I wouldn't have asked this question if I hadn't seen this image:

From this image it seems like there are reals that are neither rational nor irrational (dark blue), but is it so or is that illustration incorrect?

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    $\begingroup$ Notice that there are no numbers depicted in the blue area. $\endgroup$ – Hagen von Eitzen Sep 14 '15 at 17:27
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    $\begingroup$ It is not a very clever illustration, indeed. $\endgroup$ – uniquesolution Sep 14 '15 at 17:27
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    $\begingroup$ Never trust someone who says 0 is not a natural number ;-p $\endgroup$ – Steve Jessop Sep 14 '15 at 18:17
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    $\begingroup$ @SteveJessop never trust someone who says 0 is a natural number. Hell, I'd even say never trust someone who says 0 *has to be a number* - in many situations (mainly in actual sciences - math is not a science chuckle), 0 is just a symbol. $\endgroup$ – vaxquis Sep 14 '15 at 19:40
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    $\begingroup$ Huh? I have never, in a mathematical context, heard "whole" numbers being used to denote $\mathbb{N} \cup \{ 0 \}$. If I were to encounter it in a mathematics text, when left undefined I would probably assume it was a synonym for 'integers'. $\endgroup$ – CompuChip Sep 15 '15 at 7:37

A real number is irrational if and only if it is not rational. By definition any real number is either rational or irrational.

I suppose the creator of this image chose this representation to show that rational and irrational numbers are both part of the bigger set of real numbers. The dark blue area is actually the empty set.

This is my take on a better representation:

Subsets of real numbers

Feel free to edit and improve this representation to your liking. I've oploaded the SVG sourcecode to pastebin.

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    $\begingroup$ @Dominik a great answer (I especially like how you smartly omitted the [ambiguous] $N$ set), with one small remark from me - the distinction of "primes" is a little bit confusing, because you provide those exemplary numbers twice, once in positive ints, and a second time in "primes" - and that makes the picture a bit confusing, because you don't repeat e.g. $\pi$ in $IQ$ outside of transcendentals... maybe I'm not making myself completely clear, but I hope you know what I mean. $\endgroup$ – vaxquis Sep 15 '15 at 19:38
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    $\begingroup$ @vaxquis I know what you mean and I've thought about this when I made the image. I decided to write those numbers twice so that it is obvious where the sequence $1, 2, 3, 4, \ldots$ comes from [it might be confusing if the primes are missing]. One possible solution for this could be to write the positive integers as a sequence and then make some lines that show that the primes are in the same set ["teeth" above a rectangle that says "primes"]. But I simply thought that the current representation looks nicer. $\endgroup$ – Dominik Sep 15 '15 at 19:45
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    $\begingroup$ Aren't there infinitely more irrational numbers than rational? The diagram seems to imply something different. Although it is nice in other ways. Similarly, I think there are infinitely more transcendental numbers than algebraic ones. $\endgroup$ – user66307 Sep 15 '15 at 20:15
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    $\begingroup$ @Lembik: Yes, the diagram doesn't convey cardinalities good at all. But to do this, one would have to make all sets equally big, except for the set of transcendental numbers. This set would have to be some orders of magintude bigger. This just doesn't translate well into a graphic. $\endgroup$ – Dominik Sep 15 '15 at 20:19
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    $\begingroup$ In your new diagram, there may be integers that are neither positive nor nonpositive! :P $\endgroup$ – immibis Sep 15 '15 at 23:05

No. The definition of an irrational number is a number which is not a rational number, namely it is not the ratio between two integers.

If a real number is not rational, then by definition it is irrational.

However, if you think about algebraic numbers, which are rational numbers and irrational numbers which can be expressed as roots of polynomials with integer coefficients (like $\sqrt2$ or $\sqrt[4]{12}-\frac1{\sqrt3}$), then there are irrational numbers which are not algebraic. These are called transcendental numbers.


Irrational means not rational. Can something be not rational, and not not rational? Hint: no.

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    $\begingroup$ And yet if a set is not closed, it may still not be open. And if it is closed, it may also be open. You can't always use English language semantics to draw conclusions in math. $\endgroup$ – user4894 Sep 14 '15 at 17:34
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    $\begingroup$ @user4894 This isn't English semantics. This is logic.Your example of open sets is entirely irrelevant. $\endgroup$ – user223391 Sep 14 '15 at 17:41
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    $\begingroup$ What if you are using intuitionistic logic? $\endgroup$ – PyRulez Sep 14 '15 at 20:08
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    $\begingroup$ @PyRulez I don't even know what that means and I suspect that a discussion of the subtleties of different logics is not very helpful to someone who is uncomfortable with the difference between rational and irrational numbers. $\endgroup$ – user223391 Sep 14 '15 at 20:26
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    $\begingroup$ I may not be helpful, but as long as it's not not helpful, I think we'll be okay. $\endgroup$ – PyRulez Sep 14 '15 at 20:31

Of course, the "traditional" answer is no, there are no real numbers that are not rational nor irrational. However, being the contrarian that I am, allow me to provide an alternative interpretation which gives a different answer.

What if you are using intuitionistic logic? – PyRulez

In intuitionistic logic, where the law of excluded middle (LEM) $P\vee\lnot P$ is rejected, things become slightly more complicated. Let $x\in \Bbb Q$ mean that there are two integers $p,q$ with $x=p/q$. Then the traditional interpretation of "$x$ is irrational" is $\lnot(x\in\Bbb Q)$, but we're going to call this "$x$ is not rational" instead. The statement "$x$ is not not rational", which is $\lnot\lnot(x\in\Bbb Q)$, is implied by $x\in\Bbb Q$ but not equivalent to it.

Consider the equation $0<|x-p/q|<q^{-\mu}$ where $x$ is the real number being approximated and $p/q$ is the rational approximation, and $\mu$ is a positive real constant. We measure the accuracy of the approximation by $|x-p/q|$, but don't let the denominator (and hence also the numerator, since $p/q$ is near $x$) be too large by demanding that the approximation be within a power of $q$. The larger $\mu$ is, the fewer pairs $(p,q)$ satisfy the equation, so we can find the least upper bound of $\mu$ such that there are infinitely many coprime solutions $(p,q)$ to the equation, and this defines the irrationality measure $\mu(x)$. There is a nice theorem from number theory that says that the irrationality measure of any irrational algebraic number is $2$, and the irrationality measure of a transcendental number is $\ge2$, while the irrationality measure of any rational number is $1$.

Thus there is a measurable gap between the irrationality measures of rational and irrational numbers, and this yields an alternative "constructive" definition of irrational: let $x\in\Bbb I$, read "$x$ is irrational", if $|x-p/q|<q^{-2}$ has infinitely many coprime solutions. Then $x\in\Bbb I\to x\notin\Bbb Q$, i.e. an irrational number is not rational, and in classical logic $x\in\Bbb I\leftrightarrow x\notin\Bbb Q$, so this is equivalent to the usual definition of irrational. This is viewed as a more constructive definition because rather than asserting a negative (that $x=p/q$ yields a contradiction), it instead gives an infinite sequence of good approximations which verifies the irrationality of the number.

This approach is also similar to the continued fraction method: irrational numbers have infinite simple continued fraction representations, while rational numbers have finite ones, so given an infinite continued fraction representation you automatically know that the limit cannot be rational.

The bad news is that because intuitionistic or constructive logic is strictly weaker than classical logic, it does not prove anything that classical logic cannot prove. Since classical logic proves that every number is rational or irrational, it does not prove that there is a non-rational non-irrational number (assuming consistency), so intuitionistic logic also cannot prove the existence of a non-rational non-irrational number. It just can't prove that this is impossible (it might be true, for some sense of "might"). On the other hand, there should be a model of the reals with constructive logic + $\lnot$LEM, such that there is a non-rational non-irrational number, and I invite any constructive analysts to supply such examples in the comments.


Every real number is either rational or irrational. The picture is not a good illustration I think. Though notice that a number can not be both irrational and rational (in the picture intersection is empty)

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    $\begingroup$ Also, there are considerably more irrational numbers than rationals, something else the picture is misleading about. $\endgroup$ – RBarryYoung Sep 15 '15 at 17:06
  • $\begingroup$ As a historical comment, I believe rationals "came" first, then we proved it was reasonable to assume numbers existed which were irrational, and the "real numbers" set was defined to include both. $\endgroup$ – Cort Ammon Sep 15 '15 at 17:23
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    $\begingroup$ @CortAmmon: depends who you ask, but in some sense the continuum came first. The Pythagoreans may have believed (without proof) that all numbers were rational and discovered themselves to be wrong. Naturally the term "real" numbers wasn't needed until we started inventing "fake" (that is to say imaginary) numbers. $\endgroup$ – Steve Jessop Sep 15 '15 at 19:57

We can represents real numbers on line i.e. real line which contains rationals and irrationals. Now by completeness property of real numbers, which says that real line has no gap. So there is no real number that is neither rational nor irrational.


The set of irrational numbers is the complement of the set of rational numbers, in the set of real numbers. By definition, all real numbers must be either rational or irrational.


protected by mrf Jan 24 '16 at 0:01


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