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This question already has an answer here:

I am having an argument with a friend. I think that in a sense, the answer is no. My reasoning is that in linear algebra, a vector $(a, b)$ is not the same as a vector $(a, b, 0)$ because the first one is in $\mathbb{R}^2$, while the second is in $\mathbb{R}^3$. However I am not sure if a similar argument can be made for real vs complex numbers.

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marked as duplicate by Matthew Towers, Bungo, user137731, GEdgar, Lee Mosher Aug 1 '16 at 16:55

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  • $\begingroup$ $\mathbb{R}$ is a subfield of $\mathbb{C}$ $\endgroup$ – reuns Aug 1 '16 at 16:39
  • $\begingroup$ Recall that the set of complex numbers $\Bbb C$ is defined as $\Bbb C=\{a+ib\mid a,b\in\Bbb R\}$ where $i$ is the imaginary unit. You may see each $a+ib$ as an ordered pair $(a,b)$. What is the subset of $\Bbb C$ in which $b=0$ for every element? $\endgroup$ – learner Aug 1 '16 at 16:40
  • $\begingroup$ But (a,b,0) is in $\mathbb R^2 \times {0}$. Is $\mathbb R^2 \times {0} \ne \mathbb R^2$. In what sense can that statement be meaningful and true? Does $\mathbb C = R^2$? Not really as as $a = a + 0i \in \mathbb R$ but $a + 0i \in \mathbb C$ so ... yes, $\mathbb R \subset \mathbb C$ be any definition. $\endgroup$ – fleablood Aug 1 '16 at 16:44
  • $\begingroup$ To understand the natural embedding you need to look at the ring / algebra structure, not only the vector space structure. The Hamilton pair construction is a normalized version of $\Bbb C \cong R[x]/(x^2+1),\,$ i.e take $\,\Bbb R[x] = $ ring of polynomials with real coefficients, then work modulu $\,x^2+1.\,$ The embedding arises as a composition of two natural maps: $\,r\mapsto r\, x^0,\,$ which maps a real into a constant polynomial, then mapping that to its congruence class mod $\,x^2+1.\,$ $\endgroup$ – Bill Dubuque Aug 1 '16 at 17:00
  • $\begingroup$ @BillDubuque Not sure you need all that. The embedding $a\to (a,0)$ works rather well. $\endgroup$ – zhw. Aug 1 '16 at 17:03
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Real numbers are just complex numbers with no imaginary part. Linear algebra can get away with saying "a 2-vector is not the same as a 3-vector" because there is no sense of multiplication between vectors. However, real numbers have multiplication, and the complex numbers extend the reals by adding i.

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    $\begingroup$ "No imaginary part"? I think you mean "imaginary part = 0". Anyway, real numbers are not complex numbers with zero imaginary part, because complex numbers are elements of $\mathbb R^2. $ What is true is that the set of complex numbers of the form $(a,0), a \in \mathbb R,$ is isomorphic to $\mathbb R$ in every important mathematical sense you could bring to the table. $\endgroup$ – zhw. Aug 1 '16 at 16:57

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