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This question already has an answer here:

I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, so it seems like it should be split pretty evenly.

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marked as duplicate by Xander Henderson, ahulpke, David K, John Hughes, Parcly Taxel Feb 20 '18 at 2:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ You may want to read this question, which essentially asks the same as you do, but it is a bit more specific: When discussing infinite sets, if we say that two sets are of the same size, what we mean is a precise statement, namely, that there is a bijection between the two sets. The question I linked to suggests an erroneous way of building such a bijection (based on the same intuitive idea you describe that between two irrationals there is always a rational, and between two rationals there is always an irrational). $\endgroup$ – Andrés E. Caicedo Aug 23 '13 at 15:51
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    $\begingroup$ This might be one of those times that was made for John von Neumann's remark that in mathematics you don't understand things. You just get used to them. $\endgroup$ – MJD Aug 23 '13 at 16:02
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    $\begingroup$ I have faith that in mathematics, if you don't understand things, you just haven't found the right way of thinking about them yet. $\endgroup$ – Hammerite Aug 23 '13 at 19:23
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    $\begingroup$ Cantor's diagonal proof is pretty easy to follow and shows that there are uncountably many real numbers. There are only countably many rational numbers, so obviously there must be uncountably many irrational numbers. You note correctly that the rationals are a "dense" subset of the real number line, but they are still a countable set. $\endgroup$ – Stefan Smith Aug 24 '13 at 0:04
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    $\begingroup$ @Kaz, You can easily traverse an infinite list of infinite lists. Take the 1st element of the first list, then the 2nd element of the first list then the 1st element of the second, then the 3rd of the first, the 2nd of the second and the 1st of the third, and so on. A picture would clarify this, but I can't put pictures in comments. The point is that thinking of infinite lists of infinite lists seems plausible but is misleading. I would also suggest that the concept of infinities of different sizes is pretty well-established, and is not something you can just decide not to believe in. $\endgroup$ – Bennett McElwee Aug 24 '13 at 12:07
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I think the most intuitive explanation I have heard is to considering writing down a rational number in decimal form. This means that either it is a repeating decimal or a terminating decimal, for example $2.3737\overline{37}$ or $0.42$, which we will write as $0.4200\bar{0}$. Now, consider the probability of randomly writing down a number. So you have ten options every time you go to place a digit down. How likely is it that you will just "happen" to get a repeating decimal or a decimal where you only have zeros after a certain point? Very unlikely. Well those unlikely cases are the rational numbers and the "likely" ones are the irrational.

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    $\begingroup$ This is the best answer to the question that the OP asked in my opinion. $\endgroup$ – JP McCarthy Aug 23 '13 at 18:55
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    $\begingroup$ What an excellent way to think about it. You can even show that the probability of "rolling" a rational number is zero under the requirement of an infinite number of digits! $\endgroup$ – Greg L Aug 23 '13 at 19:22
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    $\begingroup$ However what are the chances that given a rational number you will have only $0$ after .? In any range $[n, n + 1)$ you have 1 integer and infinitely many rational numbers. Hence it would seem that there are more irrational then rational yet there is bijection. $\endgroup$ – Maciej Piechotka Aug 24 '13 at 8:09
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    $\begingroup$ The third possibility that would make the number rational is to get a repeating "set" of decimals, not just any one. The probability of repeating this set to an arbitrarily large precision given randomly-chosen independent digits is similarly small. $\endgroup$ – KeithS Aug 28 '13 at 19:48
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    $\begingroup$ @WhiteHouseFenceJumper This can help. $\endgroup$ – Pedro Jan 6 '16 at 4:47
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There is always a rational between two irrationals, and always an irrational between two rationals, so it seems like it should be split pretty evenly.

That would be true if there was always exactly one rational between two irrationals, and exactly one irrational between two rationals, but that is obviously not the case.

In fact there are more irrationals between any two (different) rationals than there are rationals between any two irrationals -- even though neither set can be empty one is still always larger than the other.

And yes, this really becomes less and less intuitive the more you think about it -- but it seems to be the only reasonable way mathematics can fit together nevertheless.

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  • $\begingroup$ Wouldn't there is exactly one irrational between two rationals be impossible? If $q_1<q_2<q_3$ were rational, and $r_1,r_2$ were irrational such that $q_1<r_1<q_2<r_2<q_3$ then we would have two irrationals between $q_1$ and $q_3$. $\endgroup$ – Stefan Hamcke Aug 23 '13 at 16:05
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    $\begingroup$ @Stefan: Yes, and that is why I say it is obviously not the case :-) $\endgroup$ – Henning Makholm Aug 23 '13 at 16:06
  • $\begingroup$ I wanted to say that one could formulate this as there is always exactly one $y\in I$ between any adjacent points in $Q.$ That is something which at least could be possible for some sets $Q$ and $R$. $\endgroup$ – Stefan Hamcke Aug 23 '13 at 16:11
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    $\begingroup$ @Stefan: Yeah ... though that just bumps the problem to the fact that there's no such thing as "adjacent points" in the sets we're looking at here. So in a sense it is true that there's exactly one irrational between to adjacent rationals, but only vacuously so because there are no adjacent rationals. $\endgroup$ – Henning Makholm Aug 23 '13 at 16:13
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The wording in the question, "it seems like it should be split pretty evenly", is quite appropriate. I think the lesson to be learned here is that sometimes when our intuition says something "seems like" it should be true, a more careful analysis shows that intuition was wrong. This happens quite often in mathematics, for example space-filling curves, cyclic voting paradoxes, and measure-concentration in high dimensions. It also happens in other situations, for example [skip the next paragraph if you want only mathematics]:

I've seen a video of a play in an (American) football game where a player, carrying the ball forward, throws it back, over his shoulder, to a teammate running behind him. The referee ruled that this was a forward pass. Intuition says that throwing the ball back over your shoulder is not "forward". But in fact, as the video shows, the teammate caught the ball at a location further forward than where the first player threw it. If you're running forward with speed $v$ and you throw the ball "backward" (relative to yourself) with speed $w<v$, then the ball is still moving forward (relative to the ground) with speed $w-v$. So the referee was quite right.

In my opinion, the possibility of contradicting intuition is one of the great benefits of mathematical reasoning. Mathematics doesn't merely confirm what we naturally know but sometimes corrects what we think we know.

Intuition is a wonderful thing. It's fast and usually gives good results, so it's very valuable when we don't have the time or energy or ability for a more thoughtful analysis. But we should bear in mind that intuition is not infallible and that more careful thought can provide new insights and corrections.

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  • $\begingroup$ The forward pass reference is really cool, I've never thought about it that way $\endgroup$ – Ovi Jan 14 '16 at 18:28
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I think the key is what Henning mentioned, that is,

There are more irrationals between any two different rationals than there are rationals between any two irrationals.

However, to gain some more intuition imagine binary expansion of a rational and irrational numbers, for example

\begin{align} r_1 &= 0.00110100010\overline{0010101010} \\ i_1 &= 0.001101000100010101010 ??? \\ r_2 &= 0.00110100010\overline{0011111010} \\ i_2 &= 0.001101000100110101010 ??? \end{align}

The expansion of rationals are periodic, so every rational can be specified with finite amount of information (bits). You can make it as long as you want, but it still will be finite. However, even if some irrationals can be specified using finite amount of information (e.g. $0.101001000100001000001\ldots$), there are infinitely many more of those that carry infinite amount of information (there are infinitely many more of those even between any two rationals). We cannot specify them$^\dagger$ (how would we?), but we know they exists and exactly those unspecifiable numbers make that there are more of irrationals than rationals.

I hope this helps $\ddot\smile$


$^\dagger$ Here I'm assuming that we are using a language with finite or even countable number of symbols. On the other hand, for example, if we were to treat physical measures like "length of this stick" or "velocity of that bird" as specifications, and given that our world is continuous instead of discrete, then, almost surely, any such definition would be one of irrational number.

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See Cantor's diagonal argument. There are infinitely more reals between each "lattice point" of a rational. This is the basis for having multiple levels of infinity.

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The between-argument just says that there exists an interesting map from $\mathbb Q\times \mathbb Q$ to $\mathbb R\setminus \mathbb Q$ and as well an interesting map from $(\mathbb R\setminus\mathbb Q)\times(\mathbb R\setminus \Bbb Q)$ to $\mathbb Q$. This does not ell us in any way thet $\mathbb R\setminus\mathbb Q$ is equinumerous to $\mathbb Q$ or anything.

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This is the most intuitive explanation I can think of. It's long, but hopefully easy to follow.

Consider a natural number, for instance $123$. This natural number is implied by our mathematical notation system to have the decimal point after the last digit, with no fractional part ($123 = 123.00...0$). However, we can take this integer number and specify that it is the stream of digits of a non-integer, by specifying that there is a decimal point at any arbitrary single position of the stream. For instance, we could transform $123$ into $12.3$, or $1.23$, or $0.123$, or $0.0123$. We do this all the time, in what's called scientific or engineering notation; $123 = 1.23*10^2$. This is also a convenient way to store these decimal quantities in computers; binary floating-point arithmetic is based on this system (using a fixed-length binary mantissa and a base-2 integer exponent). We can also use the integer negatives of the natural numbers in the same way. In fact, the ability to construct a rational number in this way is derivable* from the true definition of a rational number.

The integer mantissa (and exponent) used to construct a rational can be arbitrarily large, but both are countable numbers; theoretically, if you started counting from 0, you'd eventually reach the number used (though it may take you longer than the universe has existed or will continue to exist to do so).

Now, with irrational numbers, this changes. The mantissa is no longer any one natural number, but the concatenation of an infinite number of unique natural numbers. Some irrationals are defined as such, using a formula to define each number to be concatenated (e.g. Champernowne's constant, $0.123456789101112131415161718192021...$). For most, however, the construction is not formulaic, but it holds that any irrational number can be constructed by concatenating a subset of the natural numbers (order independent), then placing a decimal point somewhere.

The upshot is that there are only as many rational numbers as there are naturals (there's a step I'm missing to formally prove that; the exponential construction I'm using is a good start, but without further restrictions to the allowable values, it trivially allows for infinite combinations of an integer mantissa and integer exponent to produce the same rational. Cantor did it much more rigorously). However, there are as many real numbers as there are subsets of the naturals, because by appending the digits of any unique natural number to any concatenation of other numbers, and sticking a decimal place in arbitrarily, you get a whole new real number. The set of all subsets of a set is known as the power set, and its cardinality is given by $|P(A)| = 2^{|A|}$. The cardinality of the naturals is the trans-infinite number $\aleph_0$ (aleph-null or aleph-zero), and so the cardinality of the reals is $\beth_1 = 2^{\aleph_0}$ (beth-one).

The cardinality of the set of irrational numbers is the cardinality of the set of real numbers, minus the cardinality of the set of rationals (which is the same as those of the integers and the naturals), which using the trans-infinite numbers is $\beth_1 - \aleph_0$, similar to the notation of the set itself as $\bar{\mathbb Q} = R-Q$ (the set difference of the rationals from the reals, also known as the set complement of the rationals over the reals).

* - A rational number's official definition is a value that can be represented as a fraction with integer terms. For any natural number $x$, $10^x$ is a natural number (because the set of natural numbers is closed over multiplication), and $10^{-x} = 1/10^x$ and so by definition it is also rational as both terms of the fraction are naturals (which are integers). Any integer mantissa $m$, multiplied by 10 to the power of any integer $x$, is thus either $(m*10^x)/1$ or $m/10^x$ for exponents $x>0$ and $x<0$ respectively (with the trivial case of an integer $m$, including 1, being $m=m*10^0$), and so any number constructed in this way is rational as all terms of the fraction are always integers. Not all rationals can be constructed by using an integer power of 10, but by allowing any natural number base it becomes trivial to express any rational in this same exponential form; $a/b = a*b^{-1}$ for all $a \in \mathbb Q, b \in \mathbb N$.

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Not enough rep to comment yet, so I'll answer as well.

First, I love n-owad's answer as it is very intuitive. I'll +1 it when my rep is high enough.

The standard argument that I found compelling, if less intuitive to grasp, is that you can set up an enumeration of the rationals, but you can prove (by contradiction, IIRC) that the irrationals are not enumerable. If memory serves (and boy is it dusty) the enumeration works by looking at [0,1] and counting the numerators less than each denominator, so 1/1, 1/2, 1/3, 2/3, .... This pattern can be mapped to the sequence i = [1..] and thus there are the same (infinite) number of items.

As a continuation question - is there a simple argument to show that the quadratic irrationals (or all algebraic numbers) are countable? The quadratics seem obvious since they have a repeating representation (continued fractions). But the larger set isn't quite so obvious.

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  • $\begingroup$ Enough rep now - upped as promised. :) $\endgroup$ – sfjac Aug 24 '13 at 2:57
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    $\begingroup$ To show the algebraic numbers are countable, it suffices to show the set of polynomials over $\mathbf{Q}$ are countable, which is not so hard. $\endgroup$ – Maanroof Aug 1 '16 at 17:46

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