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I've always had this doubt. It's perfectly reasonable to say that, for example, 9 is bigger than 2.

But does it ever make sense to compare a real number and a complex/imaginary one?

For example, could one say that $5+2i> 3$ because the real part of $5+2i $ is bigger than the real part of $3$? Or is it just a senseless statement?

Can it be stated that, say, $20000i$ is bigger than $6$ or does the fact that one is imaginary and the other is natural make it impossible to compare their 'sizes'?

It would seem that the 'sizes' of numbers of any type (real, rational, integer, natural, irrational) can be compared, but once imaginary and complex numbers come into the picture, it becomes a bit counter-intuitive for me.

So, does it ever make sense to talk about a real number being 'more than' or 'less than' a complex/imaginary one?

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    $\begingroup$ If you can place partial orders on matrices you can certainly do the same for complex numbers! $\endgroup$ – Mehrdad Jan 23 '15 at 7:11
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    $\begingroup$ From my (programmer's) perspective, real numbers exist only in one dimension, while complex could be presented in two dimensions, hence no ordering comparison between both is possible (if we are not talking about modulus, which is a bit different). $\endgroup$ – Askar Kalykov Jan 23 '15 at 7:22
  • $\begingroup$ Actually, when I read this question, I first thought of complex numbers as vectors in a 2-D plane. North is greater than west sounds dumb, so my answer would have been no. Am I wrong? $\endgroup$ – Rohinb97 Jan 23 '15 at 7:37
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    $\begingroup$ What's bigger a coin or misery? It all depends on how you define "bigger", "coin", and "misery" $\endgroup$ – OutFall Jan 23 '15 at 20:08
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    $\begingroup$ @ziggystarman The answer I provided is something I learned in my first complex analysis class concerning the concept of order in $\mathbb{R}$ compared to trying to carry over that concept to $\mathbb{C}$. I thought it was a very helpful/explanatory way of thinking about it, but the modulus possibility explained by msteve is a fun way of considering it as well. Mine is a little more abstract I guess and not quite as "tangible" I suppose. $\endgroup$ – Daniel W. Farlow Jan 24 '15 at 4:18
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You can put (partial) orders on the complex numbers. One choice is to compare the real parts and ignore the complex ones. Another is to use the lexicographic order, comparing the real parts and then comparing the imaginary ones if the real parts are equal. Another is to use the modulus. There are many more. The distinction with the order on the reals (or subsets of the reals) is that the order relation is compatible with addition and multiplication. You can't do that in the complex numbers. The simple proof is to ask whether $i$ is greater or less than $0$. In either case, $i^2=-1$ should be greater than zero.

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    $\begingroup$ $-1 \succ 0$ is not a problem until you conclude $1 \succ 0$ at the same time. $\endgroup$ – GFauxPas Jan 23 '15 at 17:15
  • $\begingroup$ @GFauxPas: You typically want to require $z>0$ iff whenever $x>y$ then $xz>yz$. But this is only possible only if $0<1$. $\endgroup$ – cody Jan 23 '15 at 19:50
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    $\begingroup$ I'm a math novice, but I think this answer is saying "we can invent an arbitrary ordering", but there is no "logical" ordering like scalars have with each other. Would you be willing to clarify this thought in your answer? $\endgroup$ – Mooing Duck Jan 23 '15 at 20:04
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    $\begingroup$ @MooingDuck: yes that is right. Some of the orderings (like by modulus) are more natural than others, but modulus is not a total order. On the reals, there is nice interplay between the order and the operations of addition and multiplication. No order on the complex numbers has that nice interplay. $\endgroup$ – Ross Millikan Jan 24 '15 at 0:54
  • $\begingroup$ @GFauxPas If $-1 \succ 0$, then $1=(-1)^2 \succ 0^2=0$, as well. $\endgroup$ – Strants Jan 24 '15 at 3:17
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To compare two complex numbers, we usually look at their modulus: if $z = x+iy$, then the modulus of $z$ is $|z| := \sqrt{x^2 + y^2}$. Regarding $z$ as a point in the complex plane, the modulus of $z$ is the distance to the origin. We can now compare two complex numbers such as $5+2i$ and $3$: notice that $|5+2i| = \sqrt{29}$ and $|3| = 3$, so in this sense, $5+2i$ is `larger' (better to think: farther away from the origin) than $3$.

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    $\begingroup$ Beat me to it - +1+0i ;=) $\endgroup$ – HammyTheGreek Jan 23 '15 at 5:22
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    $\begingroup$ With this convention, $-9$ is "larger" than $2$ which may cause some confusion. $\endgroup$ – mrf Jan 23 '15 at 5:25
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    $\begingroup$ @mrf: it might cause confusion, the first time someone encounters that usage. More commonly it causes pedantry: someone knows what you mean and disputes it anyway. But the massive advantage of mathematics as a subject is that the speaker gets to define their terms :-) As long as you stick to saying "less than" / "greater than" when that's what you mean, it's then safe to use "small" / "large" to talk about magnitude. $\endgroup$ – Steve Jessop Jan 23 '15 at 11:39
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    $\begingroup$ Also note that this is not a total order unless you choose some (arbitrary?) convention on how to compare numbers like $5+2i$ and $2-5i$ (for example) whose distances from zero are identical. $\endgroup$ – Jeppe Stig Nielsen Jan 23 '15 at 16:04
  • $\begingroup$ $\sqrt{29}$ no? $\endgroup$ – JoeTaxpayer Jan 24 '15 at 14:37
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Since $\mathbb{R}\subset\mathbb{C}$, every $x\in\mathbb{R}$ can be written as $x + i\cdot 0$. Now if we prescribe the lexicographical (dictionary) ordering, we can compare them.

Let $z,w\in\mathbb{C}$ and $z = x+iy$ and $w=a+bi$. Then the lexicographical ordering is $z < w$ if $x<a$ or $x=a$ and $y<b$, $z = w$ if $x=a$ and $y=b$, and $z>w$ otherwise.

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Order is easy and non ambiguous in $\mathbb{R}$, because it is unidimensional. $\mathbb{C}$ on the other hand is generally seen as a plane. So you will easily define pre-order on it, that means transitive and reflexive relations, that do have sense such as the examples of Ross Millikan's answer.

But except for the lexicographic order, they are not true order relation because they are not anti-symetric : you can have $a < b$ and $b < a$ without $a = b$.

And the lexicographic order is not natural because it is not compatible with the current topology : $x+iy$ and $x +\epsilon + iy$ are topologically near, but if $\epsilon>0$, we get $x+iy < x+i(y+bigNumber) < x+epsilon+iy$ which is not natural because $x+iy$ and $x+i(y+bigNumber)$ are not topologicaly near.

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  • $\begingroup$ @MooingDuck Please use the 'edit' function rather than filling the comments with trivial and obvious mistakes. $\endgroup$ – user26486 Jan 24 '15 at 18:45
  • $\begingroup$ I beg to differ : $\mathbb R$ is of infinite dimension as a $\mathbb Q$-vector space. And $\mathbb C$ of dimension $1$ as a $\mathbb C$-vector space. ;) $\endgroup$ – Evpok Jan 24 '15 at 20:15
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Observation: Many of the properties of the real number system $\mathbb{R}$ hold in the complex number system $\mathbb{C}$, but there are some rather interesting differences as well--one of them is the concept of order. The concept of order used in $\mathbb{R}$ does not carry over to $\mathbb{C}$. That is, we cannot compare two complex numbers $z_1=a_1+ib_1,b_1\neq0$, $z_2=a_2+ib_2,b_2\neq0$, by means of inequalities. Statements such as $z_1<z_2$ or $z_2\geq z_1$ have no meaning in $\mathbb{C}$ except in the special case when the two numbers $z_1$ and $z_2$ are real. Thus, if you see a statement such as $z_1=\alpha z_2, \alpha>0$, it is implicit from the use of the inequality $\alpha>0$ that the symbol $\alpha$ represents a real number.

A number system is said to be an ordered system provided it contains a subset $P$ with the following two properties:

  1. For any nonzero number $x$ in the system, either $x$ or $-x$ is (but not both) in $P$.
  2. If $x$ and $y$ are numbers in $P$, then both $xy$ and $x+y$ are in $P$.

Question: In the real number system the set $P$ is the set of positive numbers. In the real number system we say $x$ is greater than $y$, written $x>y$, if and only if $x-y$ is in $P$. Can you see why the complex number system has no such subset $P$?

Answer: By the conditions given for an ordered system, if $i\in P$, then $i\cdot i=-1\in P$. Thus, we have that $(-1)\cdot i=-i\in P$, which is a contradiction ($i$ and $-i$ cannot both be in $P$). If $-i\in P$, then $(-i)(-i)=-1\in P$. Thus, $(-1)(-i)=i\in P$, and this is also a contradiction. Consequently, no such subset $P$ exists.

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If $s>t$ then we have $s-t>0$. If $s$ and $t$ are complex, and $s-t=u$ (u<>0), then we need $u>0$. However as u lies on a circle with +ve radius $r, u$ is always greater than $0$. This means that $-u=t-s>0$, implying $t>s$, a contradiction, so there is no order over the complex numbers.

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  • $\begingroup$ Check this out for proper typesetting on MSE (your post above clearly has some issues--even some I couldn't figure out how to resolve): meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Daniel W. Farlow Jan 24 '15 at 5:01
  • $\begingroup$ Thanks, I'll read them. May I ask what issues? $\endgroup$ – JonMark Perry Jan 24 '15 at 5:04
  • $\begingroup$ And by the way (I can't comment on your post yet!), I assume P if it exists is unique, can you prove it? $\endgroup$ – JonMark Perry Jan 24 '15 at 5:06
  • $\begingroup$ For example, what does u<>0 mean exactly? Typing it in on Wolfram|Alpha or looking somewhere else does not seem to yield anything--what exactly did you mean? The link I provided should be sufficient for doing most typesetting tasks on here, but I would recommend a basic introductory book on LaTeX to fully make use of this site and not have your posts edited all the time. $\endgroup$ – Daniel W. Farlow Jan 24 '15 at 5:07
  • $\begingroup$ u is not equal to zero. It is often left unsaid when discussing the complex plane that zero implies zero plus zero times an imaginary unit. $\endgroup$ – JonMark Perry Jan 24 '15 at 5:14
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Simple answer. No. The complex numbers can not be an ordered field. [if $a \ge 0$ then $a^2 = a*a \ge 0$. If $a < 0$ then $a^2 = a*a > 0$ so $a^2 \ge 0$ for all $a$ so $1 = 1^2 > 0$ and $-1 < 0$. If $\mathbb C$ were an ordered field, $i^2 > 0$ so $-1 > 0$. Impossible. $\mathbb C$ can not be an ordered field.]

But $\mathbb C$ can be ordered without holding to the field axioms. One simple way is the "dictionary" ordering. $a + bi > c + di$ if $a > c$ or $a = c$ and $b > d$. This is consistent with the order on R. But we can't do much with it. It doesn't follow that if $z < w$ and $v > 0$ that $zv < wv$. It doesn't follow at all.

Or we can have partial orders. $z > w$ if $|z| > |w|$. But this isn't total order. $z < w$, $z > w$, $z = w$ are not exhaustive and mutually exclusive; we can have cases where none of the three apply.

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I think that you should look at > or < relations like this: a. they categorise unequal items b. they represent some order e.g. 1<2<3<4 of the set of items A complex number is always a pair of numbers a real number is one number. So it makes no sense to compare a pair against a single item like to ask if a married couple is taller than the celebrant (without additional assumptions) To compare two complex numbers you need to define what ">" means like for two vectors. You may chose the vector's magnitude be the criteria of ordering rather than their orientation. This is a convention for what you define as criteria for order.

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