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", 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