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This is a widespread intuitive argument, asserting that $\mathrm{card} ( \mathbb{Q})=\mathrm{card}( \mathbb{Q^c})$:

Between any two rational numbers there's an irrational one and vice versa. So $\mathrm{card} ( \mathbb{Q})=\mathrm{card}( \mathbb{Q^c})$.

How can one convince the learner that this argument is invalid?

Seirios
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Behzad
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    Would I be right in thinking that this line of argument were true that it would imply that you could construct a bijection between the 2 sets? – shuttle87 Dec 09 '13 at 07:39
  • It INTUITIVELY implies it! – Behzad Dec 09 '13 at 08:38
  • There are more irrationals between any two rationals than there are rationals between any two irrationals. – user2357112 Dec 09 '13 at 09:35
  • It doesn’t hurt to point out that the intuition amounts to picturing the reals as an alternating sequence of rationals and irrationals, as if the real line looked like this (with $q$ for a rational number and $x$ for an irrational number): $$\ldots q,x,q,x,q,x,q,x,q,x,q,x,q,x,q,x,q,x\ldots$$ If $\Bbb R$ really did look like that, then we really could just pair each rational with the irrational on its right. But of course it doesn’t, because no two real numbers are adjacent. This may not actually convince someone that the intuition is wrong, but it should help a bit. – Brian M. Scott Dec 09 '13 at 21:30

5 Answers5

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It is natural to have such misconceptions, and it is also hard for someone else to help you get rid of them. You use the right concepts instead repeatedly so that after sometime your wrong intuition seems absurd to you. As von Neumann says "In mathematics, you don't understand things. You just get used to them." That is very true. I have been there.

Sameer Kulkarni
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To get an intuition of the possibility that an element of a strictly smaller set can be found between any two distinct elements of a larger set, it might be easier to consider different objects where the individual facts are easier to prove.

Consider the complete binary tree $2^{<\mathbb{N}}$ (all finite binary sequences), and the family $2^{\mathbb{N}}$ of all infinite binary sequences. Now consider the union of these sets, $2^{\leq \mathbb{N}}$, and describe a linear order on this set as follows: $$x \preceq y \; \Leftrightarrow \; \begin{cases} x\text{ is an initial segment of }y, \text{or} \\ x(i) < y(i), \text{ where }i \in \mathrm{dom}(x) \cap \mathrm{dom}(y)\text{ is least such that }x(i) \neq y(i). \end{cases}$$

  • It is easy to show that if $x , y \in 2^{\mathbb{N}}$ are such that $x \prec y$, then there is a $p \in 2^{< \mathbb{N}}$ such that $x \prec p \prec y$. (Just take $p$ to be an initial segment of $y$ long enough to witness that $x \prec y$.)
  • A simple counting argument shows that $2^{<\mathbb{N}}$ is countable.
  • The standard diagonalisation method shows that $2^{\mathbb{N}}$ is uncountable.

So even though there are more infinite binary sequences (irrational numbers), between any two you can find a finite binary sequence (rational number).

(Of course, in this example the opposite, seemingly easier, "betweenness" property is false: there is no infinite binary sequence between $\langle 0 \rangle$ and $\langle 0 , 0 \rangle$, for example.)

user642796
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  • If you're feeling sinister, you could just consider $2^{\mathbb{N}}$, and take as your counterpart of the rationals all sequences which are eventually constantly $0$. – user642796 Dec 09 '13 at 09:01
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If the cardinality of the 2 sets was equal then there must exist a bijection between the 2 sets. I would suggest asking them to find such a bijection.

The other thing I would point out is that for the argument "Between any two rational numbers there's an irrational one and vice versa" to be valid it implies that there must be a cap on the number of irrational numbers between any two rational numbers. When I have talked to people about this I think this implicit assumption has caused people a lot of problems understanding the flaw in the argument. Specifically an injective function is not enough here but I think the troubles in understanding the flaw in the argument come from the distinction between an injective and bijective function here.

shuttle87
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Here is an overkill, but it works.

Show that every countable dense linear orders without endpoints is isomorphic to $\Bbb Q$. This can be done using just regular induction going back-and-forth between $\Bbb Q$ and the countable order.

Since $\Bbb R$ satisfies the completeness axiom, it cannot possibly be isomorphic to $\Bbb Q$, and therefore uncountable.

Finally, the union of two countable sets is countable, and so $\Bbb{R\setminus Q}$ is uncountable.

One can point out that this fails because given $q_1<q_2$ from $\Bbb Q$ and an irrational number $x$ between them, there is some rational numbers $p_1<p_2$ such that $q_1<p_1<x<p_2<q_2$. So what would be the rational value we assign to $x$?

Also, and perhaps more importantly, it is imperative to tell new students that their intuition sucks and severely underdeveloped.

This can be, and should be, a great reason for forcing them to work step by step from the definitions. Not to rely on intuition that they don't have.

After a few weeks, solving exercises and reading theorems, a good student will develop some intuition, weak students won't. If that has happened, then one can present the statement again and ask for the intuitive reason. It's unlikely to be given, but strong students might recognize that there are too many irrational numbers between two rational numbers, so we can't assign them like that.

Asaf Karagila
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And yeah, such counterintuitive results are to be very much expected when you are dealing with infinity. You'll see many such examples in due course. For now you just have think that there is no bijection between rationals and irrationals. The proof is standard.

Sameer Kulkarni
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