This is not a deep question, but if there is a definite answer then here is the place where I will find it.

Is it justified to say that $i =\sqrt{-1}$ is rational?

The origin of this question lies in a regular discussion I have over this t-shirt of mine:

Get Real / Be Rational

While obvious $\pi$'s comment is completely legit, $\sqrt{-1}$'s might be hypocritical, if the rationality of $\sqrt{-1}$ is questionable.

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    $\begingroup$ $i$ is not rational. Note that tee-shirts are not peer-reviewed. $\endgroup$ – Will Jagy Sep 21 '12 at 20:28
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    $\begingroup$ There are too many real hypocrites in the world for us to worry about imaginary ones... $\endgroup$ – Trevor Wilson Sep 21 '12 at 20:38
  • $\begingroup$ Decided to change the accepted answer. I intentionally did not ask if $\sqrt{.1}$ is a rational number, which it is not, of course. But if it is rational, which could be seen as a property orthogonal to the real/imaginary axis. $\endgroup$ – DoubleMalt Sep 21 '12 at 21:01
  • $\begingroup$ Think you didn't understand the t-shirt.$\pi$ is not rational,then $\sqrt -1$ aks pi to be rational,and how $\sqrt -1$ is complex,not real, $\pi$ asks it to "get real". $\endgroup$ – HipsterMathematician Sep 21 '12 at 21:04
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    $\begingroup$ Note that though $i$ may not be rational, it can be properly described as integral since it is a root of a monic polynomial with integer coefficients. Needless to say this is a generalisation of the notion of integer ... $\endgroup$ – Mark Bennet Sep 21 '12 at 21:59

It is a Gaussian rational number, but it is not rational in the conventional sense of the word because rational numbers are real.

  • $\begingroup$ That's the answer I wanted to give. $\endgroup$ – M Turgeon Sep 21 '12 at 20:31
  • $\begingroup$ Rational numbers are real numbers, are gaussian rational numbers algo gaussian real numbers? $\endgroup$ – Billy Rubina Dec 29 '12 at 13:34
  • $\begingroup$ @GustavoBandeira: What do you mean by "Gaussian real number"? If it follows the pattern of "Gaussian rational numbers" and "Gaussian integers", i.e. complex numbers whose real and imaginary parts are both rational/integral (respectively), then a Gaussian real number is precisely just a complex number. So I guess the answer to your question is yes. $\endgroup$ – Clive Newstead Dec 29 '12 at 14:16

The number $\sqrt{-1}$ is not real. Since the rationals are just a particular type of real number, it cannot be rational, either.

Another way to look at it: Were there integers $a$ and $b$ such that $\sqrt{-1} = \frac{a}{b}$ then $$ -1 = \frac{a^2}{b^2}, $$ and so $$ a^2 = -b^2. $$ Since $b^2$ is certainly positive, that means $a^2$ is certainly negative, which is impossible.


$\sqrt{-1} = i$ is an imaginary number, not lying anywhere on the real number line. Therefore as others have said, it is neither rational nor irrational in the usual senses of those words.

To comment on the specific grammar of the shirt: $\pi$'s imperative is 'Get real' which I would say has the connotation of 'join [the speaker's] group'. Clearly this makes sense, as $\pi$ is a real number. $i$, on the other hand, says only 'be rational' which I would say lacks the connotation of joining the speakers group; instead, $i$ is only imploring $\pi$ to join the rationals, with no implication of $i$'s membership.

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    $\begingroup$ While technically it is true that the connotation of joining the speaker's group is not explicit in $\sqrt{-1}$'s case, under most circumstances the demand to adhere to something the speaker him- or herself does not will be considered extremely hypocritical ;) $\endgroup$ – DoubleMalt Sep 21 '12 at 20:44

Rather $\sqrt{-1}$ is the algebraic number, but not rational.


Echoing what other people have already said: no $i= \sqrt{-1}$ is not a rational number.

You have $$ \begin{align} &\mathbb{C} \;\text{ the complex numbers}\\ &\cup \\ &\mathbb{R} \;\text{ the real numbers}\\ &\cup \\ &\mathbb{Q} \;\text{ the rational numbers}\\ &\cup \\ &\mathbb{Z} \;\text{ the integers}\\ &\cup \\ &\mathbb{N} \;\text{ the natural numbers} \end{align} $$ Here the $\cup$ denotes that the lower is contained in the upper. So for example all real numbers are complex numbers. And: an integer is a real number. Note that for example not all complex numbers are real numbers. Not all complex numbers are rational numbers. Not all integers are natural numbers.

So the question now is whether $i = \sqrt{-1}$ (which is a complex number) is a rational number. And here we first note that the rational numbers are those numbers that can be expressed as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers (so belongs to $\mathbb{Z}$) ($b\neq 0$). So is $i = \frac{a}{b}$ for any integers $a$, and $b$? As provided in Austin's fine answer, one can show that indeed the answer is no.

To have something to think about, maybe you can answer the question: Is $\sqrt{2}$ a rational number? (You can find the answer here on M.SE).

  • $\begingroup$ You did not answer my question ;). I know that $\sqrt{-1}$ is not an element of the rational numbers. The question was if it is rational in a more abstract sense. $\endgroup$ – DoubleMalt Sep 21 '12 at 21:03

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