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

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?

Solution 1:

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.

Solution 2:

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 $3$.

Solution 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.

Solution 4:

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.