The question is$$\text{Is }\mathbb{R}^2\text{ a subspace of }\mathbb{C}^2?$$My first thing to think about it now is $$\text{Is }\mathbb{R}^2\text{ a subset of }\mathbb{C}^2?$$ I think no because what is $\mathbb{R}^2$ it's like all those pairs $(0,1)$ right? and what is $\mathbb{C}^2$ it's like euh $((a,b),(c,d))$ (because a complex number is a pair of two numbers with multiplication defined as $(a,b)(c,d)=(ac-bd,ad+bc)$ and addition $(a,b)+(c,d)=(a+c,b+d)$) so they aren't the same$$$$another point is of course $\mathbb{R}^2$ is a vector space over $\mathbb{R}$ with specific operations, and $\mathbb{C}^2$ is a vector space over $\mathbb{C}$ with different operations, so my gut feeling tells me that they can't be the same;;;;;;;; in general if $V$ is a vector space over $D$ with operations $+,\times$ and $U$ is a vector space over $B$ with operations $\oplus,\otimes$ and $B\ne D$ can it be the case that $U$ is a subset of $V$? Over what?
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$\newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}}$ Well, if $\RR$ is a subset of $\CC$, then $\RR \times \RR$ can be seen as a subset of $\CC \times \CC$, as the pairs of points that only involve real numbers. – Pedro M. Jan 28 '15 at 15:11
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@PedroM. But the operations are different, also I don't think $\mathbb{R}$ is a subset of $\mathbb{C}$ since this very last one is a pair of two reals. and $\text{real number}\ne(\text{real number , 0})$ – SQL Injection Jan 28 '15 at 15:15
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In Mathematics, we often identify things that are "pretty much equal'', even though they are technically not equal. In this particular case, the function $x \in \mathbb{R} \to (x,0) \in \mathbb{C}$ is continuous, respects sums and products etc: in short, working with (complex) $(x,0)$ is the same as working with (real) $x$, so we say that $\mathbb{R}$ is a subset of $\mathbb{C}$. – Pedro M. Jan 28 '15 at 15:20
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The title of the post is so bad... – Troy Woo Jan 28 '15 at 15:49
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@TroyWoo i'm so sorry my dear, can you change it? – SQL Injection Jan 28 '15 at 15:52
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@SQLInjection I think you should improve it yourself. – Troy Woo Jan 28 '15 at 15:53
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@TroyWoo i don't know, in general it's rather subjective, so if you don't like it you are the best one to improve it – SQL Injection Jan 28 '15 at 15:54
2 Answers
Usually we say that $\mathbb R$ is a subset of $\mathbb C$, because any real number, such as $0$ or $2$ or $\pi$ is also a member of $\mathbb C$. There are various schools of thought about how to justify this formally, but that doesn't need to concern us for the purpose of this question -- just so long as we accept that, say, $\pi\in\mathbb C$ is a true statement, which is definitely the case in ordinary everyday mathematics.
In this case it is indeed the case that $\mathbb R^2\subseteq \mathbb C^2$ because every element of the former is also in the latter.
For example $(\sqrt 2,\pi)\in \mathbb R^2$ is also in $\mathbb C^2$ because $(\sqrt2,\pi)=(\sqrt2+0i,\pi+0i)\in\mathbb C^2$.
So $\mathbb R^2$ is certainly a subset of $\mathbb C^2$. Whether it is also a subspace depends on which kind of vector space we consider $\mathbb C^2$. If we say $\mathbb C^2$ is a complex vector space, then $\mathbb R^2$ is not a subspace because it fails to be closed under scalar multiplication. But if $\mathbb C^2$ is a real vector space (with the obvious operations), then $\mathbb R^2$ is a subspace.
So the real answer is that the question is ambiguous.
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Thanks, very instructive answer, clears up my misconception that $\text{real number}\ne(\text{real number , 0})$. What about my very last question? – SQL Injection Jan 28 '15 at 15:19
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@SQLInjection: In order to be subspaces the two vector spaces need to be over the same field. But if $(V,{+},{\times})$ and $(U,{\oplus},{\otimes})$ are vector spaces over the same field $F$, then we can say that $V$ is a subspace of $U$ if $V\subseteq U$ and $+$ equals the restriction of $\oplus$ to $V\times V$ and $\times$ equals the restriction of $\otimes$ to $F\times V$. – hmakholm left over Monica Jan 28 '15 at 15:21
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If $V$ is a vector space over $\mathbb C$, we can derive a vector space over $\mathbb R$ from it, simply by restricting the multiplication operator to $\mathbb R\times V$. But we usually talk about the resulting space as a different vector space, because the scalar field is different. – hmakholm left over Monica Jan 28 '15 at 15:23
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@HenningMakholm: Regarding your comment, I agree with your choice to have two different sums for $U$ and $V$, but there is no need for two cartesian products, as this refers simply to the standard set-theoretical notion. – Pedro M. Jan 28 '15 at 15:24
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@PedroM: Note that the $\times$ and $\otimes$ in the above comments are meant to be the scalar products of the relevant vector spaces, not cartesian products. (This is not really standard notation, but it's what the OP used at the end of the question, and I didn't want to confuse the OP further by quibbling over his notation). – hmakholm left over Monica Jan 28 '15 at 15:26
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@Henning: I think you are missing the point here. A real number, as usually defined, is not a complex number, as usually defined. You are so used to the natural embedding $x \mapsto x+0i$ that you are forgetting that it is not an identity. – TonyK Jan 28 '15 at 15:43
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@TonyK: It can be an identity if you want to (see the "various schools of thought" link), and it definitely is an identity the way mathematics is usually spoken of. – hmakholm left over Monica Jan 28 '15 at 15:51
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I totally disagree with your definitely. Perhaps it's how we were raised, but if somebody says "$\mathbb R$" to me, I think of Dedekind cuts, or equivalence classes of Cauchy sequences. I definitely don't think of $\mathbb C$, unless it becomes relevant later on. Then I perform the embedding. By the way, does your identity extend to the quaternions too? – TonyK Jan 28 '15 at 17:08
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@TonyK: Nobody says you should think of complex number when someone says $\mathbb R$, but when someones says $\mathbb C$, you're certainly supposed to think of something that has $\mathbb R$ as a subset. You can define quaternions to be an actual superset of $\mathbb C$ if you want to, sure. – hmakholm left over Monica Jan 28 '15 at 17:37
According to most of the usual definitions, $\mathbb R$ is not a subset of $\mathbb C$. You are absolutely correct about this. In the same way, $\mathbb N$ is not a subset of $\mathbb Z$, which is not a subset of $\mathbb Q$, which is not a subset of $\mathbb R$.
However, $\mathbb N$ has a natural embedding in $\mathbb Z$, which has a natural embedding in $\mathbb Q$, etc. For instance, the integer $n \in \mathbb Z$ is mapped by this natural embedding to (the equivalence class of) the rational $\dfrac{n}{1} \in \mathbb Q$. And the real number $x \in \mathbb R$ is mapped to the complex number $x+0i \in \mathbb C$.
Often $-$ in fact, almost always $-$ this natural embedding is left implicit. So when $\mathbb R^2$ is treated as if it were a subset of $\mathbb C^2$, this is to be understood as $\mathbb R'^2$ being a subset of $\mathbb C^2$, where $\mathbb R'$ is the image of the natural embedding $\mathbb R \to \mathbb C$.
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I didn't know about natural embeddings thanks, but what about the last question? – SQL Injection Jan 28 '15 at 15:39