23

On the one hand, $i(=\sqrt{-1})$ cannot be expressed as a ratio of integers, so, by definition, $i$ is not rational $\iff i$ is irrational.

However, the set of irrational numbers, $\mathbb{J}=\mathbb{R}\setminus\mathbb{Q}$ is defined to be the set of all real numbers that are not in $\mathbb{Q},$ but, clearly, $i \notin \mathbb{R}$, so it must be that $i \notin \mathbb{J}$, so $i$ is not irrational.

Clearly, the first two paragraphs seem to contradict each other, so I'm asking: is $i$ irrational or rational?

beep-boop
  • 11,595
  • 10
    It's an integer ;) (it's a zero of the monic polynomial $X^2 +1 \in \mathbb{Z}[X]$, so it's an algebraic integer) – Daniel Fischer Jun 07 '14 at 14:56
  • 7
    If is it not real, it cannot be rational [no joke] http://www.gweigt.net/ – AlainD Jun 07 '14 at 14:56
  • 2
    If you define irrationals to be $\mathbb{R}\setminus \mathbb{Q}$ then to be an irrational number one must be in $\mathbb{R}$. So if you stick to your definition then you must not count it as an irrational number. – user50948 Jun 07 '14 at 14:56
  • 14
    In a sense it is like asking whether a tree is irrational, on the one hand it cannot be expressed as a ratio of integers, one the other hand... (i'm not trying to be smart, just illustrating the importance of having a clear idea of what one is talking about). – user50948 Jun 07 '14 at 14:58
  • 2
    @user50948 No, it's not. It's a perfectly reasonable question to ask whether a number is rational- it's an unrelated and irrelevant question to ask whether a tree has the properties of a number. – beep-boop Jun 08 '14 at 00:56

6 Answers6

19

It depends on convention whether or not the scope of "irrational" includes non-real numbers. As you say, if the irrationals are defined as $\,\Bbb R\setminus \Bbb Q\,$ then $\,i\,$ is neither irrational nor rational. However, many authors use "irrational" to mean "not rational", i.e. $\,\not\in \Bbb Q,\,$ therefore $\,i\,$ is irrational. This is quite common usage in university-level algebra.

For example, in my 2006/3/8 sci.math post I remarked that if one searches books.google.com for "irrational algebraic" one finds such usage by many eminent mathematicians: e.g. John Conway, Gelfond, Manin, Ribenboim, Shafarevich, Waldschmidt (esp. in diophantine approximation, e.g. Thue-Siegel-Roth theorem, Gelfond-Schneider theorem, etc). See also other posts in that sci.math thread titled "Is $\,i\,$ irrational"?

Out of curiosity, I ran another Google Books search on "real irrational" numbers. Authors using such terminology presumably employ the more general definition of irrational numbers. Among such authors I found the following eminent mathematicians: Bombieri, Davenport, Dedekind, Euler, Hurwitz, Kronecker, Kirilov, Mahler, Lang, Ostrowski, Ribenboim, Weil.

However, it is not easy to find an explicit definition since higher-level textbooks assume the reader already knows basic terminology. I vaguely recall that someone (Gerry Myerson?) once posted to sci.math some links to definitions which make it unquestionably clear that the author employs the more general definition of "irrational". Perhaps someone can dig those up, or locate others.

In any case, it is a matter of convention. In most cases one can quickly infer the intended denotation from the context, so there is little chance for confusion.

Bill Dubuque
  • 272,048
  • Could this concept be wxtwbded to other integer rings? Say I am working with $Z[\sqrt{-2}]$ and I divide $1+\sqrt{-2}$ by $2+\sqrt{-2}$. The thing I get does not belong the the algebraic integer ring, but might it be called "rational" with respect to the ring? – Oscar Lanzi Dec 31 '16 at 23:28
  • 1
    @OscarLanzi Yes, fractions over a domain $D$ are sometimes called $D$-rational - see my comment on Tom's answer. – Bill Dubuque Jan 03 '17 at 16:01
  • One context in which it's not immediately clear which convention is being used is the Gelfond-Schneider Theorem - it pertains to values $a^b$ where $a$ and $b$ are algebraic and $b$ is irrational, but it's not clear without knowing the details of the proof whether $i$ counts as irrational or not (it does). Personally I'd want to call nonreal numbers "non-rational" or something to avoid confusion, and leave "irrational" to mean real and non-rational. – Ken Williams Feb 14 '17 at 04:27
  • Under this definition my dog, refrigerator, and 2015 Chevrolet Camaro are irrational numbers. – Jonathan Hebert Dec 29 '19 at 20:43
  • 1
    @Jonathan The definition applies to elements of the ambient ring (universe) containing $\Bbb Q,,$ just like nonintegral, nonzero, etc. – Bill Dubuque Dec 29 '19 at 20:49
12

It is neither. Just as it's neither positive nor negative. The real numbers are partitioned into rational and irrational numbers; but things which aren't real numbers don't have to be one or the other.

Your conundrum is similar to saying "A $2 \times 2$ matrix is not rational. So it must be irrational. But it's not irrational; a contradiction." Does it seem more clear why that's a false dichotomy?

Nick Matteo
  • 9,006
7

If I tell you that for lunch you can either have pizza or pasta, and you see someone eating a burger, does that negate my statement? Or reality?

Context is not absolute in mathematics, and it's important to remember that. It is true that in the context of the real numbers we have rationals and irrationals, and whatever is not rational is irrational. But in the context of the real numbers we also don't have $\sqrt{-1}$.

In the context of $\Bbb C$ we often talk less about irrational numbers so the context can be taken to both direction. It is possible to declare "In the context of the complex numbers, the irrational numbers are $\Bbb{C\setminus Q}$" in which case $i$ is certainly irrational, but it is also possible to make other declarations, like "The irrational numbers are still $\Bbb{R\setminus Q}$", in which case $i$ is not irrational -- but it doesn't make it rational either.

It's a question of context and it's not true that there's always a conventional and agreed upon context. This very question is a good example for that.

Asaf Karagila
  • 393,674
2

$\sqrt{-1} \notin \mathbb{R}$ and $\mathbb{N} \subsetneq \mathbb{Z} \subsetneq \mathbb{Q} \subsetneq \mathbb{R} \subsetneq \mathbb{C}.$

Note that $i$ is a number, and for any number, the question to ask whether it is rational or not is makes sense. It is not rational, since it is not a ratio of two integers. Hence, it is irrational, as irrational numbers are the complement of the rational ones (complement depending on context, either reals or complex numbers).

Per Alexandersson
  • 3,626
2

Several previous comments and answers apply, but no one yet seems to have mentioned Gaussian Integers, (see http://en.wikipedia.org/wiki/Gaussian_integer) - complex numbers whose real and imaginary parts are both (real) integers. By this definition, i is a Gaussian Integer. the concept extends to Gaussian Rationals (see http://en.wikipedia.org/wiki/Gaussian_rational), so that i is also a Gaussian Rational.

Tom Collinge
  • 7,981
  • 25
  • 59
  • But this is employing a different convention, namely the convention of referring to the elements of the fraction field of a a domain $,D,$ as $,D$-rational. By that convention, any number can be considered "rational" relative to an appropriate domain (e.g. often used to distinguish such rationals from higher degree algebraic numbers over $,D).\ \ $ – Bill Dubuque Jun 07 '14 at 16:03
  • @BillDubuque Well the issue in the OP question is one of mixing conventions, i.e. using terms that normally describe real numbers to create a confusion about an imaginary number: my answer is simply pointing out that there is a convention which does address rationality in the complex numbers. – Tom Collinge Jun 07 '14 at 16:12
0

Actual definition of irrational number says that an irrational number is any real number that cannot be expressed as a ratio of integers. So your first line is infact wrong.

RE60K
  • 17,716