rational number "equal" to an irrational you would need to know all of the digits. No matter how far you go and what sort of pattern you think you see there is no pattern. You can think of it in terms of probability, in order to fit the pattern I have seen so far I need the next digit to be 3, $P(X_n=3) = \frac1{10}$ and since the number is irrational all the probabilities are independent so you have $\prod_{n=0}^\infty \frac1{10}$ which goes to $0$. – kleineg Jan 31 '14 at 15:22
  • The trick is that each digit is independent, if they were not then they would be rational. – kleineg Jan 31 '14 at 15:23
  • Nice explanation. By this argument between any two real numbers there are more irrational numbers than rational numbers . Roughly speaking, can this be considered a problem of infinities? – Sriram Natarajan Jan 31 '14 at 15:46
  • Late reply, but that is exactly it. There are far more irrationals than rationals. Even though you can find a rational between any two irrationals. Like you said, it has to do with the rank, or size, or the infinities. The irrationals have a higher rank of Infinity, it kind of relates to "depth", but since you can't order them you can't do the trick to count all the rationals. – kleineg May 09 '19 at 11:43