We have the set-theoretic definition for pairs as: (x , y) = {{x}, {x, y}} Also we have the definition: complex 1 = (1, 0) So if real 1 = complex 1 we would have: 1 = (1, 0) = {{1}, {1, 0}} Which seems paradoxical. Am I missing a point?
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1In the complex "space" the real $1$ is the complex $(1,0)$. – Mauro ALLEGRANZA Jan 31 '20 at 13:32
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Nice question, +1. Ultimately I think the point you are 'missing' is that in reality no-one thinks about numbers (or other mathematical objects) in terms of their set-theoretic definitions. But of course this is not an answer to your question in the strict sense. – Vincent Jan 31 '20 at 13:36
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If we think in terms of sets, we have that the set of complex numbers has a subset that "looks like" the set of real, that is not exactly the same as considering the reals a subset of the complex. – Mauro ALLEGRANZA Jan 31 '20 at 13:37
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Real $1$ is *identified* with complex $1$ but not identical – J. W. Tanner Jan 31 '20 at 13:44
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Yes, technically each formal theory (real line, complex plane, set theory) has its own realization of $1$, but this is glossed over when one theory is embedded into another. Without the gloss we get what is called ["junk theorems"](https://mathoverflow.net/questions/90820/set-theories-without-junk-theorems), like $1\in2$ in set theory. – Conifold Jan 31 '20 at 13:45
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2Real $1$ and complex $1+0i$ are elements of different fields so you cannot really compare them. – Vasili Jan 31 '20 at 13:58
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This is not an *exact* duplicate, but the answers are the same. This is about setting up context for inclusion, and whether or not $\Bbb{R\subseteq C}$ or $\Bbb{N\nsubseteq Q}$ is a matter of convention. In either case we do have *canonical* embeddings, so we *may* identify the canonical copy inside each "larger" structure. Whether or not we choose to do so is up to what is more convenient. – Asaf Karagila Jan 31 '20 at 14:09