2
$\begingroup$

I have a small point to be clarified.We all know $ i^2 = -1 $ when we define complex numbers, and we usually take the positive square root of $-1$ as the value of "$i$" , i.e, $i = (-1)^{1/2} $.

I guess it's just a convention that has been accepted in maths and the value $i = -[(-1)^{1/2}] $ is neglected as I have never seen this value of "$i$" being used. What I wanted to know is, if we will use $i = -[(-1)^{1/2}] $ instead of $ (-1)^{1/2} $, would we be doing anything wrong? My guess is there is nothing wrong in it as far as the fundamentals of maths goes. Just wanted to clarify it with you guys. Thanks.

$\endgroup$
  • 2
    $\begingroup$ See here (as well as this recent question). Actually, possible duplicate of the former... $\endgroup$ – Zev Chonoles May 20 '13 at 12:07
  • $\begingroup$ Using your notation I think we are only switch the role of $z \leftrightarrow \overline{z}$ ? Because you are defining $i'=-i$ ($i'$ is your new imaginary unit) $\endgroup$ – Riccardo May 20 '13 at 12:11
  • 2
    $\begingroup$ The imaginary number isn't strictly defined as either the positive or negative square root of $-1$. It's defined literally by $i^2=-1$, and is otherwise merely a symbol. $\endgroup$ – Glen O May 20 '13 at 12:19
  • 1
    $\begingroup$ This question: math.stackexchange.com/questions/396384/what-is-i-exactly looks almost the same as the question here. Do we think it's a duplicate? @Zev Chonoles $\endgroup$ – ncmathsadist May 20 '13 at 12:25
  • $\begingroup$ ...and we usually take the positive square root of $-1$ as the value of "$i$"... Game Over $\endgroup$ – Matemáticos Chibchas May 20 '13 at 12:31
10
$\begingroup$

... and we usually take the positive square root of −1 as the value of "$i$".

There's no such thing as a positive or negative complex number. By treating $\mathbb{R}$ as a subset of $\mathbb{C}$ you can call some complex numbers positive or negative, but only those which actually are real numbers. There's no way to define a total order on $\mathbb{C}$ which would behave like the usual order on $\mathbb{R}$ does.

What happens is that one simply picks any root of $-1$, i.e. any solution of $x^2 + 1 = 0$, and calls it $i$. There's then always a second solution, and that solution is $-i$. You can't even say which solution you picked for $i$ - the two solutions of $x^2 + 1=0$ are algebraically indistinguishable, i.e. you cannot tell them apart with algebraic means.

Imagine a friend hands you a bag containing two balls, of equal size and material. You can pick one arbitrarily and call it "ball 1", and take a pen and mark it with a big "1". The other is then "balls 2", and gets marked with a big "2". Now, imagine you had picked the other ball. Would you end up in a different situation? No! You'd still have two balls, one marked "1" and one marked "2", and indistinguishable otherwise. Now, after you've marked the balls, they are of course different. You can now for example put both back into the bag, let your friend pick one, and ask "Which ball have you picked?". But that only works after you marked them!

$\endgroup$
  • $\begingroup$ ,thanks for your feedback. So,one can define i = $ -(-1)^{1/2} $ as a solution to $ i^2 $ = -1 and can take -i = $ (-1)^{1/2} $ ... right? there is nothing wrong in it... $\endgroup$ – under root May 20 '13 at 12:58
  • $\begingroup$ i mean only if one wants to...though its better to follow what has been followed since years... $\endgroup$ – under root May 20 '13 at 12:59
  • 2
    $\begingroup$ @under-root Well... no, not really. I suggest you re-read my last paragraph. The whole point is that $\sqrt{-1}$ isn't a uniquely defined thing. It's a charaterization of a thing, as a solution of $x^2 = -1$. But there are two things which fullfill this equation, and both are completely indistinguishable. $i$ is an arbitrary one. $-i$ is then the other one. You can't say which one $i$ is, same as you can't say which ball you marked with "1". $\endgroup$ – fgp May 20 '13 at 13:29
  • $\begingroup$ @under-root You can, of course, first pick an arbitrary one and call it $\sqrt{-1}$. Then you can set $i = -\sqrt{-1}$, i.e. call the other one $i$. $\endgroup$ – fgp May 20 '13 at 13:31
  • 1
    $\begingroup$ @under-root The point is, $\sqrt{-1}$ isn't any particular solution. It's an arbitrary one. $i$ is an arbitrary one too. This leaves you with the choice of whether $\sqrt{-1}=i$ or $\sqrt{-1}=-i$. Usually one picks the former, but you can of course pick the latter if you're so inclined... $\endgroup$ – fgp May 20 '13 at 13:34
1
$\begingroup$

The square roots with the properties we know are used only for Positive Real numbers.

We say that $i$ is the square root of $-1$ but this is a convention. You cannot perform operations with the usual properties of radicals if you are dealing with complex numbers, rather than positive reals.

It is a fact that, $-1$ has two complex square roots. We just define one of them to be $i$.

You have to regard the expression $i=\sqrt {-1}$ just as a symbol, and not do operations.

Counterexample: $$\dfrac{-1}{1}=\dfrac{1}{-1}\Rightarrow$$ $$\sqrt{\dfrac{-1}{1}}=\sqrt{\dfrac{1}{-1}}\Rightarrow $$ $$\dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}\Rightarrow$$ $$\dfrac{i}{1}=\dfrac{1}{i}\Rightarrow$$ $$i^2=1\text { ,a contradiction}$$

$\endgroup$
  • $\begingroup$ Could you please be more specific as to what operations you are referring to that we cannot do? $\endgroup$ – Jonas Meyer May 20 '13 at 12:11
  • $\begingroup$ @JonasMeyer does the example I added supplement this? $\endgroup$ – Dimitris May 20 '13 at 12:20
  • 1
    $\begingroup$ I agree with the part about having to define $i$ to be one of the square roots of $-1$ and sticking with the choice. However, I don't see how the part about being careful with operations is particularly relevant to the question. You say "counterexample," but it is not a counterexample to anything in the question as far as I can see, but rather to the property $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ which would hold for the usual square root function if $a\geq 0$ and $b>0$. $\endgroup$ – Jonas Meyer May 20 '13 at 12:21
  • $\begingroup$ I want to make clear the fact that $i=\sqrt{-1}$ cannot be used that way in operations. So, does it make sense to take the other root $i = -[(-1)^{1/2}]$? $\endgroup$ – Dimitris May 20 '13 at 12:24
  • 1
    $\begingroup$ Yes, I completely agree with that. So, is there a real difference between $i = -[(-1)^{1/2}] $, and $i = [(-1)^{1/2}] $? We sometimes think in the sense of Real numbers and operations. That's what I wanted to point out. $\endgroup$ – Dimitris May 20 '13 at 12:28
0
$\begingroup$

You are correct that $\sqrt{-1} = -i$, but you are not correct when you say that this fact is neglected--it's used all the time. E.g. the quadratic formula:

$$x^2+x+1=0$$

$$x=\frac{1}{2(1)}+ \frac{\sqrt{1^2-4(1)(1)}}{2(1)}$$

$$x=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}$$

The plus-or-minus is there because $\sqrt{-3}$ has two roots: $i\sqrt{3}$ and $-i\sqrt{3}$.

$\endgroup$
  • 1
    $\begingroup$ If $\sqrt{-1}$ and $i$ are used to mean the same thing, as they are in many contexts including the beginning of the question above, then the equation $\sqrt{-1}=-i$ is incorrect. You seem to be using the square root symbol to denote a 2-valued function, but this does not clear up the confusion in the question as far as I can see. $\endgroup$ – Jonas Meyer May 20 '13 at 12:15
  • $\begingroup$ Strictly speaking, the square root operation (that is, $\sqrt{x}$) always takes the "positive" value. This is why the $\pm$ is usually included in the quadratic formula. $\endgroup$ – Glen O May 20 '13 at 12:17
  • 1
    $\begingroup$ Unfortunately $i$ is not positive in any senseful meaning. $\endgroup$ – Andrea Mori May 20 '13 at 12:21
  • $\begingroup$ @GlenO Ok, I see what the confusion is now, but as I understand it $y=\sqrt{x}$ is anything that satisfies $y^2=x$, unless it's explicitly stated that $\sqrt{x}$ is intended to be used as a function. $\endgroup$ – Ataraxia May 20 '13 at 12:24
  • 1
    $\begingroup$ @Glen0: What you just wrote only makes sense for square roots of nonnegative numbers. In particular, note that $|x|$ is not a square root of $x^2$ unless $x$ is real. $\endgroup$ – Jonas Meyer May 20 '13 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.