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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.

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By 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. –  Gerry Myerson May 9 '12 at 7:27
    
@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. –  yiyi May 9 '12 at 8:49
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It's not hard to understand the statement of the Gelfond-Schneider Theorem. Understanding the proof, that's a different matter. –  Gerry Myerson May 9 '12 at 13:06
    
@GerryMyerson rarely is a proof an easy matter. –  yiyi May 9 '12 at 15:27
    
It 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. –  Pete L. Clark May 10 '12 at 17:26

2 Answers 2

up vote 8 down vote accepted

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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It always pays to read the definitions! –  Grumpy Parsnip May 9 '12 at 2:44
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To be fair, I went somewhat further than the definition given in the paper, which required the reader know what a complex integer is. –  Alex Becker May 9 '12 at 2:47
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Doesn't the first sentence of the paper define complex integers? –  Grumpy Parsnip May 9 '12 at 2:49
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Note 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 $\mathbb Q(i)$. –  lhf May 9 '12 at 3:09
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But 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 $5$. –  Lubin May 9 '12 at 3:14

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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I 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. –  process91 May 9 '12 at 7:19
    
thanks, it was helpful. –  yiyi May 9 '12 at 8:47

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