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On Wikipedia > Number > Classification there is this diagram:

enter image description here

A. This diagram implies that the irrational numbers are a subset of solely the real numbers. But afaik, that's not true, because any number, which is not rational, is an irrational number, and thus all imaginary numbers are irrational, because they are not rational.

I think the diagram suffers from the fact that some irrational numbers are real numbers and some are complex numbers with non-zero imaginary part. So irrational numbers are neither a subset of solely the real numbers nor solely of $\mathbb{C}\setminus\mathbb{R}$.

B. Moreover, the diagram seems to imply that the complex numbers consist solely of real numbers and purely imaginary numbers. Afaik, all complex numbers which have both, a non-zero real part and a non-zero imaginary part, are neither real nor imaginary, so they are missing in the diagram.

Is this diagram misleading? Or is it even wrong?

Daniel S.
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  • Probably connected to https://math.stackexchange.com/questions/823970/is-i-irrational . – Daniel S. Aug 01 '22 at 11:21
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    People differ as to whether or not non-real complex numbers should be called irrational. Many writers say that only reals can be irrational. Context should make clear which convention is being followed. – lulu Aug 01 '22 at 11:37
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    A: In general, the standard definition of the irrational numbers is $\mathbb{R} \setminus \mathbb{Q}$. Some people don't love that definition, but it seems to be the most commonly used one. So the diagram isn't wrong. B. The diagram is correct but vaguely misleading. $\mathbb{C}$ contains $\mathbb{R}$ as well as the imaginary numbers, as well as complex numbers. This diagram shows only the hierarchy, rather than the "amount encompassed" by each set, otherwise there would be huge empty spaces. – Eric Snyder Aug 01 '22 at 12:15
  • @ryang Alright, I will keep it. feel free to answer. – Daniel S. Aug 01 '22 at 15:39
  • Evidently the chart only shows subsets, not partitions, but except for dyadic (subset of finite decimal) and complex (missing $\mathbb C \setminus (\mathbb R \cup i\mathbb R)$) it would be a partition graph, which may induce some confusion in a reader trying to glean the relation. – Angelica Aug 01 '22 at 17:25
  • "thus all imaginary numbers are irrational, because they are not rational." In my opinion that bankrupts the concept of irrational and rational. Step back a notch and the question can be "Is $-5$ and integer?" What's the definition of integer any way? If $-5$ is an integer then why isn't $5i$ and integer. And if $5i$ is an integer then $\frac 53i$ is rational. – fleablood Aug 01 '22 at 18:48

1 Answers1

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all imaginary numbers are irrational, because they are not rational.

Does ‘insensitive’ refer to every entity that isn't sensitive, or is this adjective understood to apply only to humans? This is how I persuade myself that it's not illogical to define the irrationals as $\mathbb{R} {\setminus} \mathbb{Q}.$

Similarly, ‘nonpositive’ is conventionally defined on the reals (so, $i$ isn't considered a nonpositive number). Even so, however, when it's not perfectly clear that the context is strictly real, to avoid ambiguity, I find myself writing “nonpositive real number”.

the diagram seems to imply that the complex numbers consist solely of real numbers and purely imaginary numbers.

The diagram is a little misleading, but it's technically correct: the labels are merely category names, and the illustration does not assert that Imaginary numbers and Real numbers form a partition of the Complex numbers, only that the former two are disjoint subsets of the latter.

ryang
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