# Can a complex number ever be considered 'bigger' or 'smaller' than a real number, or vice versa?

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?

#### Solutions Collecting From Web of "Can a complex number ever be considered 'bigger' or 'smaller' than a real number, or vice versa?"

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.

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 $3$.

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.

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.

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.

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.

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.