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A teacher of mine says to me that the real numbers is NOT a subset of complex numbers, but that complex numbers has an isomorphic copy of real numbers with the isomorphism $x$ $\rightarrow$ $(x,0)$. I can see this if we consider the complex numbers as $$C=\{(a,b):a,b \in \mathbb{R} \}$$ because the real numbers are not ordered couples.

But what if I consider $$C=\{x+iy:x,y \in \mathbb{R}, i^2=-1\}?$$ Then $\mathbb{R}$ is a subset of $\mathbb{C}$ because $x= x+0$ for all $x \in \mathbb{R}$.

So... $\mathbb{R}$ is a subset of $\mathbb{C}$ or not?

Any help would be appreciated.

NOTE: My question is slightly different from the one "$a=a+0i$?" because I'm taking two definitions of $\mathbb{C}$ but there is more sets isomorphic to $\mathbb{C}$ and I want to know if always the reals are a subset of the complex numbers or not.


marked as duplicate by MJD, Dietrich Burde, Andrew D. Hwang, Siminore, Aloizio Macedo Oct 3 '15 at 19:56

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Well, it is not really true that the complex numbers are ordered pairs. Ordered pairs just give a specific model of the complex numbers.

The difference is best seen when considering the construction of the real numbers. If you look at the linked Wikipedia page, you'll find quite a lot of different constructions. For example, there's the original Dedekind construction, where every real number is a pair of sets of rational numbers. Then there's the modern version, where instead of a pair, only a single set of rational numbers is used. Then there's the construction as equivalence class of Cauchy sequences; in that construction, the real numbers are sets of functions from the natural numbers to the rational numbers.

In all those constructions, the real numbers end up as very different sets. So which of them are the "real" real numbers? Well, all of them. All those construction give different models for the same thing, namely the real numbers. You can show that by doing an 1:1 mapping from one model to another, that preserves the structure (addition, multiplication, etc.). That is, "the real numbers" is not any single of those constructions, it's the structure for which all those constructions generate a model.

Let me suggest yet another, non-standard construction of the real numbers: First, I define the complex numbers as equivalence classes of Cauchy sequences of pairs of rational numbers using e.g. the Manhattan distance. Note that in this construction, the complex numbers are not at all pairs of real numbers (indeed, I bypassed the definition of the real numbers so far), but equivalence classes of sequences of pairs of rational numbers. Then I define the real numbers as complex numbers with zero imaginary part.

Note that in this definition, the real numbers are by construction a subset of the complex numbers. And if done correctly (I've only sketched the way, of course) these thus defined real and complex numbers have exactly the same properties as the real and complex numbers defined in any other valid way.

  • $\begingroup$ Or see Chapters 1 and 2 of Hardy's Pure Mathematics as well, but it's a pretty in-depth text. $\endgroup$ – jm324354 Oct 3 '15 at 18:00
  • $\begingroup$ You might want to add when you talk about 1:1 maps between different versions of the real numbers, that those maps also preserve the structure you have in each version. $\endgroup$ – Asaf Karagila Oct 3 '15 at 18:07
  • $\begingroup$ @AsafKaragila: You're right, it's better to explicitly spell it out. I'll edit my answer accordingly. $\endgroup$ – celtschk Oct 3 '15 at 18:13

As you have mentioned, it depends on how one defines the complex numbers. From a purely set-theoretic point of view, $\mathbb{R} \not \subset \mathbb{C}$, rather the set of things in $\mathbb{C}$ which resemble real numbers is isomorphic to $\mathbb{R}$. For all intents and purposes however, one can consider $\mathbb{R} \subset \mathbb{C}$.

  • 4
    $\begingroup$ One can also decide to define $\Bbb C$ in one way or another, then define $\Bbb R$ to be literally the subset of $\Bbb C$ which satisfies $z=\bar z$ (where $\bar z$ is the complex conjugation). Then $\Bbb R$ is by definition a subset of $\Bbb C$. $\endgroup$ – Asaf Karagila Oct 3 '15 at 17:49

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