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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.

Sebastiano
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    I'd say that's the definition – Exodd Sep 16 '23 at 11:54
  • @Exodd I asked this question because a student asked why it must be $i^2=-1$? I have replied that it is an assertion. Thank you for your comment. – Sebastiano Sep 16 '23 at 11:56
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    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. – Jam Sep 16 '23 at 11:59
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    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. – lulu Sep 16 '23 at 12:00
  • @Jam Thank you very much for your contibute. – Sebastiano Sep 16 '23 at 12:02
  • @lulu Unfortunately, I am rusty. Could you give me a complete answer please? I did not have understood your comment. – Sebastiano Sep 16 '23 at 12:03
  • Well, $(a+b\sqrt 2)+(c+d\sqrt 2)=(a+c)+(b+d)\sqrt 2$. And $(a+b\sqrt 2)\times (c+d \sqrt 2)=(ac+2bd)+(ad+bc)\sqrt 2)$. Division is similar to the complex numbers. We define the norm $|a+b\sqrt 2|=(a+b\sqrt 2)(a-b\sqrt 2)=a^2-2b^2$ and use that (just as we divide one complex number by another). – lulu Sep 16 '23 at 12:06
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    The simplest polynomial equation which does not have a real root (solution) is $x^2+1=0$ . So we define $i$ as a root of this polynomial then we can extend this for any polynomial, which does not have an any real root. – lone student Sep 16 '23 at 12:06
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    Please see What allows us to use imaginary numbers? – Jam Sep 16 '23 at 12:07
  • The point is that you can sometimes "extend" some basic algebraic structures by adjoining roots of polynomials to the system. That's how we get from the reals to the complex numbers, for example. It's an amazing fact that just adjoining $\sqrt {-1}$ is enough to get us roots to all polynomials with real coefficients. – lulu Sep 16 '23 at 12:07
  • @lulu The previous comment is not very clear to me. Sorry. – Sebastiano Sep 16 '23 at 12:09
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    @Sebastiano Re Lulu's comment, the algebra of $\mathbb{C}$ (ie, $a+bi$) operates very similarly to the algebra of the set of $a+b\sqrt{2}$. Students may be more comfortable with the latter since it exclusively uses real values, but the algebra is almost identical. For both, addition brings values to the $a$ and $b$ parts respectively, while multiplication brings $a\cdot a$ and $b\cdot b$ terms to the $a$ part and $a\cdot b$ terms to the $b$ part. – Jam Sep 16 '23 at 12:11
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    @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. – Sebastiano Sep 16 '23 at 12:13
  • @Jam I did not know this fact. Even in the university I don't think I studied it. – Sebastiano Sep 16 '23 at 12:15

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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.

Michał Dobranowski
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