4
$\begingroup$

I'm reading "Understanding Analysis" by Abbott, and I'm confused about the density of $Q$ in $R$ and how that ties to the cardinality of rational vs irrational numbers.

First, on page 20, Theorem 1.4.3 "Density of $Q$ in $R$" Abbot states:

For every two real numbers a and b with a < b, there exists a rational number r satisfying a < r < b.

For which he provides a proof.

Later, on page 22, in the section titled "Countable and Uncountable Sets" he states:

Mentally, there is a temptation to think of $Q$ and $I$ as being intricately mixed together in equal proportions, but this turns out not to be the case...the irrational numbers far outnumber the rational numbers in making up the real line.

My question is: how are these two statements not in direct contradiction? Given any closed interval of irrational numbers of cardinality $X$, $A$, shouldn't be the case that we would have corresponding set of $X-1$ rational numbers, $B$, where each rational in $B$ falls "between" two other irrationals in $A$?

If this is not the case, how do we have so many more irrationals than rationals while still satisfying our theorem that between every two reals there is a rational number?

I know there are other questions similar to this, but I haven't found an answer that explains this very well, and none that address this (perceived) contradiction.

$\endgroup$
  • 8
    $\begingroup$ You touch on one of the most counter-intuitive aspects of infinity and the continuum. Even crazier is how for all $\epsilon>0$ you can put $\Bbb Q$ inside an OPEN set whose total size is less than $\epsilon$. You can't do that with the irrationals. $\Bbb Q$ really is small, even though it is dense. Get used to it. $\endgroup$ – Gregory Grant Jan 4 '16 at 22:05
  • 2
    $\begingroup$ Infinity is funny. What you say makes sense in a domain where each number can have a "next number" (a successor, in other words). Then, since between each pair of rationals is (at least) an irrational, and between each pair of irrationals is (at least) a rational, they would have to alternate. ¶ But in fact, there is no next number in either the rationals or the reals (not with respect to the usual relation $<$, anyway). $\endgroup$ – Brian Tung Jan 4 '16 at 22:18
  • 1
    $\begingroup$ What contradiction? You have not given any reason for your contention that the fact that there are rationals between any two irrationals means that there can't be "more" irrationals than rationals. Is that supposed to be obvious? $\endgroup$ – bof Jan 4 '16 at 22:20
  • $\begingroup$ @GregoryGrant: That is one of the weirdest things to me. I know it's true, and I can even construct a suitable set. What I can't do is visualize what that set looks like. It's much more alien to my intuition than the notion that $\mathbb{Q}$ is countable. $\endgroup$ – Brian Tung Jan 4 '16 at 22:21
  • 1
    $\begingroup$ The irrationals are dense, too.... See math.stackexchange.com/questions/46822/density-of-irrationals $\endgroup$ – symplectomorphic Jan 4 '16 at 22:28
3
$\begingroup$

Given any closed interval of irrational numbers of cardinality $X$, $A$, shouldn't be the case that we would have corresponding set of $X-1$ rational numbers, $B$, where each rational in $B$ falls "between" two other irrationals in $A$?

That will certainly be true if you change the word "interval" to "set" and stipulate that $X$ is a finite integer.

Consider a finite set $A$ containing $X$ distinct irrational numbers and nothing else, where $X \in \mathbb Z$. Then you can arrange the members of $A$ in increasing sequence, that is, write $A = {a_i}, 1 \leq i \leq X$ such that $a_i > a_{i-1}$ when $i > 1$. And then you can insert $X - 1$ rational numbers in the "gaps" between the consecutive members of ${a_i}$.

The problem with this in the more general case is that there are more than a finite number of irrational numbers in any closed interval in $\mathbb R$. In fact, there are more than a countable number of them. You can't just go and insert a rational number between each consecutive pair of irrational numbers, because there is no such thing as a consecutive pair of irrational numbers in an interval. In fact, take any two irrational numbers $r, s$ in the interval; there will be an uncountably infinite number of irrational numbers between $r$ and $s$.

We do indeed have a rational number $q$ that falls between $r$ and $s$, in fact a countably infinite set of such numbers; but we also have an uncountably infinite set of irrational numbers that fall between $r$ and $s$. There is no way to organize these numbers into an increasing sequence of alternating irrational and rational numbers, like this:

$$ r_1 < q_1 < r_2 < q_2 < r_3 < \cdots < r_{X-1} < q_{X-1} < r_X, $$

so any counting argument based on imagining such a sequence is incorrect.

$\endgroup$
  • $\begingroup$ I actually realized this with this exact line of reasoning about 10 minutes before you posted this, but an excellent explanation either way. To re-iterate another comment I made: just because you have "infinite" of something and "more infinite" of something doesn't mean you will ever encounter an interval of the "more infinite" thing that doesn't have an element from the "infinite" thing between it. They are both still infinite. At least, that's the way I'm thinking of it. $\endgroup$ – user1105224 Jan 4 '16 at 23:47
4
$\begingroup$

Given any closed interval of irrational numbers of cardinality $X$, $A$ shouldn't be the case that we would have corresponding set of $X−1$ rational numbers, $B$, where each rational in $B$ falls "between" two other irrationals in A?

Abbott's Theorem 1.4.3 is true, but could be better written in a way that didn't encourage thinking of numbers discretely. A more complete version of the theorem (which you should try proving) might be

For every two real numbers $a$ and $b$ with $a < b$, there exist infinitely many rational numbers $r$ satisfying $a < r < b$.

As a result the idea that you can line up the rationals and say 'here is the one between this irrational and that irrational, and here is the one between that irrational and that other one' doesn't hold up because it doesn't clearly specify a number. Even worse, an interval of $\mathbb{R}$ with some finite $X$ (ir)rationals is simply impossible.

A better way to intuitively see that there are more irrationals than rationals (but infinitely many of both) between any two numbers is the following.

Instead of points on the number line, imagine points on a half circle. Now draw a right triangle with legs of length $a,b\in(\mathbb{Z}\cup(\mathbb{R}-\mathbb{Q}))$ so that leg $a$ and the hypotenuse meet at the center of the half-circle. The angle of the triangle inside the semicircle is $\arctan(b/a)$. Drawing this picture out, it may be easier to convince yourself that there are many fewer rational points (corresponding to the case where $a$ and $b$ are integers, though the angle itself needn't be and probably isn't rational) than irrational ones (At least one of $a$ and $b$ not an integer).

$\endgroup$
  • 1
    $\begingroup$ This made me realize I didn't completely understand the definitions. I thought if there were more irrationals than rationals then by taking smaller intervals you would reach an interval no rationals between the irrationals. But there are infinite intervals between irrationals with infinite rationals in those intervals, and infinite intervals between even smaller irrationals with infinite rationals between them. So showing that irrationals are "more infinite" doesn't mean you'll run into an interval of irrationals with no rationals between them. $\endgroup$ – user1105224 Jan 4 '16 at 23:44
3
$\begingroup$

As some of the comments have indicated, there is no logical contradiction here, in the sense that you can prove both one thing and its negation. I think what you mean is that the state of affairs that you have laid out does violence to your intuition, which is another thing entirely.

I think it is quite common to have one's intuition thrown for a loop by this aspect of infinity. If these sets were finite, or even infinite in the same way as the integers (so that each value had a unique successor), then your argument would have merit. Between each pair of distinct rationals would be an irrational, and between each pair of distinct irrationals would be a rational. The only way these statements could both be true is if they alternated (and therefore their cardinalities would be essentially the same).

But that is not the way that either the rationals or the irrationals or the reals work. They do not have unique successors; you cannot name the next rational value after $\pi$. Equivalently, between any two distinct values, there is not only one rational value, but an infinite number of them. Also, between any two distinct values, there is not only one irrational value, but an infinite number of them. Thus, one cannot rely on that kind of straightforward one-to-one mapping to establish an equivalence between the cardinalities of the two sets.

Instead, one must use arguments such as Cantor's diagonalization to establish a difference between them, to show that the rationals are countable (can be placed in a one-to-one correspondence with the natural numbers) and the irrationals are uncountable (cannot be placed in such a correspondence), so that there are many more (infinitely more) irrationals than rationals.

I am not sure there is a way to see for the first time, intuitively, that there are infinitely more irrationals than there are rationals. After a while, it becomes more natural to countenance, but I wouldn't equate that familiarity with obviousness, necessarily.

$\endgroup$
1
$\begingroup$

Given any closed interval of irrational numbers of cardinality XX, AA, shouldn't be the case that we would have corresponding set of X−1X−1 rational numbers, BB, where each rational in BB falls "between" two other irrationals in AA?

You just proposed a map from pairs of irrational numbers to rational numbers (i.e. for irrationals $a < b$, choose $q$ in between them). This map will not be one-to-one, which is the flaw in your proof that there are as many rationals as pairs of irrationals.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.