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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?

Glorfindel
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Alex Azazel
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    Notice that there are no numbers depicted in the blue area. – Hagen von Eitzen Sep 14 '15 at 17:27
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    It is not a very clever illustration, indeed. – uniquesolution Sep 14 '15 at 17:27
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    Never trust someone who says 0 is not a natural number ;-p – Steve Jessop Sep 14 '15 at 18:17
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    @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. –  Sep 14 '15 at 19:40
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    nb I am aware of Peano axioms - still, the choice of whether $0 \in N$ or not is arbitrary - in many situations it's easier to deal with $N_0$, in many - with $N_+$... If you need e.g. additive identity - go with $N = N_0$; if you need e.g. to be able to have a well-defined $a/b \in Q$ or $a^b \in N$ $\forall a,b \in N$ - go with $N = N_+$... –  Sep 14 '15 at 19:50
  • No, but there are many numbers that we can describe for which we do not know how to determine, today, whether or not they are rational. E.g. $\pi+e$. – Mark Adler Sep 14 '15 at 23:10
  • @MarkAdler And worse are numbers whose rationality is not only unknown, but independent of one's logical system. (Although there's nothing stopping $\pi+e$ from being such a number) – Milo Brandt Sep 15 '15 at 03:35
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    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'. – CompuChip Sep 15 '15 at 07:37
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    @vaxquis: sure, it's just a matter of how you define addition in your Peano system, specifically whether the base case is $x + b = x$ or $x + b = S(x)$, with $b$ the base element and $S$ the successor function. If you're going to use the former definition you call the base element $0$, if the latter then $1$. It's completely arbitrary based on convenience. All I'm saying is there's something shifty about the latter types. Their eyes are too small or something ;-) – Steve Jessop Sep 15 '15 at 08:23
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    A more intuitive view, natural numbers are numbers that are used for counting and ordering. Well $0\in\mathbb{N}$ because $0$ is the number of dinosaurs on the planet. It's not a matter of convention at all. If you define $0\not\in\mathbb{N}$, you are stripping something fundamental from $\mathbb{N}$. And for me, $x$ is positive means $x\geq 0$ and $x$ is strictly positive means $x>0$ (some people call it "nonnegative"!). – user5402 Sep 15 '15 at 17:07
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    @whatever calling $0$ "positive" is IMO a severe abuse of language, and will make your statements ambiguous and misunderstood at best; $0$, by definition, is neither positive nor negative. By definition $x : "positive" \iff x > 0 ; x : "negative" \iff x < 0$ - thus, $0$ is neither. Still, if we define $N$ as set of every possible cardinality of a finite & countable set - $N = N_0$ ; if we define $N$ as set of every possible cardinality of a finite, countable & nonempty set, $N = N_+$ –  Sep 15 '15 at 19:29
  • @SteveJessop aah, the shifty eyes... that's completely true! Erm, well, now I can strictly agree with your statement. ^_^ –  Sep 15 '15 at 19:32
  • @vaxquis Positive means $\geq 0$ not $>0$. If $x>0$, we say $x$ is strictly positive. The adjective "strictly" is a lot better than "nonnegative". $0$ is at the same time positive and negative. – user5402 Sep 15 '15 at 19:47
  • @vaxquis Also $\mathbb{R}+=[0,+\infty[$ and $\mathbb{R}+^*=]0,+\infty[$. I don't know how you denote it. – user5402 Sep 15 '15 at 19:48
  • @SteveJessop You clearly didn't get my point. Saying that $x$ is positive means that $x$ may be $0$ but saying that $x$ is srictly positive means that it can't be $0$ and the same thing whith (negative, stricly negative). Adding that one adjective "srictly" clarifies everything. I'm saying that the terminology $(positive, strictly positive)$ is better than $(nonnegative, positive)$. – user5402 Sep 15 '15 at 19:55
  • @HagenvonEitzen Of course, one can have a set of real numbers without being able to explicitly name particular numbers within that set. – Kyle Strand Sep 15 '15 at 23:49
  • @whatever to sum up the now-removed discussion I can only say this: if you consider allowing something to be positive & negative at the same time "a lot better solution" that not allowing it, as well as being a proponent of using terminology that's alien to most actual mathematicians worldwide because, in your opinion, it's a better solution and because "there are different types of english [sic!]" - and if, at the same time, you say that the definition of particular sets in mathematics is not a matter of convention at all - then I simply rest my case and fare you well in the future. –  Sep 17 '15 at 21:00
  • @vaxquis Why do you say increasing/strictly increasing but you don't say positive/strictly poitive? Double standards?! – user5402 Sep 17 '15 at 21:09

7 Answers7

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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.

Anixx
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Dominik
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    @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. –  Sep 15 '15 at 19:38
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    @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. – Dominik Sep 15 '15 at 19:45
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    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. –  Sep 15 '15 at 20:15
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    @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. – Dominik Sep 15 '15 at 20:19
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    @Lembik Yes, in fact, all of the boxes except "transcendental numbers" should be exactly the same size ($\aleph_0$), and "transcendental numbers" ($\mathfrak{c} = 2^{\aleph_0}$) should be infinitely larger than all the rest put together. – zwol Sep 15 '15 at 20:22
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    In your new diagram, there may be integers that are neither positive nor nonpositive! :P – user253751 Sep 15 '15 at 23:05
  • Ideas for an extension that handles some more oddballs among the reals: $\text{real root of }x^5-x+1$ is algebraic but has no closed form. $\Omega$ (Chaitin's constant) is non-computable, but well-defined. $0.78391434\stackrel{?}{\dots}$ (a specific "random" real number) is undefinable. – Lynn Sep 16 '15 at 02:12
  • @zwol: How can the transcendental set be larger than the irrational set which contains it? – MichaelS Sep 16 '15 at 04:38
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    @MichaelS because the image is wrong. The transcedental set is the "biggest" of them all. – Jakob Sep 16 '15 at 07:57
  • @zwol No, the set of irrational numbers is also uncountable. – Daniel R Sep 16 '15 at 14:02
  • @Jakob The image is correct, the set of irrational numbers is also uncountable. – Daniel R Sep 16 '15 at 14:02
  • @Jakob If I'm reading irrational and transcendental numbers correctly, Irrational is the slightly "larger" set, almost all of which are also Transendental. The non-transcendental irrationals being the Algebraic Numbers (examples of which are shown outside the transcendental set above) and which are countable. Probably the Transcendentals and Algebraic numbers should be shown as side-by-side subsets of the Irrationals, as is done for irrationals/rationals as subsets of reals. – TripeHound Sep 16 '15 at 14:09
  • @Dominik I really like the "teeth" idea, that's about what I had in mind as a possible solution. –  Sep 16 '15 at 14:32
  • @MichaelS I meant the size of the irrational-but-not-transcendental box. The algebraics are still countable. (And of course the "reals" box still contains everything.) – zwol Sep 16 '15 at 15:27
  • @TripeHound This won't work, as all rational numbers are algebraic. The algebraic numbers would have to span the rationals, some irrationals, be disjoint to the transcendental numbers and still the irrationals would have to consist of the transcendental numbers and a subset of the algebraic numbers. I don't think it's really feasible to cram all this information into a single diagram. – Dominik Sep 16 '15 at 15:27
  • @Dominik On reflection, I think I agree, the diagram would be harmed by trying to scale everything to cardinalities. – zwol Sep 16 '15 at 15:29
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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.

Asaf Karagila
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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.

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

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    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. – user4894 Sep 14 '15 at 17:34
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    @user4894 This isn't English semantics. This is logic.Your example of open sets is entirely irrelevant. –  Sep 14 '15 at 17:41
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    What if you are using intuitionistic logic? – Christopher King Sep 14 '15 at 20:08
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    @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. –  Sep 14 '15 at 20:26
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    I may not be helpful, but as long as it's not not helpful, I think we'll be okay. – Christopher King Sep 14 '15 at 20:31
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    @PyRulez I chuckled. :) –  Sep 14 '15 at 20:33
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    @avid19 I think user4894's point is that mathematics uses very precise definitions which generally fit alongside their English meanings, but every now and then words that we presume are opposites in English are not defined to be opposites in mathematics. User4894's example is one case where, in English, the concept of "open" and "closed" are treated as opposites, but in set theory they are not. If we encourage questioners to resolve this question with intuitive semantics using English wordings, they may get in trouble later, such as in set theory. – Cort Ammon Sep 15 '15 at 17:27
  • As a English semantics line of reasoning regarding opposites, I am compelled to quote Pirates of the Caribbean: Will: "This is either crazy, or brilliant." / Sparrow: "Remarkable how often those two traits coincide." – Cort Ammon Sep 15 '15 at 17:29
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    @PyRulez I wrote an answer for you discussing not-not-rationality. – Mario Carneiro Sep 15 '15 at 22:46
  • @MarioCarneiro I was going to write a new, higher level question, but thanks. – Christopher King Sep 16 '15 at 00:24
  • @Cort Ammom: The worst example of a disconnect that I know of between ordinary-language (“English”) semantics and mathematical meaning is whether R has divisors of zero. We certainly agree that, say, 3 is a divisor of 0, and so the natural take-away is that R has divisors of 0. Also, it’s not really news that antonyms can, because of context, be used synonymously. For example, I could have called this the “best” example of a disconnect instead of the “worst”. – Mike Jones Oct 31 '15 at 13:22
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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)

usermath
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    Also, there are considerably more irrational numbers than rationals, something else the picture is misleading about. – RBarryYoung Sep 15 '15 at 17:06
  • 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. – Cort Ammon Sep 15 '15 at 17:23
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    @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. – Steve Jessop Sep 15 '15 at 19:57
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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.

neelkanth
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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.

adamcatto
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