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I have read this PDF from ocw.mit.edu about complex numbers. There is one interesting question: Imagine yourself at the time, when complex numbers had to be invented yet. How to show that $$\sqrt[3]{-1+\sqrt{-7}}+\sqrt[3]{-1-\sqrt{-7}}$$ is a real number.

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    $\begingroup$ You are right: if you don't assume the existence of complex numbers, this expression is meaningless. But this is exactly how complex numbers came to be invented. $\endgroup$
    – TonyK
    Jan 29, 2015 at 20:12
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    $\begingroup$ fiddle with Cardano's method and see what equation of type $x^3 + p x + q = 0$ this expression solves. I'm not saying this really leaves complex numbers out of the picture... $\endgroup$
    – Will Jagy
    Jan 29, 2015 at 20:19
  • $\begingroup$ @TonyK I didn't express myself clearly. I was imagining myself in the situation before the invention of complex numbers. At that time negative roots did not make sense. Then, how would I know that this expression above makes sense at all (i.e. the result is a real number)? $\endgroup$
    – user50224
    Jan 29, 2015 at 20:23
  • $\begingroup$ I don't see any misunderstanding between us! Cardano didn't think it made sense either, but it gave results, so he used it. It reminds me of the 20th-century quantum field theorists who discovered that they could reproduce experimental results by subtracting infinity from infinity to leave a finite quantity. (This last problem still hasn't been resolved, as far as I know.) $\endgroup$
    – TonyK
    Jan 29, 2015 at 22:46

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The polynomial $x^3-6x+2=0$ has three real roots, which is easy to prove. By Cardano's method, one of the roots is just the above expression. So it must be real. If we do not believe the existence of complex numbers though, the expression is not real, because it does not exist.

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    $\begingroup$ Right. Note that this is a case of en.wikipedia.org/wiki/Casus_irreducibilis so that there is no way to re-write this using entirely real numbers. Finding the real and imaginary parts of the cube root of a complex involves more cubic polynomials with three real roots; all very cyclical, there is no way to win. $\endgroup$
    – Will Jagy
    Jan 29, 2015 at 20:29
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    $\begingroup$ The Fundamental Theorem of Algebra wasn't available in Fontana (Tartaglia) / Cardano / Ferrari's time [see en.wikipedia.org/wiki/Fundamental_theorem_of_algebra ] . At best, it was thought that a cubic polynomial could have three solutions, but the arguments we'd give now weren't yet known. Some of Cardano's examples simply had evident "real" solutions (that is, you could see the integer that works), but the formula gives this complicated equivalent expression. $\endgroup$ Jan 29, 2015 at 20:32
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    $\begingroup$ @user50224, this is the best you can do. Suggest you read other historical discussions for the ins and outs between the first perception of a need for complex numbers and their invention. $\endgroup$
    – Will Jagy
    Jan 29, 2015 at 20:33
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    $\begingroup$ amazon.com/An-Imaginary-Tale-Princeton-Science/dp/0691146004 $\endgroup$
    – Will Jagy
    Jan 29, 2015 at 20:38
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    $\begingroup$ Bombelli [ en.wikipedia.org/wiki/Rafael_Bombelli ] had begun to sort out the concept of complex numbers pretty closely on the heels of the work on cubic and quartic polynomials, but it was quite some time before it really made an impression in mathematical thinking. The marvel of this is that all of these people are doing this without modern algebraic notation! (Viete's work is still decades in the future...) For historical discussion, I'll chime in with Mario Livio's The Equation that Couldn't Be Solved. $\endgroup$ Jan 29, 2015 at 20:39
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Let $$r = (-1 + \sqrt{-7})^{1/3} + (-1 - \sqrt{-7})^{1/3},$$ where by $\sqrt{-7}$ we mean some number $x$ whose square is $-7$; i.e., $x^2 = -7$. Recall the identity $$(a^{1/3} + b^{1/3})^3 = a + b + 3(ab)^{1/3}(a^{1/3} + b^{1/3}).$$ Consequently, $$r^3 = -2 + 3((-1)^2 + 7)^{1/3}r = 6r - 2.$$ Now consider the trigonometric identity $$\begin{align*} \cos 3\theta &= \cos \theta \cos 2\theta - \sin \theta \sin 2\theta \\ &= \cos^3 \theta - \cos \theta \sin^2 \theta - 2 \sin^2 \theta \cos \theta \\ &= \cos^3 \theta - 3 \cos\theta \sin^2 \theta \\ &= \cos^3 \theta - 3 \cos\theta (1 - \cos^2 \theta) \\ &= 4 \cos^3 \theta - 3 \cos \theta. \end{align*}$$ This suggests the choice $r = 2 \sqrt{2} \cos \theta$ gives $$\begin{align*} r^3 - 6r + 2 &= 16 \sqrt{2} \cos^3 \theta - 12 \sqrt{2} \cos \theta + 2 \\ &= 4 \sqrt{2} (4 \cos^3 \theta - 3 \cos \theta) + 2 \\ &= 4 \sqrt{2} \cos 3\theta + 2 = 0. \end{align*}$$ Consequently, $$\theta = \frac{1}{3}\cos^{-1}\left( - \frac{1}{2 \sqrt{2}} \right),$$ hence $$r \in 2 \sqrt{2} \cos \left( \frac{1}{3} \cos^{-1} \left( - \frac{1}{2\sqrt{2}} \right) + \frac{2\pi k}{3}\right), \quad k = 0, 1, 2,$$ all of which are real values. Of course, this result presupposes that the rules that allow us to work with real numbers also work for numbers like $\sqrt{-7}$. To be completely rigorous, in my opinion, requires an axiomatic treatment of the complex numbers. The above really only demonstrates that $r$ is real if the rules we use for arithmetic are extended to such values.

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I think it's a stretch to claim that this result can be proven without complex numbers, especially because of the presence of $\sqrt{-7}$. Anyway, here's a way to look at it. Let $A = \sqrt[3]{-1 + \sqrt{-7}}$ and $B = \sqrt[3]{-1 - \sqrt{-7}}$. If $t = A + B$, then $$t^3 = A^3 + B^3 + 3AB(A + B) = (-1 + \sqrt{-7}) + (-1 - \sqrt{-7}) + 3t\sqrt[3]{1 - (-7)} = -2 + 6t$$ So $t$ is a root of the cubic $x^3 - 6x + 2$. Let $f(x) = x^3 - 6x + 2$. Then $f(-3) = -7 < 0$, $f(0) = 2 > 0$, $f(1) = -3 < 0$, and $f(3) = 11 > 0$. Hence, by the intermediate value theorem, $f$ has three real roots. Therefore, $t$ is real.

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