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The imaginary number $i$ is defined as the solution to the equation $x²=-1$. Since the solution to this equation could be either positive or negative, does $i$ have two possible values? The original question asked if $i$ was a variable, however I had not understood the mathematical meaning of the word "variable".

ToMath
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    $i$ is neither positive nor negative in the field of complex numbers (which does not have an order compatible with field operations); it is a solution to $x^2=-1$ and not a variable; there are real numbers, and there are complex numbers that are not real numbers – J. W. Tanner Aug 10 '21 at 20:55
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    Define "variable". – azif00 Aug 10 '21 at 20:56
  • +1 to your question, taking it from $-1$ to $0$. I reversed the downvote because your posting is obviously an interpretation question, so I don't think that the normal protocol applies here. For what it's worth, there is no quality control on downvotes, especially anonymous downvotes. – user2661923 Aug 10 '21 at 20:57
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    $i$ cannot be both positive and negative. $i$ is neither positive nor negative. However, there are two square roots of $-1,$ and we do pick one arbitrarily. The other one is then $-i.$ But that doesn't mean $i$ is a variable. – Thomas Andrews Aug 10 '21 at 20:58
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    It is a constant, like $\pi$. – David G. Stork Aug 10 '21 at 21:00
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    @azif00. I'll try.. an object which represents either an unknown value or a group of values.. – ToMath Aug 10 '21 at 21:01
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    If we want to be very precise, we can define a complex number to be an ordered pair of real numbers, and then $i$ is defined to be the ordered pair $(0,1)$. – littleO Aug 10 '21 at 21:02
  • @J.W.Tanner I thought i was the vertical axis in the complex plane, and it could occur there as both +i and -i, above or below the axis. Should I understand that in isolation, as a result of its definition, i is neither positive nor negative, but as a part of a complex number such as 0 -i or 0 + i, it can be expressed as being signed? – ToMath Aug 10 '21 at 21:21
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    @ToMath It's a choice/convention that the imaginary unit $i$ is the one pointing in the positive $y$ direction, and it's neither positive nor negative. We could also have chosen what we call $-i$ to be the imaginary unit. Complex conjugation is how you flip between these conventions. – Kyle Miller Aug 10 '21 at 21:23
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    @ToMath Maybe that shows that a priori you might not expect side lengths of negative-area squares to be positive or negative (a stronger fact is indeed true that there is no good way to to think of them as being positive or negative). But instead, I like to think about multiplication by $i$ as a 90 degree rotation. It doesn't really make sense to say a rotation is positive or negative. Even if you consider angles, it's also a -270 degree rotation. – Kyle Miller Aug 10 '21 at 21:59
  • @KyleMiller I haven't done enough work to understand complex numbers, but I kind of don't feel good about doing work that I don't understand. I have a vague sense that the complex plane is somehow composing something similar to area, only one of the factors is negative. The rotation completes a turn when the negative side is equal to the positive side, and it then becomes a scaled version of the positive side. I'll go and watch what happens some more. – ToMath Aug 10 '21 at 22:12
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    @ToMath Just keep working, and if you pay attention to what you're doing and keep learning more math, you'll develop a deeper understanding. My geometric intuition of complex numbers is that they represent compositions of planar rotations and dilations (this is what polar form is all about). While multiplying by $-1$ for the real numbers could be viewed as reflecting through the origin, the complex numbers reveal it's "actually" a 180 degree rotation through a new 2nd dimension. Now you can do fractional rotations. The "square root" of a 180 degree rotation is a 90 degree rotation. – Kyle Miller Aug 10 '21 at 22:19
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    @ToMath (That point of view is to illustrate how multiplication works, since to multiply the complex numbers you compose the transformations. This makes seeing how addition works more complicated, though. If you know about adding and multiplying $2\times 2$ matrices, then what I'm saying is that $a+bi$ is essentially the same thing as $\left[\begin{smallmatrix}a&-b\b&a\end{smallmatrix}\right]$. Really, $i$ can be thought of as a more compact way of writing $\left[\begin{smallmatrix}0&-1\1&0\end{smallmatrix}\right]$.) – Kyle Miller Aug 10 '21 at 22:25
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    @ToMath: $i$ is neither positive nor negative, but it can be multiplied by $-1$ to get $-i$ (which is also neither positive nor negative) – J. W. Tanner Aug 11 '21 at 00:21
  • @J.W.Tanner. How would you say -i if you had to say it out loud? I'm expecting the answer to be "negative i" or "minus i". So I just say "negative i" and remember it is not actually negative? – ToMath Aug 11 '21 at 06:51
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    @ToMath well, even for a real number $a$, "negative $a$" is not necessarily negative. For instance if you take $a = -1$, then "negative $a$" is $1$, which is positive. So it's not really that unnatural that "negative $i$" is not a negative number. When we say "negative $a$", what we really mean is "the number that you have to add to $a$ to get $0$" - it has nothing to do with whether the thing is actually positive or negative. – Sebastian Monnet Aug 11 '21 at 11:14
  • @SebastianMonnet thanks. I seem to be encountering the fact that maths is intentionally abstract, whereas I would like to try to understand it by my experience in the real world. So I guess I would perceive a as an unknown among the reals until it is found or defined, so I would perceive -a as negative and it would become the product of -1 and its value when it is found or defined. This doesn't appear to be how real mathematicians do things. – ToMath Aug 11 '21 at 12:13
  • @ToMath What do you mean when you say you would perceive $-a$ as negative? I think you're implicitly assuming that $a$ is somehow "positive by default", and the possibility of it being negative is some pesky case that you just deal with if it happens. – Sebastian Monnet Aug 11 '21 at 12:17
  • @SebastianMonnet that is exactly what I am doing. My experience of reality is made up of things which exist. So if I had dinner set for four, and I only had three eggs an no desire to cut them up, I would have a negative egg on one of the plates in the given contxt of "I need four eggs". I will need quite some experience before I will be able to develop intuitions from purely abstract notions. I may be in the wrong place to carry that mindset. – ToMath Aug 11 '21 at 12:22

2 Answers2

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There are two square roots of $-1$: they are $i$ and $-i$. There is no sensible way of defining what a "positive" and "negative" complex number is. Because of this, the notation $i=\sqrt{-1}$ is misleading, as it suggests that $i$ is the positive square root of $-1$. It is better to think of $i$ as one of the solutions to the equation $x^2=-1$. Since $i$ and $-i$ are both solutions to the equation $x^2=-1$, in a very rough sense they "cannot be distinguished from each other". This does not mean that $i$ is a variable. As David G. Stork mentions in the comments, $i$ is a mathematical constant like $\pi$.

Joe
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  • is the rough sense that they cannot be distinguished from each other related to some quirk like if you choose -i and just reverse the rotations in the complex plane, you will end up with the same result? – ToMath Aug 10 '21 at 21:37
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    @ToMath Yeah, the complex numbers can't "see" which direction the y-axis points. They don't intrinsically know that multiplication by $i$ is a 90 degree counterclockwise rotation versus a clockwise one. The properties of the complex conjugate show that you can do all the basic operations by flipping the convention, doing the operation, then flipping back. – Kyle Miller Aug 10 '21 at 21:49
  • $i=\sqrt{-1}$ is true. the square root function is well defined on all of $\mathbb C$. –  Aug 12 '21 at 15:00
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    @river: Whether $i=\sqrt{-1}$ is true depends on if and how you are defining the function $f:\Bbb{C}\to\Bbb{C}$ given by $f(z)=\sqrt{z}$. For the reason described in my post, I think it is misleading (especially for beginners) to use the notation $\sqrt{-1}$, as the complex numbers are not an ordered field. – Joe Aug 12 '21 at 15:05
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No, $i = \sqrt{-1}$ is a constant not a variable.

Edit: The question was edited to change its meaning. To answer the new question:

$i$ is a single number, it is not two numbers.

  • @egglog No, $i$ is not defined by $i^2=-1$ since $-i$ also satisfies this relation. – saulspatz Aug 10 '21 at 21:08
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    @egglog That's not an actually, you're just saying that the solutions to $x^2=-1$ are $x=\pm i$. All $i$ "is" is one of the two formal solutions to this. – Kyle Miller Aug 10 '21 at 21:09
  • @egglog No, it doesn't. $i$ is a single value, no plus or minus anything. – saulspatz Aug 10 '21 at 21:09
  • @egglog You're close to seeing the absurdity of your claim: now $i=-i$ implies $2i=0$ implies $i=0$. – Kyle Miller Aug 10 '21 at 21:11
  • @egglog: You haven't defined what $\sqrt{x}$ means when $x$ is a negative number. It is common to abstain from defining $\sqrt{-1}$ as there is no obvious choice principal square root of $-1$. – Joe Aug 10 '21 at 21:12
  • @river I apologise if I changed the meaning. I'm having a hard time tryng to respond to automated encouragment to reformulate closed questions while not really understanding mathematical terminology. – ToMath Aug 14 '21 at 18:01
  • @ToMath, no its fine. but the system didn't give me an alert that it was edited and people downvoted my answer a lot so I had to change it. –  Aug 14 '21 at 18:15