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$$ ((-1)^{3/2})^2=(-1) ^3=(-1)(-1)(-1)=-1 $$

I haven't done any further research nothing came up on google for this, my calculator said non real but all the working checks out, I checked the exponent rules again and again before this post

edit: I understand that (-1)^3/2 isn't exactly a real number number but how can we prove it? I was always told that -1 didn't have any square roots and that's why it was written as i

why is it the case that (-1)^3/2 isn't a real number? maybe you could do what I did to -1 again? how do we know this process won't have infinitely many steps?

Arsh Dixit
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    $(-1)^{3/2} = -i ; \text{OR} ;i$ is an imaginary number. – Enforce Sep 23 '21 at 21:14
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    And what real number would $$(-1)^{3/2} $$ be? – justabit Sep 23 '21 at 21:15
  • @ArshDixit Accept an answer? Or do you want to delete the question? If you want to close the question (which is again something else), I believe you can flag your own question for closure. – mrtaurho Sep 23 '21 at 21:18
  • I mean you could do the same thing I did for -1 to (-1)^3/2? maybe at some point you would @justabit why is it the case that this will go on forever can we prove this is the case? – Arsh Dixit Sep 23 '21 at 21:19
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    If you just want to get rid of the question, delete it using the Delete button under the question body. I don't think you need any special privileges to delete your own questions. – Rob Arthan Sep 23 '21 at 21:24

2 Answers2

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Good question! I think you squared this one too many times:

$$ (-1)^\frac{3}{2}=(-1\cdot-1\cdot-1)^\frac{1}{2}=(-1)^\frac{1}{2}=\sqrt{-1}=i $$

mrtaurho
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    He did not square it one too many times; he applied $(a^b)^c=(a^c)^b$ which is simply not valid for all complex numbers! – mrtaurho Sep 23 '21 at 21:17
  • why is it not valid for all complex numbers? that's very interesting I didn't know that. – Arsh Dixit Sep 23 '21 at 21:28
  • @mrtaurho, thanks--I missed that! ArshDixit, the answer from Luis Felipe answer why it is invalid. As stated above, we know the domain of $\sqrt{x}$ is $0\leq x<\infty$, so $-1$ is not a member of its domain. –  Sep 23 '21 at 21:36
  • @integer You can't (sadly I might add) refer to two people in one comment using @. Responding to two different people is hence better done by posting two separate comments (although this may appear to as ineffective). – mrtaurho Sep 23 '21 at 21:37
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You are having an error of understanding complex numbers. Several authors construct the complex numbers with the base that $i^2 = -1$. This is different that saying that $\sqrt{-1}=i$, because the function $\sqrt{\cdot}~~$ doesnt have the $-1$ in its domain.

A good book can help you, the complex numbers are seeing as the field $\mathbb{C} =(\mathbb{R}^2,\oplus,\odot)$ where $\oplus,\odot$ are the usual operations on $\mathbb{C}$,mapped on $\mathbb{R}^2$

L F
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  • Would saying $\sqrt{-1}:=i$ be a better representation than $\sqrt{-1}=i$? I teach/tutor college-level algebra and precalculus, so I want to be sure to accurately represent imaginary numbers to my students. –  Sep 23 '21 at 21:34
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    I am a former mathematician. Its better to teach as $i^2 = -1$. You can start for pedagogical reasons wih this: On the set N, we ask what number can solve the equation x+2 = 1, so we create Z. On Z we ask what number can solve the equation $3\cdot x = 4$, so we create Q. Then we ask what numbers are given by geometry like $\pi, e$, so we need the irrational numbers and thus we have R. So the next question is: what number solves x^2 = -1?. We call that number as $i$ – L F Sep 23 '21 at 21:38
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    @integer this will be better, because on university they will be able to understand the idea of $C=(R^2,⊕,⊙)$ , because in this scenario, if you make the draw of $R^2$, you can see that $i=(0,1)$ is the unit on the imaginary axis – L F Sep 23 '21 at 21:40
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    Thank you for the suggestions. Looking back over my comment and your responses, $i^2=-1$ is way more intuitive than the $:=$ notation since using imaginary numbers to provide an answer to $x^2+1=0$ provides a strong foundation (as you rightly pointed out) for many lesson directions. Thank you! –  Sep 24 '21 at 01:03
  • would it then be safe to say that a possible value of i is -1^3/2 ?@LuisFelipe – Arsh Dixit Sep 24 '21 at 14:58
  • @ArshDixit formally speaking, no. – L F Sep 24 '21 at 16:28
  • why? we know that -1^3/2 squared is -1 and that i^2 = -1 therefore it's equal to i, what about my reasoning is wrong? @LuisFelipe – Arsh Dixit Sep 26 '21 at 11:43
  • Read again. What is the domain of square root function? Can you apply to a negative value? – L F Sep 27 '21 at 14:42