I am asking because I was reading this and the mathematics is a little over my head. The title of the paper is Rational Approximations to Irrational Complex Number, and I didn't think that complex irrational numbers could exist.
Are irrational complex numbers possible?
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6By the way, the definition in that paper is convenient for use in that paper, but by the standard definition, *every* complex number with non-zero imaginary part is irrational. Irrational simply means not rational, and the rationals are a subset of the reals, so if it's complex and not real it's irrational. For example, $i$ is irrational. This definition comes in handy when stating advanced results like the Gelfond-Schneider Theorem in transcendence theory. – 2012-05-09
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0@GerryMyerson I think I have a ways to go before I could understand Gelfond-Schneider Thm in transcendence theory. I'll look for it, it always fun to try to understand what I don't understand. – 2012-05-09
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1It's not hard to understand the statement of the Gelfond-Schneider Theorem. Understanding the proof, that's a different matter. – 2012-05-09
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0@GerryMyerson rarely is a proof an easy matter. – 2012-05-09
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0It is worth pointing out that the paper linked to by the OP is almost one hundred years old. In general that's more than enough time for terminology to slide around a bit. Of course if you're interested in the paper you'll want to know what the title means, but I wouldn't draw any conclusions about current nomenclature / terminology from it. – 2012-05-10
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0@PeteL.Clark thanks, didn't realize how old the paper was, just trying to find some papers that I can understand but still push me to learn more. also, I ask alot of stupid questions, hehe. – 2012-05-12
2 Answers
The paper defines rational complex numbers as numbers of the form $$x=\frac{a+bi}{c+di} \text{ where } a,b,c,d\in\mathbb Z.$$ It is easy to see that irrational complex numbers (complex numbers not of the above form) exist. For one thing, any number of the above form has norm of the form $$\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}$$ which is either rational or quadratic irrational yet (for example) $|\sqrt[3]{2}+i|=\sqrt{\sqrt[3]{2}+1}$ which is of algebraic degree $6$.
Edit: Corrected statment about norms. Thank you Michael Boratko for pointing out my error.
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3It always pays to read the definitions! – 2012-05-09
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1To be fair, I went somewhat further than the definition given in the paper, which required the reader know what a complex integer is. – 2012-05-09
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2Doesn't the first sentence of the paper define complex integers? – 2012-05-09
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0@AlexBecker I understand and get the "rational complex number", or i think i do. Could you give me an example of a complex number that can not be written as a fraction of two complex numbers? – 2012-05-09
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0@Mao: any complex number can be written in the form $\dfrac{x+iy}{1+0i}$. Now if you meant to add restrictions... – 2012-05-09
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0@JimConant Oh, it does. I guess I was context-blind after jumping to the definition of rational complex number. – 2012-05-09
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0@MaoYiyi The example $\sqrt{2}+i$ cannot be written as a ratio of complex *integers*, as demonstrated. – 2012-05-09
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0@JimConant I get that m & n are integers, oh. I get it. Because m&n are limited to being just integers then there can be. Question for you: if m & n are not limited to integers, is there any complex number that can not be written as a ratio of two complex numbers? – 2012-05-09
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1@MaoYiyi: See J.M.'s comment. – 2012-05-09
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0@MaoYiyi: Every complex number $z$ can be written $z/1$ or $(iz)/i$ or any of an infinite list of fractions of complex numbers... – 2012-05-09
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5Note that rational complex numbers are the same as the complex numbers of the form $p+qi$ with $p,q\in\mathbb Q$, a set usually denoted $\mathbb Q(i)$. – 2012-05-09
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2But if your definition of the “norm” of a complex number $a+bi$ is its absolute value $(a^2+b^2)^{1/2}$, as seems to be @AlexBecker’s intention, then the statement about norms being rational is not true: the absolute value of $2+i$ is the square root of $5$. – 2012-05-09
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0@MaoYiyi: Even in the reals, the rationals are countable and the reals are not, so there must be reals that are not rationals. Similarly, the complex numbers that can be written in your form are countable, as they are in correspondence with a subset of $\mathbb Z^4$ so there are lots of (most) reals that cannot be expressed that way. – 2012-05-09
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0@JimConant I thought that was true, but was afraid I was missing something in my thinking. – 2012-05-09
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0@RossMillikan is that because i^n is like Z mod 4? – 2012-05-09
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1@MaoYiyi No, it's because every rational complex number can be written using 4 integers. – 2012-05-09
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1@AlexBecker Does a rational complex number always have a rational norm? What about $1+i$? – 2012-05-09
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0@MichaelBoratko Oops, that's a very good point. I will edit to reflect. – 2012-05-09
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0@RossMillikan oh, so its "4-D shape", could you point me to information that proves that. That is wild. – 2012-05-09
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0@MaoYiyi 4-D shape is the wrong way to think of it. What Ross is saying is that there is an obvious bijection between the set of rational complex numbers and the set $\mathbb Z^4$, in the same way that there is an obvious bijection between the set of rational real numbers and $\mathbb Z^2$. – 2012-05-09
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0@AlexBecker I am not understanding $Z^2$ was thinking of it like $R^2$ which is a plane and $R^3$ is 3D space...and being $Z^2$ would be a complex plane, etc. Am I missing something? – 2012-05-09
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0$\mathbb Z^2$ is the set of pairs of natural numbers. It has dimension $0$, as does $\mathbb Z^n$ for all $n$. It looks like [this](http://mathworld.wolfram.com/images/eps-gif/LatticePoints_1000.gif). – 2012-05-09
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0@AlexBecker how is that not a plane? – 2012-05-09
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0let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3381/discussion-between-alex-becker-and-maoyiyi) – 2012-05-09
Taking the definition of a complex rational number $z$ as any number which can be represented by the form:
$$z= \frac {a+bi}{c+di}\qquad \text{for } a,b,c,d\in\mathbb Z$$
it is easy to see that if $z$ is any number with irrational parts (say, for instance, $\sqrt 2 + i$ as Alex mentions, or even just an irrational number like $\pi$) then it is also an irrational complex number. This is because we have $$\frac {a+bi}{c+di}=\frac {(a+bi)(c-di)}{(c+di)(c-di)}=\frac{ac+bd}{c^2+d^2}+\frac{cd-da}{c^2+d^2}i\, ,$$ and thus when we equate real and imaginary parts...
Indeed, one can verify that all complex rational numbers are just the points in the complex plane with purely rational coordinates. The definition may just as well have been all numbers $z$ which can be written in the form $$z=\frac a b + \frac c d i \qquad \text {for } a, b, c, d \in \mathbb Z$$ or just $$z=p+qi\qquad \text {for } p,q \in \mathbb Q$$
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0I added this explanation even though this was thoroughly addressed in the comments of Alex's answer because this particular point hadn't been made yet, and also because the reasoning that any rational complex number has a rational norm is incorrect - for instance (as Lubin points out) $2+i$ is a rational complex number with an irrational norm. – 2012-05-09