If I'm correct, a complex number can be interpreted as a set in the following manner:

$$ \forall x, y \in \mathbb{R}, x + yi = \{(x,\ y)\}.\ \mathbf{(1)} $$

My question is, is it technically correct to say:

$$ \forall a \in \mathbb{R},\ ( b = 0 \implies a + bi = a)?\ \mathbf{(2)} $$

Since a real number and complex numbers are different objects, would it not be "irelevant" to ask something as $\mathbf{(2)}?$

In asking this, I'm taking a cue from the reasoning presented in Whitehead's and Russell's Principia Mathematica. Thank you all in advance.

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    $\begingroup$ Your question is not clear. You don't put quantifiers at the end of a sentence, you put them at the beginning. If you're asking if real numbers can be interpreted as complex numbers with zero imaginary part, then yes. $\endgroup$ Commented Feb 25, 2012 at 22:27
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    $\begingroup$ It took me a while to understand statement (2). If I understand correctly, it's saying: $\forall a \in \mathbb{R}, (b = 0 \implies a + bi = a)$. $\endgroup$ Commented Feb 25, 2012 at 22:28
  • $\begingroup$ I'll just fix that. $\endgroup$ Commented Feb 25, 2012 at 22:31
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    $\begingroup$ I think what you are asking whether it is meaningful to talk about the real numbers as a subset of the complex numbers (those with imaginary part equal to 0). It is perfectly valid to think of the real numbers this way, and in many contexts it is very useful to do so. $\endgroup$ Commented Feb 25, 2012 at 22:32
  • $\begingroup$ Indeed Principia Mathematica is far from the latest word in mathematical reasoning. It was a considerable achievement for its time, but it was followed by several decades of intensive developments that built upon and refined its ideas, and replaced some of them that were found to be lacking. Thus, modern formal reasoning is considerably simpler, better understood, more regular and worn by actual use than the PM system. $\endgroup$ Commented Feb 25, 2012 at 22:56

3 Answers 3


If you define a complex number as an ordered pair $(x,y)$ of real numbers, then the set of real numbers is not a subset of the set of complex numbers.

That is a problem that is easy to fix. The complex numbers of the form $(x,0)$ are a "copy" of the reals in the complex numbers. We identify this copy of the reals with the real reals, and then don't worry about it. The way that the addition and multiplication are defined on the complex numbers makes the arithmetic of the ordered pairs $(x,0)$ the same as the arithmetic of the reals.

In technical language, there is a natural isomorphism between the reals and their arithmetic and the complex numbers of the shape $(x,0)$, with the arithmetic inherited from the definition of sum and product of complex numbers.

By the way, the sum of two complex numbers $(u,v)$ and $(x,y)$ is defined to be $(u+x, v+y)$. The product of these two complex numbers is defined to be $(ux-vy, uy+vx)$. Calculate $(0,1)$ times $(0,1)$. You will find something interesting!

  • $\begingroup$ That answers my question nicely André. Thanks! $\endgroup$ Commented Feb 26, 2012 at 0:04

Even your first part is not quite correct. $x + yi = \{(x,\ y)\}$ seems to say that $x+yi$ is the set whose only element is the ordered pair $(x,y)$ -- but usually one defines a complex number to be that ordered pair itself. That is, if one uses the definition of complex numbers as ordered pairs in the first place, which is common but not the only way to go. One could also define complex numbers as residue classes in the quotient ring $\mathbb R[X]/(X^2+1)$ or as real matrices of the form $\pmatrix{a&b\\-b&a}$, for example.

The second part of your question is more interesting. (Your use of symbols is rather off, as others have remarked, but I'll not dwell on that). The convention that real numbers "are equal to" certain complex numbers looks rather suspect when those complex numbers themselves are defined as structures that contain the real numbers in the first place.

What one usually does about this is sweep it under the rug as much as possible. At best, the textbook will contain a remark that what we really mean is that there is an injection $\mathbb R\to\mathbb C$ that preserves the behavior of the real numbers. From that point un, the reader is tacitly assumed to keep track of what is what and mentally insert "invisible" applications of that injection into formulas as required to get them to make sense.

This seems to work fine in practice -- at least inasmuch as working mathematicians don't think it confuses them and are generally able to pinpoint exactly where the reinterpretations are needed, provided that you can communicate to them what it is you want, because the request is so rare as to be "strange". But nobody seriously doubts that mathematics as it is commonly done can be formalized with explicit conversions between $\mathbb R$ and $\mathbb C$ if one cares to put in the effort.

Still, I don't think it is really satisfactory that nobody seems to want to put in the effort of formalizing some rules to govern this implicit reinterpretation. I think it would be quite doable, borrowing form the quite powerful techniques that has been developed in computer science for type systems for programming languages. But that's a hobby horse of my own and not something that appears to bother mathematics in general.

By the way, it's not just complex numbers that exhibit the problem. We run into exactly the same situation when we use the rational numbers to construct the reals (using Dedekind cuts or Cauchy sequences) and then declare that some of those real numbers equal certain rationals. Or when we use integers to construct the rationals (which contain the integers), or use the natural numbers to construct the integers (which contain the naturals).

  • $\begingroup$ Very interesting. I got to learn a few things reading your response! I'll also ask you for a book where I can read up on writing Mathematics. The book on proofs I read didn't have much to say on the matter. haha $\endgroup$ Commented Feb 26, 2012 at 0:03

What's happening in (2) is that you are actually identifying $\mathbb{R}$ with the subset (subfield, actually) $$ \{ a+i 0: \ a\in\mathbb{R}\}\subset\mathbb{C}. $$ This identification carries absolutely every property of $\mathbb{R}$ with itself, so it is completely harmless to forget about it and consider $\mathbb{R}$ as a subset of $\mathbb{C}$. Incidentally, the same happens with any of the inclusions $$ \mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}. $$

  • $\begingroup$ Every property ? Also the order properties ? $\endgroup$
    – Kasper
    Commented Mar 12, 2014 at 17:14
  • $\begingroup$ Yes. The function $a\mapsto a+i0$ is a ring monomorphism that preserves order. $\endgroup$ Commented Mar 12, 2014 at 21:47

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