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I was working on this problem from my Complex Analysis book:

First, I assume that $a, b, c$ are not literarly elements of $\mathbb{R}$; we know that $\mathbb{R} \not \subset \mathbb{C}$ and my book hasn't defined multiplication between elements of $\mathbb{R}$ and $\mathbb{C}$, so I'll assume it meant $a, b, c \in \mathbb{C}$ with Im $a, b, c = 0$.

Now if Re$(b^2-4ac, 0)$ is a negative real number, then we can check that $(0, 1) \cdot (\sqrt{-b^2+4ac}, 0)$ is a solution to $z^2 = (b^2-4ac, 0)$. So why do we need to define $\sqrt {b^2-4ac} = i \sqrt {-b^2+4ac}$ if $b^2-4ac < 0$, when in fact it couldn't even be anything else, if it was to remain consistent with the rules of multiplication already given?

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Ovi
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    I don't know that $\mathbb{R} \not\subset \mathbb{C}$. I almost always work in a setting where $\mathbb{R} \subset \mathbb{C}$. – Daniel Fischer Jan 31 '17 at 15:42
  • @DanielFischer In the linked question, I asked weather $\mathbb{R} \subset \mathbb{C}$. The answer is "no, not literarly", because $\mathbb{C}$ is defined as ${(x, y) : x,y \in \mathbb{R} }$. The elements of $\mathbb{C}$ are ordered pairs of real numbers, wile the elements of $\mathbb{R}$ are real numbers. – Ovi Jan 31 '17 at 15:46
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    Are you sure it couldn't be anything else? There are two solutions to $z^2=b^2-4ac$. It's similar to how you define $\sqrt{2}$ to be the positive one, not the negative one. – user160738 Jan 31 '17 at 15:49
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    It depends on how $\mathbb{C}$ and $\mathbb{R}$ are defined. If the book defines $\mathbb{C}$ as $\mathbb{R}^2$, then indeed $\mathbb{R}\not \subset \mathbb{C}$. But one can define things differently, and it's usually much more convenient to have actual inclusions $\mathbb{N}\subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$. – Daniel Fischer Jan 31 '17 at 15:51
  • @DanielFischer Could you point me in the direction of some of those definitions please? I couldn't find them with a quick google search – Ovi Jan 31 '17 at 16:03
  • Henning Makholm linked to this question in the comments at your question, his answer there contains one construction. That construction is however rather ugly in some respects, so I prefer to start defining $\tilde{\mathbb{N}},\dotsc, \tilde{\mathbb{R}}$, with the corresponding embeddings, then define $\mathbb{C}$, for example as $\tilde{\mathbb{R}}^2$, and finally define $\mathbb{R}$ etc. as the image of the embeddings into $\mathbb{C}$. – Daniel Fischer Jan 31 '17 at 16:16
  • @DanielFischer: That's a fair point -- though I personally am willing to suffer a bit of ugliness such that $\mathbb N$ can equal the set-theoretic $\omega$. – hmakholm left over Monica Feb 03 '17 at 11:15
  • @HenningMakholm It depends on what one is doing. If I'm doing set theory, sure, $\omega$ rules. If I'm doing complex analysis, I'm not (or at least very very rarely) interested in $\omega$. And if I'm doing number theory, it's even often convenient to say $0 \notin \mathbb{N}$. I take the liberty to tack the label $\mathbb{N}$ on whatever is the most convenient version for the current purpose. – Daniel Fischer Feb 03 '17 at 11:45

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I believe this author resorts to 'define' because the author takes as given that the square-root symbol is only valid for non-negative real inputs and only produces non-negative real outputs, so strictly speaking it cannot be directly applied to a negative value; hence the notation must be extended to handle the case of a negative discriminant properly. Most authors would gloss over this point.

PMar
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The reason is quite simple. The $\sqrt{\cdot}$ function needs to be defined, for example for real positive $x$ there are two $y$-solutions to $x^2=y$, the convention being that $\sqrt x$ will be the positive solution.

Similarly when $x<0$ the equation $x^2=y$ has two solutions : $i\sqrt{-x}$ and $-i\sqrt{-x}$. So you pick one. Here the convention is that $\sqrt{x}=i\sqrt{-x}$.

ps. In your question, $x=b^2-4\,a\,c$.

A.G.
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Say $b^2-4ac=-3$.then what is $\sqrt{-3}$? It is $\sqrt3i$ right? ($i$ being $\sqrt{-1}$)

Then $\sqrt{b^2-4ac}$ becomes $\sqrt3i$ ,that is, $i\sqrt{(-b^2+4ac)}$.

MR_BD
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jimm
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  • put $ $ around mathmatical equations. Go to http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference for more details –  Jan 31 '17 at 16:05