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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

Sri Krishna
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    This is typically the definition of $i$, so a proof is inappropriate. Was the Olympiad held on April 1st, by any chance ?? :-) – bubba May 07 '13 at 12:23
  • There must be context missing, because as it is, the question is either trivial or nonsensical. – Najib Idrissi May 07 '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 $\sqrt[2]{0}=\delta$. –  May 07 '13 at 13:01
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    +1 just for the "Closed as not a real question" part. – Joeytje50 Nov 25 '14 at 22:16

3 Answers3

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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.

Manuel Eberl
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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 07 '13 at 12:57
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Something similar was discussed here:

What's the thing with i

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

Community
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ArminB
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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 $(-i)^2 = (-1)^2 i^2 = -1$).

Stefan
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