Is $\mathbb{R}^2$ the same as my dear $\mathbb{C}^2$? 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?
 A: 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.
A: 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 $\mathbb R \to \mathbb C$.
