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These days I am explaining radicals to my 15-year-old students, and I wanted to add that square roots or roots with even index and negative radical can be solved with complex numbers. I know that n mathematics, the imaginary unit $i$ (sometimes represented by the Greek letter ($\iota$) makes it possible to extend the range of field numbers $\mathbb {R}$ to the field of complex numbers $\mathbb {C}$. The imaginary unit is characterized by being a number whose square is equal to $-1$. The powers of $i$ repeat periodically (they are cyclic with period $4$).

Is there a real reason why $$i^2=-1\,?$$

Or it is like a postulate that we must assume. Sorry for the trivial question.

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    $\begingroup$ I'd say that's the definition $\endgroup$ Commented Sep 16, 2023 at 11:54
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    $\begingroup$ As Exodd says, this is the definition of the imaginary unit. That is the symbol we give to the object with that asserted property. So it makes no more sense to ask why than it would to ask why a triangle is a shape with three sides -- that is simply what the term means. $\endgroup$ Commented Sep 16, 2023 at 11:59
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    $\begingroup$ You might consider showing your students that much the same thing can be done with similar definitions. If $\alpha^2=2$ then you can look at the set of all combinations of the form $a+b\alpha$ where $a,b$ are rational. You can add and multiply (and divide) in such a set. $\endgroup$ Commented Sep 16, 2023 at 12:00
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    $\begingroup$ Please see What allows us to use imaginary numbers? $\endgroup$ Commented Sep 16, 2023 at 12:07
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    $\begingroup$ @Jam I have understood the part of your link where $(0,1)\cdot(0,1)=(-1,0)$ where $(0,1)=i$ and for the rule of the product $(a,b)\cdot(a',b')=(aa'-bb',ab'+a'b)$ we obtain $-1$. Now it is more clear. $\endgroup$ Commented Sep 16, 2023 at 12:13

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This is simply the definition of complex numbers. But I would say that this is just one of the available extensions of the real numbers. We can denote $\mathbb{C} = \mathbb{R}[i] = \{a + bi: a, b \in \mathbb{R}\}$. Similarly:

  • dual numbers: $\mathbb{R[\varepsilon]} = \{a + b\varepsilon: a, b \in \mathbb{R}\}$, where $\varepsilon^2 = 0, \varepsilon \neq 0$;
  • split complex numbers: $\mathbb{R}[j] = \{a + bj: a, b \in \mathbb{R}\}$, where $j^2 = 1, j \notin \mathbb{R}$.

So why do we use complex numbers and not the other two? One sensible explanation is that the complex numbers are the only ones of the three to form a field.

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    $\begingroup$ I don't think this is appropriate for 15 yo students $\endgroup$ Commented Sep 16, 2023 at 12:32
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    $\begingroup$ If they learn about complex numbers, why they shouldn't learn about dual numbers? I think it's an interesting extra-curricular extension of the lesson. $\endgroup$ Commented Sep 16, 2023 at 12:35
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    $\begingroup$ definitely interesting, but too complex to a high school student... $\endgroup$ Commented Sep 16, 2023 at 13:43
  • $\begingroup$ @SineoftheTime dual and split-complex numbers are simpler than complex numbers. $\endgroup$ Commented Oct 13, 2023 at 20:38

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