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If we really don't know which is bigger if $ i $ is greater or $ 2i $ or so on then why do we plot $ i $ first then $ 2i $ and so on, on the imaginary axis of the Argand plane? My teacher said that imaginary numbers are just points and all are dimensionless so they are incomparable and the distance really doesn't matter. I want to get this more clear

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  • $\begingroup$ Numbers like $-8$, $\pi$ and $\sqrt{2}$ are also dimensionless. Are you aware of the difference between complex numbers and imaginary numbers? $\endgroup$ – Frank Vel Feb 9 '18 at 19:42
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    $\begingroup$ Yes I do imaginary number is the one written in $ i.b $ form where b can be any real number while. Complex number is of the form $ a + i.b $ where $ a , b $ are real numbers.and $ i $ is the square root of $ -1 $. I hope this is not false. $\endgroup$ – user508298 Feb 9 '18 at 20:08
  • $\begingroup$ Technically it's defined by $i^2 = -1$ ($-1$ can have several square roots, so $i$ is a square root of $-1$), but your other definitions are correct. I was only concerned because your question is about imaginary numbers, but the many of the answers are only about complex numbers, which are related, but slightly different. $\endgroup$ – Frank Vel Feb 9 '18 at 20:24
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  1. There is no way to order complex numbers, in a way that preserves operations in a sensible way. The precise term is ordered field; among their properties, $x^2 \ge 0$ for every $x$. Since $i^2=-1$, we would need $-1\ge 0$, which is impossible.

  2. However, if you want to measure distance, you can do that with a norm. In the complex numbers, this is calculated as $|a+bi|=\sqrt{a^2+b^2}$. Hence, it is correct to say that $2i$ is twice as far from the origin as $i$ is.

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    $\begingroup$ Pam asks about ordering just imaginary numbers, not all complex numbers. $\endgroup$ – CiaPan Feb 8 '18 at 12:56
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    $\begingroup$ @Joshua $i>0 \Rightarrow i\cdot i\cdot i>0 \Rightarrow -i > 0 \Rightarrow i < 0$ $\endgroup$ – Leonhard Feb 8 '18 at 14:14
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    $\begingroup$ @Leonhard Within the pure imaginary numbers, wouldn't $i \cdot i$ not be defined, meaning we couldn't consider $i \cdot i \cdot i$? Of course, this means they can't be a field in the first place, at least with the standard definition of multiplication. $\endgroup$ – el duderino Feb 8 '18 at 15:05
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    $\begingroup$ @elduderino You can define $i \cdot i = i$ so that the imaginary numbers are isomorphic to the reals via $ai \mapsto a$. But these imaginaries are not a subfield of $\mathbb{C}$. $\endgroup$ – Solomonoff's Secret Feb 8 '18 at 19:47
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    $\begingroup$ @pam: Somebody deleted my original comment. Trivial ordering exists for pure imaginary numbers because a trivial mapping exists between imaginary integers and integers. $\endgroup$ – Joshua Feb 9 '18 at 19:17
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There is a good reason to avoid defining "orderings" like $i > -i$, or any similar comparisons, even if you don't care about ordered fields. That is: there is no way to distinguish between $i$ and $-i$, except for the symbols used to denote them. $i$ is defined as a solution of $x^2 +1 = 0$. But there are two solutions, so how do you decide which one gets the privilege to be bigger than the other?

If you take any sentence in the language of complex numbers in which $i$ occurs, and replace it everywhere with $-i$, then the truth of the sentence doesn't change. That isn't true anymore if you introduce definitions like $i > -i$.

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    $\begingroup$ $i$ and $-i$ are distinct numbers. This is obvious if we switch to the equality operators $=$ and $\neq$. $-1$ and $1$ are both solutions to $x^2 - 1 = 0$. Does that mean there is no way to distinguish between $-1$ and $1$. $\endgroup$ – Kaz Feb 9 '18 at 20:08
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    $\begingroup$ @Kaz the point is that it's arbitrary which is which. $\endgroup$ – leftaroundabout Feb 9 '18 at 20:09
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    $\begingroup$ @Kaz there's an axiomatic reason to not treat $-1$ and $1$ the same. Namely, $1$ is the neutral element of multiplication, $-1$ is not. But neither $i$ nor $-i$ are neutral elements of any fundamental operation. $\endgroup$ – leftaroundabout Feb 9 '18 at 20:15
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    $\begingroup$ Suppose you encounter some alien who uses strange symbols for integers. You want to explain your ordering to them, with some examples. They write "A" and "B". You don't know what the symbols mean. They show you: $A*A = A$, $B*B = A$. Okay, now you can translate: A is 1 and B is -1. So you can show them: $A > B$ (or $A < B$, if that's your ordering). Next they show you C and D, and that $C*C = B$ and $D*D = B$. What kind of question could you ask them to figure out which of $C$ or $D$ is $i$? It is not possible. That's what I meant by indistinguishable. $\endgroup$ – Jonny Lomond Feb 9 '18 at 23:05
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    $\begingroup$ @leftroundabout True; there is a symmetry between conjugates that isn't there between additive inverses. E.g. we can see this in the Mandelbrot set (symmetric around number line, but not around imaginary axis) and numerous other situations. $\endgroup$ – Kaz Feb 13 '18 at 18:32
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  1. We plot $i$ before $2i$ and so on purely by convention (and maybe that our brains are acquainted to working with numbers from smaller magnitudes to larger magnitudes). On a computer screen, for example, $2i$ would have lower $y$ coordinate than $i$ and thus is plotted first.

  2. There is no way of imposing a total ordering on $\mathbb C$ that would give you the same nice properties as one on $\mathbb R$, for example, having $x^2 \geq 0$ for all $x$, is not possible.

  3. You can compare the magnitudes of complex numbers. That is, their absolute values. Distance in this case does matter because distance is a real number and is thus comparable. It is just that the complex points themselves are not comparable.

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  • $\begingroup$ I understood! Thanks $\endgroup$ – user508298 Feb 7 '18 at 22:08
  • $\begingroup$ I thought 2i would have a higher y coordinate? $\endgroup$ – theonlygusti Feb 9 '18 at 12:34
  • $\begingroup$ @theonlygusti The top-left corner is usually $[0,0]$. Since the "positive-imaginary" axis points up, $2i$ has lower $y$ coordinate. $\endgroup$ – Henricus V. Feb 9 '18 at 12:35
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    $\begingroup$ I can't agree with 1. There is a very obviously natural ordering on $\left\{ix\colon x\in \mathbb R\right\}$ and that ordering, to me, is convincing enough to plot "$i$ before $2i$". The problem only arises when one compares numbers not on the axis, where there's no natural choice. $\endgroup$ – Git Gud Feb 9 '18 at 22:19
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    $\begingroup$ @GitGud it seems obvious, until one considers that the distinction between $i$ and -$i$ is arbitrary in a way that the distinction between 1 and -1 is not - see comments on Jonny Lomond's post for more discussion about this. Once we notice that $i$ and -$i$ are interchangeable in any statement about complex numbers, then it becomes equally valid to argue that -2$i$ is greater than -$i$. $\endgroup$ – Geoffrey Brent Feb 10 '18 at 9:21
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For purely imaginary numbers, the ones with zero real part, i.e. $z = bi$ (where $b$ is a real), it's trivial to set an ordering: just consider the imaginary parts and compare those as reals. That's what we do implicitly when drawing the imaginary axis or choosing the "$y$" coordinate when plotting complex numbers on a plane.

For complex numbers in general: $z = a+bi$, where both $a$ and $b$ can be nonzero, there's no clear order. E.g. any ordering between all four of $(\pm 1 \pm i)$ would be arbitrary.

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In an ordered field, you can prove that $x^2 \geq 0$. In $\mathbb{C}$, we have $i^2=-1<0$ so you cannot put an order on it.

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    $\begingroup$ Perhaps you can put an order on the complex numbers as a set, but you lose some useful properties $\endgroup$ – Henry Feb 7 '18 at 22:01
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    $\begingroup$ @Henry: Sure. For example, you can identify $\mathbb{C}$ with $\mathbb{R}^2$ and put lex-order on it: $x+iy < x' + iy'$ if either $x < x'$ or $x = x', y < y'$. Of course, it's not particularly well-behaved under the usual operations, you wind up with a different topology, etc. $\endgroup$ – anomaly Feb 8 '18 at 2:34
  • $\begingroup$ there is an oder as well see Guy Fsone answer's $\endgroup$ – Guy Fsone Apr 1 '18 at 5:19
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Imaginary numbers are comparable

As you said we can order the imaginary numbers like $-i < 0 < i < 2i < \ldots$ and in general write $ai \le bi$ to mean $a \le b$ for real numbers $a,b$. There is nothing wrong with this.

Some complex numbers are comparable

If you want to order all the complex numbers we can define $a+bi \le c+di$ to mean that $a \le c$ and $b \le d$ as real numbers. This gives what's called a partial order on $\mathbb C$. This order is unlike the order on the real (or imaginary numbers) in that there exist pairs of elements that are incomparable. For example $1+2i$ and $2+1i$ are incomparable.

Edit: We cannot compare the complex numbers $1+2i$ and $2+1i$.

This is because we defined $1+2i \le 2+i$ to mean $1\le 2$ and $2 \le 1$ which is untrue.

And we defined $2+i \le 1+2i$ to mean $2 \le 1$ and $1 \le 2$ which is untrue.

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    $\begingroup$ It's worth mentioning that this order on the imaginary numbers doesn't work quite like the standard ordering on the reals - because if $a$ is real and $a>0$ then $a^3$>0, but in this ordering $i>0$ but $i^3<0$. $\endgroup$ – psmears Feb 8 '18 at 14:47
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    $\begingroup$ How can you write $ i> -i $ if you dont know that $ i $ is positive or negative. I think we can just compare the magnitudes and not the 'imaginary' number themselves. Can you please explain why you said that $ 1 + 2i $ and $ 2 +i $ are incomparable. One can also argue that $ 2 + i - (1+2 i) >0 $ $\endgroup$ – user508298 Feb 8 '18 at 20:03
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    $\begingroup$ The partial order on complex numbers doesn't respect multiplication either. Both of these are really more of an ordering on $\mathbb R^2$ rather than $\mathbb C$, I think. $\endgroup$ – Ben Millwood Feb 8 '18 at 20:27
  • $\begingroup$ I never claimed either of $i$ is either positive or negative. What I said is that there's a way of comparing imaginary numbers. This is not the same thing. Sure there's a way of comparing different movies, and we don't say some movies are 'positive' or 'negative' whatever that might mean . . . . $\endgroup$ – Daron Feb 11 '18 at 16:06
  • $\begingroup$ Hold your horses Ben! I never claimed the orderings respect multiplication. $\endgroup$ – Daron Feb 11 '18 at 16:06
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Here is one way to compare them With undermentioned orders we have Example see that $$\color{blue}{i \lesssim 2i~~~~~and~~~~i \lessapprox 2i}$$

With the usual oder on $\Bbb R$ it is not possible since we have the obstruction $i^2 = -1<0$ which directly drive us to contradiction.

However,given to couple $(x,y)$ and $(a,b)$ whose complex representations are $ x+iy$ and $a+ib$ respectively we can define the following lexical oders

We say that $$(x,y)\lesssim (a,b)~~~~\mbox{ iff}~~~x<a ~~~or (~~x=a~~~and ~~~y< b.)$$

$$(x,y)\lessapprox (a,b)~~~~\mbox{ iff}~~~x\le a ~~~or (~~x=a~~~and ~~~y\le b.)$$

you can check these are actually on $\Bbb C$.

Example see that $$\color{blue}{i \lesssim 2i~~~~~and~~~~i \lessapprox 2i}$$

Be aware that they do not preserve classical well known properties of the oders $<$ and $\le $ such as $$Z\lesssim W \not\implies ZZ'\lesssim WZ'$$

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  • $\begingroup$ It is a big shame to down vote without any explanation $\endgroup$ – Guy Fsone Feb 10 '18 at 13:15
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The well-ordering theorem tells us that if we accept the Axiom of Choice then any set, obviously including the complex numbers, can be ordered. But it doesn't tell us how to construct such an ordering, and the ordering won't have the useful properties of the ordering of the real numbers (the properties of an ordered field). So while we can in principle order the complex numbers, in practice it is both difficult and pointless.

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  • $\begingroup$ What does this has to do with the question? Do you need the axiom of choice to linearly order $\Bbb C$? No. You don't even need to put any effort if you only compare complex numbers of the form $iy$ for $y\in\Bbb R$. $\endgroup$ – Asaf Karagila Feb 10 '18 at 11:06
  • $\begingroup$ This is a misleading answer. OP wasn't really talking about well ordering at all. Bringing in AC is tangential at best. It is trivial to define linear orders on C, not "difficult" as you suggest. $\endgroup$ – John Coleman Feb 10 '18 at 14:28
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I've come to think of i as a rotation operator, so ordering may not make any sense. But these are the thoughts of an old man, and I'm not sure there is any basis to this? This probably ought to be a comment, but you can't comment on a question until you have a reputation level of 50+.

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  • $\begingroup$ Since you haven't defined "rotation operator" and you haven't said why ordering them doesn't make sense, this is at best a vague intuition. On the level of vague intuitions -- we frequently do compare rotations (e.g. "this picture is more rotated than that picture") so why state without explanation that ordering doesn't make sense? Something like this might make a decent comment, but it really isn't an answer. $\endgroup$ – John Coleman Feb 10 '18 at 16:44
  • $\begingroup$ It is a vague intuition as I intended to imply. So, I edited the "answer" because I cannot make a comment to the original question. I don't see that a compare operation would imply any order, but you are right, there could be order after a rotation. I also view -1 as a rotation operation on N that creates Z and still has order I think? $\endgroup$ – Terry Shannon Feb 11 '18 at 23:04

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