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I've read the post Can square roots be negative? and there doesn't seem to be a definitive answer, so I'll ask it a different way: can imaginary numbers be negative? That is, if you were to use imaginary numbers in a coordinate system, would the axis extend into negative numbers?

EDIT: I understand that imaginary numbers don't behave like real numbers, so perhaps the right way to phrase this question is: if you had an imaginary coordinate axis, would this axis be symmetrical about the origin?

Quark Soup
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    What do you mean by "negative" here? If you are asking if in the complex plane whether or not you can go in the "opposite direction" as $i$... the answer is of course yes. Going "down" in the plane vertically, you go towards $-i$. – JMoravitz Oct 16 '21 at 00:30
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    Now... "positive" and "negative" have a very specific definition in abstract algebra. In an ordered ring, you have the "positive cone" which are those values "greater" than zero according to the order, and the negative cone which are those values less than zero, noting that two negatives multiplied result in a positive. Recognize that the complex numbers are not ordered. "Greater than" has no meaning here, so trying to define a positive cone or negative cone here can not be done. – JMoravitz Oct 16 '21 at 00:32
  • @JMoravitz - If you are asking if in the complex plane whether or not you can go in the "opposite direction No, I'm not asking this question. – Quark Soup Oct 16 '21 at 01:46
  • @JMoravitz - so trying to define a positive cone or negative cone here can not be done That is not sufficient. If I can build a coordinate system using imaginary numbers, then the question is valid: Is this coordinate system symmetrical about the origin? If I can't build a coordinate system using imaginary numbers, I would like to know why I can't. – Quark Soup Oct 16 '21 at 01:49
  • Then I haven't got a clue what you are asking about here. If you are asking if there exists some way of adding and subtracting real multiples of $i$ such that the result is a negative real number (noting that zero does not count as a negative real number) then the answer is obviously no... but surely you wouldn't be asking such a basic question. You talk about symmetry about the origin and you talk about positive vs negative. These are different concepts. You need to be clearer what you mean. – JMoravitz Oct 16 '21 at 02:37
  • @JMoravitz - You need to be clearer what you mean Can I build a coordinate system using imaginary numbers? If so, is this coordinate system symmetrical about the origin? – Quark Soup Oct 16 '21 at 15:59
  • You keep using those words. I do not think you really know what they mean. As for "can square roots be negative" there is a definitive answer. The principal square root is never negative, by definition. As for imaginary coordinate axis... how does that differ from the traditional complex plane and how does my first comment not address this already? – JMoravitz Oct 16 '21 at 16:52
  • @JMoravitz - What words am I using, specifically, that are causing you confusion? And I'm not talking about a complex plane, I'm trying to build a plane with two imaginary axes. Will these axes be symmetrical about their origins? – Quark Soup Oct 16 '21 at 17:01

2 Answers2

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In $\Bbb R$ there is a relationship between the arithmetic operations $+,\times$ and the order $<$ as follows:

For any $a,b,c\in\Bbb R$

$(i).$ if $a<b$ then $a+c<b+c$,

$(ii).$ if $a<b$ and if $0<c$ then $ac<bc$.

In $\Bbb C$ we cannot define a linear order $<$ that satisfies $(i)$ and $(ii)$ for all $a,b,c\in\Bbb C.$ Suppose we try.

First, we must have $0<1$. Otherwise, if $1<0$ then by $(i)$ with $a=1,\,b=0,\,c=-1$ we get $0<-1,$ but then by $(ii)$ with $a=1,\,b=0,\,c=-1$ we get $-1<0$, so we would get $0<-1<0.$

Second, let $i^2=-1.$

Suppose $i<0.$ Then by $(i)$ with $a=i,\,b=0,\,c=-i$ we get $0<-i.$ But then by $(ii)$ with $a=i,\,b=0,\,c=-i$ we get $1=-(-1)=-(i^2)<0,$ but we must have $0<1.$

Suppose instead that $0<i.$ Then by $(ii)$ with $a=0,\,b=i,\,c=i$ we get $0<i^2=-1,$ but then by $(i)$ with $a=0,\,b=-1,\,c=1$ we get $1<0$ again.

So it can't work.

DanielWainfleet
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  • I don't understand your explanation. You give us (i) if $a<b$, then $a+c<b+c$ then in the examples you give $a=1,b=0,c=-1$, so without even getting into the imaginary part, you seem to have violated the rules of (i) because $a>b$. – Quark Soup Oct 16 '21 at 16:10
  • Quarkly that example is used to demonstrate a contradiction. – Boris Oct 16 '21 at 16:17
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It appears the answer is 'yes', imaginary numbers on a coordinate axis can be negative. The Argand Diagram displays the answer to my question:

Argand Diagram

Quark Soup
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    Compare that to my very first comment... And, it isn't so much that the imaginary numbers are negative (that word is reserved for use only in the real numbers or another ordered ring) so much as the direction of $-y$ is the opposite direction as $y$. The word "negative" has a very specific meaning which does not work for imaginary numbers. – JMoravitz Oct 16 '21 at 20:35