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This question already has an answer here:

I know three questions (that gained momentum) that have been posted asking a question which seems the same, but answers to none of them answer the following very well.

Please jump to point 2 & 3 for immediate addressing to the problem.

  1. Knowledge that I currently have: I was introduced to imaginary numbers a while back where just the square root of -1 seemed to be called iota. This made a whole lot of 'things' definable/analyze-able. I have currently studied the two dimensional complex plane, the rotation, Euler form, defining locus/area which is indefinitely better than an algebraic equation/expression, etc.

  2. My Question: However, I have failed to grasp that how imaginary numbers even make sense. Let's take an example, let there be a quadratic polynomial not intersecting the x-axis. However, it does with two imaginary numbers. How does it intersect the x-axis when the equation of the quadratic dictates that it doesn't intersect the x-axis (discriminant<0)?

  3. My current understanding:Several sites led me to the conclusion that imaginary-axis acts as the 'z' axis and gets the quadratic to intersect with the x-axis somehow. This pushes the curve form a 2-D to a 3-D curve existing in the complex space when we start feeding complex values into the function with the x-axis as the intersection the the two, Cartesian and Complex plane. This 3-D structure now intersects somewhere on the complex plane with the x-axis giving us the complex roots of the quadratic.

I however fail to understand that how this curve transitions from the Cartesian plane to the complex space while being continuous. Does it transform to a surface instead of a curve? If it transforms to a surface then it would have infinite roots with the x-axis(as seen in the video mentioned in the footnote)

I understand that the complex numbers were 'invented' rather than discovered (I'm saying this because other numbers; irrational, rational, negatives actually mean something, hence discovered).

Thank you very much for your valuable time. I really appreciate it.

FOOTNOTE : I have been to the following questions/sites :

1

2

3

4

5 This video here is very good

6

EDIT: The question was marked as a duplicate though it did not in a way adress the same question as the question that was marked as 'original'. I have therefore, changed the title and content to make the question as precise and understandable as possible.

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marked as duplicate by Asaf Karagila, Sean Roberson, Thomas Andrews, Did, TheGeekGreek Jul 25 '17 at 20:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The fact that you've seen some previous questions and read their answers is one thing. Another is explaining what you don't understand about these answers. if the answer to that is "more or less everything", then I'm sorry to be the one to tell you, but math is hard work. There are no shortcuts. And it's hard to explain things to people who did not walk these many miles themselves. At least not in a way which is satisfactory to anyone. $\endgroup$ – Asaf Karagila Jul 25 '17 at 19:24
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    $\begingroup$ Maths really is wizardry. But in this case most people can be wizards to some extent. $\endgroup$ – Robert Israel Jul 25 '17 at 19:27
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    $\begingroup$ Also, let me give you a free piece of advice, title "tags" and SELECTIVE CAPITAL LETTERS make it seem as if you don't control the tone of your voice, and you're shifting from speaking to screaming. I'm sure it makes sense to you, but for the most of us, it just looks weird. $\endgroup$ – Asaf Karagila Jul 25 '17 at 19:36
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    $\begingroup$ Complex numbers make sense in precisely the manner in which they are defined. You can not expect them to make sense like natural numbers (used for counting) or real numbers (used for measuring). Complex numbers can be used to represent two dimensional vectors and they prove very useful especially in dealing with plane transformations like scaling, rotation, translation. $\endgroup$ – Paramanand Singh Jul 25 '17 at 19:43
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    $\begingroup$ The curve actually becomes a 4-D thing $\endgroup$ – Akiva Weinberger Jul 25 '17 at 21:53
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Your problem seems to be that you are trying to use the same intuition that you use for real numbers.

How does it intersect the x-axis when the equation of the quadratic dictates that it doesn't intersect the x-axis

It doesn't intersect the $x$-axis: the $x$-axis consists of real numbers. However, there are solutions that are complex numbers. These are not on the $x$-axis, because they are not real numbers.

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  • $\begingroup$ Yes sir. I shouldn't confuse complex numbers with the real numbers. The very example that you have addressed to, I understand this : The x-axis is just the intersection of the complex plane and the Cartesian plane. The imaginary axis adds another dimension (just like z-axis) and transforms the curve to a 3-d curve/surface when we feed complex numbers instead of real numbers into the function (or the quadratic equation). Is this correct? Also, a new question stems, is this new 3-d curve continuous, I mean, Does the new 3-d curve /surface continuous even after the introduction of complex plane? $\endgroup$ – Gaurav Agarwal Jul 25 '17 at 19:59
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A number system where we can do algebra needs to have addition, multiplication, subtraction and division and they need to behave well (for example $ab=0$ needs to imply $a=0$ or $b=0$, etc...). Such a system is called a "field". One such field is $\Bbb Q$ the rational numbers. Another is $\Bbb R$ the real numbers. Yet another is $\Bbb C$. And in fact $\Bbb Q\subseteq \Bbb R\subseteq \Bbb C$. It's just a set with four operations (two operations and their inverses really). Even though you cannot find $\Bbb C$ in your back yard, it's still a legitimate field that contains $\Bbb R$ and as such is a valid place to do algebra.

Now it turns out even if all you care about is $\Bbb Q$ it can be extremely useful to have the elbow room you get from thinking of $\Bbb Q$ as a subset of $\Bbb R$. And similarly in some cases (many cases) it's useful to have the additional elbow room you get from $\Bbb C$. So for example, Fermat's Last Theorem is a theorem about $\Bbb Q$ but if you didn't give yourself the elbow room of embedding $\Bbb Q$ into $\Bbb C$ then you would never be able to prove FLT.

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Anyway did you ever see an irrational number?

This being said, when one asserts the parabola, say $y=x^2+x+1$ intersects the $x$-axis, one assumes $x$ and $y$ are complex numbers, i.e; the curve lives in $\mathbf C^2$, not in $\mathbf R^2$.

It is an algebraic variety of dimension $1$ over $\mathbf C$. In other words, from the real point of view, it is a surface embedded in a space of dimension $4$.

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    $\begingroup$ I saw an irrational number, it was the diagonal of a square of side length one. $\endgroup$ – Gregory Grant Jul 25 '17 at 19:59
  • $\begingroup$ Did you transcendental numbers, then? $\endgroup$ – Bernard Jul 25 '17 at 20:41
  • $\begingroup$ Yes, it was the circumference of a circle of radius one. Or the radius of a circle of circumference one. $\endgroup$ – Gregory Grant Jul 25 '17 at 20:49
  • $\begingroup$ None else? They're supposed not to be countable… $\endgroup$ – Bernard Jul 25 '17 at 20:51
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    $\begingroup$ They're not countable, but I only claimed to have seen two of them. $\endgroup$ – Gregory Grant Jul 25 '17 at 21:02

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