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.
2026-04-06 11:35:59.1775475359
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Intuitive explanation for how could there be "more" irrational numbers than rational?
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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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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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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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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 $v-w$. 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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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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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.
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 $2.\overline{37}$ or $0.42$, which we will write as $0.42\bar{0}$. 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.