-3
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$a \in \mathbb{C}$

Which is the correct option:
1) $a = 1$
2) $a = 42$
3) $a = i =\sqrt{-1}$
4) $0^a \in \mathbb{C}$
5) $a^3 = 1$

Hint 1:

Read the title.

Hint 2:

If only one option is right, the other options must be wrong.

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closed as off-topic by Rand al'Thor, Peregrine Rook, Chowzen, QuantumTwinkie, hexomino Mar 18 at 14:29

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  • 1
    $\begingroup$ You should really write $i$ instead of $\sqrt {-1}$ which is a very misleading notation (there is just one step from there to thinking that $i$ is the complex number whose square is $-1$). $\endgroup$ – Arnaud Mortier Mar 8 at 20:20
  • $\begingroup$ Thank You for the feedback, but I do not understand. I wrote $\sqrt{-1}$ as I felt that it would be more clear because not everyone uses $i$ as the complex part of a number. Also, I am confused, because $i$ is defined as the number that on squaring equals to $-1$. $\endgroup$ – Kartik Soneji Mar 9 at 6:47
  • 3
    $\begingroup$ @JonMarkPerry That is very much not true! $i$ is not positive! $i$ is defined as a number whose square is $-1$. To use $\sqrt{\cdot}$ with complex numbers, you need to figure out how to define it, which isn't as easy as you'd think. (You have to choose a branch cut: say, $\sqrt{\cdot}$ gives the number whose argument is in $[0,\pi)$.) And it doesn't interact as nicely with exponents anymore. $\endgroup$ – Deusovi Mar 9 at 9:02
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    $\begingroup$ @JonMarkPerry But $i$ is not defined in terms of itself; it is simply defined as a number that squares to $-1$. Just because you can express that in terms of itself doesn't mean that's how the definition works. $\endgroup$ – Deusovi Mar 9 at 10:07
  • 3
    $\begingroup$ @JonMarkPerry There is no intrinsic way to make a difference between the two roots of the polynomial $X^2+1$. That is the core idea behind Galois theory. In other words, $i$ is a symbol to denote one of them, and it should always be referred to as a root of $-1$ (as in all of Deusovi's comments). $\endgroup$ – Arnaud Mortier Mar 9 at 10:33
3
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I suppose any of

3, 4, or 5 could be correct.

because

both $i$ and $-i$ work for 3

and

any real number that isn't 1 or 42 works for 4

and

$\frac{1}{2} + i\frac{\sqrt{3}}{2}$ or $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$ works for 5

Note:

According to Wolfram, $0^i$ is undefined

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  • $\begingroup$ Your last comment is the reason why the problem is ill-posed. $\endgroup$ – Arnaud Mortier Mar 8 at 20:26
  • $\begingroup$ Added the Solution. $\endgroup$ – Kartik Soneji Mar 9 at 7:34
  • $\begingroup$ There is one correct answer that makes all the other options wrong. $\endgroup$ – Kartik Soneji Mar 9 at 8:27
1
$\begingroup$

If $a=1$ then

it is also true that $0^a\in \mathbb{C}$ and that $a^3=1$, so option 1 cannot be the only correct one.

In fact,

$0^a\in \mathbb{C}$ is straightforwardly true for any nonzero $a$, which means that none of options 1,2,3,5 can possibly be the only correct one.

Therefore, if "only one option is correct" then

it must be option 4, which will e.g. be the only correct option if $a=2$.

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  • $\begingroup$ Added the Solution. $\endgroup$ – Kartik Soneji Mar 9 at 7:34
  • $\begingroup$ But if Option 4 is correct, then Option 1, Option 2 and Option 5 may also be correct. Note that they may be correct, and do not have to be correct, but there is only one option correct, Thus, Option 4 must be wrong. $\endgroup$ – Kartik Soneji Mar 9 at 8:26
  • $\begingroup$ Bleh. I don't like this question, but I do have to retract my statement that $0^a=0$ is straightforwardly true for any nonzero $a$. I had it in my head that as $\varepsilon\rightarrow0$, all the (in general infinitely many) possible values of $\varepsilon^a$ have modulus $\rightarrow0$, but actually that's entirely untrue and I agree with Kartik that the best convention is to leave $0^{\textrm{anything}}$ undefined, at least if the "anything" isn't real. $\endgroup$ – Gareth McCaughan Mar 9 at 11:48
-6
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Solution:

Assuming: Option 1 is correct.
That means Option 4, and Option 5 must also be correct.
But only one option can be correct, so Option 1 cannot be correct.

Assuming: Option 2 is correct.
That means Option 4 must also be correct.
But only one option can be correct, so Option 2 cannot be correct.

Assuming: Option 4 is correct.
That means Option 1, Option 2, and Option 3 may be correct.
But only one option may be correct, so Option 4 cannot be correct.

Assuming: Option 5 is correct.
That means Option 1 must also be correct.
But only one option can be correct, so Option 5 cannot be correct.


Assuming: Option 3 is correct.
Then:
Option 1 must be wrong(contradiction).
Option 2 must be wrong(contradiction).
Option 4 must be wrong, because
$0^i$ is undefined, and $\notin \mathbb{C}$.
Option 5 must be wrong, because $i^3 = -i$.
So, Option 3 is the right answer.

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  • 1
    $\begingroup$ If option 3 is correct, then option 4 is correct; your manipulations there are invalid for multiple reasons. For starters, $a^{(b^c)}$ is not the same as $(a^b)^c$. $\endgroup$ – Deusovi Mar 9 at 8:41
  • $\begingroup$ @Deusovi Fair point. I have edited the answer. $\endgroup$ – Kartik Soneji Mar 9 at 8:48
  • 1
    $\begingroup$ I think the major confusion here is that what you MEAN is: "The correct answer is the one that is mutually exclusive of all the others." but what you SAID is: "If only one option is right, the other options must be wrong." ... those don't exactly mean the same thing. If you're relying on exact interpretation, you need to give exact formulation. $\endgroup$ – Rubio Mar 9 at 9:07
  • 1
    $\begingroup$ But it's possible for 4 to be correct and the others wrong. It's true that "if exactly one option is correct, and option X is correct, then all other options are wrong". It's not true that "if exactly one option is correct, and without that restriction it's possible for X and Y to both be correct (or X individually), then both X and Y are wrong". $\endgroup$ – Deusovi Mar 9 at 10:02
  • 2
    $\begingroup$ @Deusovi There is not only the $0^a$ issue, in the OP's solution it is claimed that $a^3=1$ implies $a=1$. There should be an option to vote for closing questions for being nonsensical. $\endgroup$ – Arnaud Mortier Mar 9 at 11:38

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