Rudin says in page fourtheen in theorem 1.29 : If $a$ and $b$ are real, then $(a,b)=a+bi$. Proof he gives: $$a+bi=(a,0)+(b,0)(0,1)\\=(a,0)+(0,b)=(a,b)$$ of course this is correct (if we accept $a=(a,0)$), however i'm interested in how to justify that $a=(a,0)$, it seems entirely unrigorous to me, on one hand we have a pair, so it will be equal to $\{\{a\},\{a,0\}\}$ and on the other we have only one number, so how to make that rigorous?
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Earlier he proves that that $\mathbb{R} \subseteq \mathbb{C}$ by the embedding $x \mapsto (x,0)$; he also defines $i$ to be $(0,1)$ and shows that $i^2 = -1$. Theorem 1.29 simply says that an arbitrary element $(a,b) \in \mathbb{C}$ (indeed we defined $\mathbb{C}$ to be pairs of reals with certain operations making it a field) can be expressed as $a+b i$, where '$a$' denotes $(a,0)$, the canonical embedding of the real $a$ into $\mathbb{C}$.
It's a completely trivial result that only serves to justify the common notion of writing complex numbers as $a+bi$, $(a,b \in \mathbb{R})$. I believe you are really overthinking this as there's no reason to consider the set-theoretic definition of an ordered pair.
basket
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so in $a+bi$, $a$ is simply a notation for $(a,0)$ and not the real number equaling the pair? – user153330 Feb 05 '16 at 09:42
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Yes. As I said, we proved $\mathbb{R}$ is embedded as a field in $\mathbb{C}$ and so just writting $a$ is short for $(a,0)$. – basket Feb 05 '16 at 09:44
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so if i wanted to be a rigorous extremist then the proof to me would be $$\mathbf{a}+\mathbf{b}i=(a,0)+(b,0)(0,1)=(a,0)+(0,b)=(a,b)$$, correct? – user153330 Feb 05 '16 at 09:47
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That is rigorous. It is also exactly the same as Rudin's proof... although it seems more motivated to say any $(a,b) \in \mathbb{C}$ can be expressed as $a+bi$. – basket Feb 05 '16 at 09:49
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but that's only a proof that $(a,0)+(b,0)(0,1)=(a,b)$ as the quantity $a+bi$ would be undefined if $a=$ a real number and $b=(1,0)$ for example – user153330 Feb 05 '16 at 09:52
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$a$ as a real number (in $\mathbb{C}$) is $(a,0)$. We can think of $a$ as a real, an element of $\mathbb{R}$, or as $(a,0) \in \mathbb{C}$, where in this context we mean the latter. But these are of course different objects in ZFC and they belong to different sets. – basket Feb 05 '16 at 10:00
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Rudin is deliberately blurring the distinction to get us used to thinking of $\mathbb{R}$ as embedded in $\mathbb{C}$, even though they are not subsets. – basket Feb 05 '16 at 10:06
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what do you mean by "in $\mathbb{C}$ $a$ is $(a,0)$" formally? v plus if $a=(a,0)$ then $(a,0)=((a,0),0)$ then $((a,0),0)=(((a,0),0),0)$ and so on, which is highly dubious – user153330 Feb 06 '16 at 12:02
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Let us continue this discussion in chat. – basket Feb 06 '16 at 14:23
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okay, i'm wiaiting for ur response there – user153330 Feb 06 '16 at 14:50
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1I don't know how to explain this any more plainly. First note we have a field homomorphism, and embedding $\mathbb{R} \hookrightarrow \mathbb{C}$ given by $a \mapsto (a,0)$. We could call the image of $\mathbb{R}$ in $\mathbb{C}$ $\bar{R}$ considering that $\mathbb{R}$ and $\bar{R}$ are isomorphic yet still different sets. However, this is just nonsense and a pointless formal consession. Why not just call $\bar{R}$ $\mathbb{R}$? Think of $\mathbb{R}$ as contained in $\mathbb{C}$ and $(a,0)$ as a real. Lay off the set theory for a sec – basket Feb 06 '16 at 15:42
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In short, writing $a$ instead of the correct $(a,0)$ can be thought of as convenience or laziness, if that would help you accept it. Really though this 'mistake' deliberately suggests that we should think of $\mathbb{R}$ as a subset, embedded in $\mathbb{C}$, even though the underlying sets are different. – basket Feb 06 '16 at 15:47
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okay, that's much clearer for now, thx so much for your patience and help =) – user153330 Feb 07 '16 at 21:16
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A definition: A complex number is a number of the form $a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit, satisfying $i^2 = −1$.
$a$ is the real part of the complex number $a + bi$; the real number $b$ is the imaginary part of $a + bi$.
So $a = a+0\cdot i = (a,0)$.
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- that's not the definition provided by rudin 2) i'm well aware that $a$ is called the real part of $a+bi$ etc.. 3) what i'm trying to do is to justify that step $a=(a,0)$ as it seems totally unrigorous to me
– user153330 Feb 05 '16 at 09:40 -
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i don't accept that $a=a+0i$ just as i don't accept that $a=a+0\times\begin{pmatrix}0&0&0\0&0&0\0&0&0\end{pmatrix}$ – user153330 Feb 05 '16 at 09:44
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Well of course you don't, you have not defined a suitable multiplication. – fosho Feb 05 '16 at 09:55
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but even if we did then we can't add a real number to an ordered pair, can you do $6+(1,8)$ or so that it becomes clear to you $8+(9,1,7)$? – user153330 Feb 05 '16 at 09:56
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but here the adition operation $+$ in $\mathbb{R}$ is from $\mathbb R^2\longrightarrow\mathbb R$ – user153330 Feb 05 '16 at 09:58
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When you say you don't accept that $a=a+0\times\begin{pmatrix}0&0&0\0&0&0\0&0&0\end{pmatrix}$ you would surely accept that $\begin{pmatrix}a&0&0\0&a&0\0&0&a\end{pmatrix} =\begin{pmatrix}a&0&0\0&a&0\0&0&a\end{pmatrix}+0\times\begin{pmatrix}0&0&0\0&0&0\0&0&0\end{pmatrix}$. When Rudin writes $a$ he really means $(a,0)$... – basket Feb 05 '16 at 10:10