# prove that : $i = \sqrt {-1}\$ [closed]

i have a pretty nasty question. i was glancing through a few olympiad papers and stumbled upon this question:

prove that

$i = \sqrt {-1}\$.

i tried the conventional methods namely euler's formula but could not figure what to do next. how do you actually proceed further? i have checked a few books and a proof is not available.are there any proof available for this

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## closed as not a real question by J. M., Amzoti, vonbrand, vadim123, Lord_FarinMay 7 '13 at 13:27

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

This is typically the definition of $i$, so a proof is inappropriate. Was the Olympiad held on April 1st, by any chance ?? :-) – bubba May 7 '13 at 12:23
There must be context missing, because as it is, the question is either trivial or nonsensical. – Najib Idrissi May 7 '13 at 12:34
What were the assumptions? We assume there's a mysterious element $i$ that can be added or multiplied, subtracted or divided by reals numbers and combinations form $a+bi$ and $i^2=-1$. It can be proved it doesn't prove contradiction and we define $\sqrt{-1}=i$. Instead you can add another mysterious element $\delta$ with same addition and multiplication properties and $\delta\ne 0$ and $\delta^2=0$. It does not prove the contradiction too and you can define $\sqrt[2]{0}=\delta$. – user59671 May 7 '13 at 13:01
+1 just for the "Closed as not a real question" part. – Joeytje50 Nov 25 '14 at 22:16

The square root is typically only defined for real numbers. This has a number of reasons, for instance, a number of simplification rules for the square root only work if the argument is nonnegative, such as

$$\sqrt{ab} = \sqrt{a}\sqrt{b}$$

If we assume this also holds for negative numbers, we get:

$$1 = \sqrt{1} = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\cdot \sqrt{-1} = \left(\sqrt{-1}\right)^2 = -1$$

This is obviously false. Another thing is that $\sqrt{\ \ \ }$ is defined as always returning the non-negative root when there are two, i.e. $\sqrt{9}=3$ and not $-3$, even though $-3$ is also a solution of $x^2=9$. Complex numbers, however, are not ordered; therefore the definition gets a bit tricky. Is $\sqrt{-2i}=1-i$ or $i-1$?

Furthermore, as a consequence of this, there is no way of distinguishing $i$ and $-i$ except on a formal level. You say that $i$ is, formally, some number that, when squared, yields $-1$; but $-i$ also fulfils that requirement, so if you mail-order the $i$ number, i.e. a number that when squared gives you $-1$, I could take it out, negate it and put it back in and you would never be able to tell the difference.

So, to summarise: defining square roots of negative numbers is probably not a good idea.

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Very nice answer (+1). I was going to answer, but I was thinking only of part of what you have said. – robjohn May 7 '13 at 12:57

Something similar was discussed here:

What's the thing with i

Afaik. $i$ is usually +defined as $i^2 = -1$, which is different from $i = \sqrt{-1}$.

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the problem is that its just a question on why and not the exact proof.. – Sri Krishna May 7 '13 at 12:31

You may want to show that $i$ solves the equation $z^2 = -1$. The square root of x is typically defined as the (positive) solution of the equation $z^2 = x$. If we are talking about roots in $\mathbb C$, one usually avoids to write $i = \sqrt{-1}$, because there are multiple solutions to this (consider $(-i)^2 = (-1)^2 i^2 = -1$).

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