No, the issue is algebraic. The usual notion of “ordering” includes the fact that the product of two positive numbers is positive.

Without that, you can define whatever you want.

Ah i see, so thats the reason i cant tell whether 100+100i is bigger than 1+i

is there any theorem for the "usual notion of ordering"?

Look up the definition of an ordered field.

You can define an ordering (many) just fine, but it won’t have algebraic properties.

Ah thanks Ted for the keywords to google

Arg... twice, now, Ted has said what I was typing, but managed to get it into the chat before I can paste in my links.

Sure. Also google dictionary (or lexicographic) order.

haha you can still shoot your links

Any set can be well-ordered, as a set. However, there is no ordering on the complex numbers which is compatible with the algebraic structure of the complex numbers.

I’m on the ipad, so snail slow, Xander.

@TedShifrin Yeah, but I don't have the incredible recall that you have, and have to Google for basic results from time to time, which is more time consuming than just remembering things.

I forget more than you think.

@TedShifrin tried, didnt understand a single word wikipedia and so i promptly closed the page in 2 seconds

@TedShifrin Yeah, but I was never that smart to begin with. I'm sure that you've forgotten more than I'll ever know. :D

@XanderHenderson ahh this puts things in perspective nicely

$a+bi<c+di$ if $a<c$ or if $a=c$ and $b<d$.

@AdilMohammed Think about how words are ordered in the dictionary. All the words that start with "a" come before all the words that start with "b". Imagine doing something similar with the complex numbers.

you cant, cause that doesnt make sense on the argand plane...

All of the numbers which start with $1$ are bigger than all of the numbers that start with $0$. So, for example $0 + 10^{10000} i < 1 + 0i$.

I should have said that!

Sure it makes sense.

it is not nice and orderly

Nothing to do with modulus, but so what? You threw that out already.

@AdilMohammed It is perfectly nice and orderly.

It just doesn't agree with the field structure of the complex numbers.

It’s totally nice and orderly .

@Koro Suppose we try that on $\Bbb R^2$

<— slow

Let $i=(1,0)$ and $j=(0,1)$, though it doesn't really matter

@Koro Then $\langle i,i\rangle=\langle i,i+j\rangle-\langle i,j\rangle=0-0=0$

But one of the properties of inner products is that $\langle v,v\rangle=0$ iff $v=0$.

So this is not an inner product.

@XanderHenderson ah the complex field structure, i see...

*quietly puts it down in list of words to look up later

@Koro There is no canonical inner product on a vector space.

Once you choose a basis, sure. (But there is no canonical basis of an arbitrary vector space either.)

@AkivaWeinberger Yeah, but there is a cannonical inner product. It involves gunpowder, though, so you have to be careful.

smacks Xander

Similarly, if $V$ is a finite-dimensional vector space, there is no canonical isomorphism $V\to V^*$ where $V^*$ is a dual, because this gives you an inner product

@AdilMohammed It just means that the ordering is not compatible with the addition and multiplication operations. For example, as Ted pointed out above, if $a, b < 0$, then it must be the case the $ab > 0$. This fails pretty badly for lexicographic order on the complex numbers.

ah that does kinda defy common sense... see not nice and orderly

so thats the algebraic issue of comparing complex numbers

YUM!

Thanks Ted and Xander, now i know what to say when i become a math teacher one day

You can define an ordering on any set*, but there's no ordering on the complex numbers that satisfies the properties we want

It’s a good question.

(*I think you need to use the axiom of choice to do this on the power set of R or something. Not relevant to this stuff though)

@TedShifrin totally, my teacher just told "you cant compare, cause it doesnt make sense" (assuming you meant mine was a good question)

@AkivaWeinberger right

@AkivaWeinberger Lies. The Axiom of Choice is obvious nonsense.

@AkivaWeinberger thanks :)

I looked it up and it is in fact true that, without choice, it is consistent that $\mathcal P(\Bbb R)\simeq\mathcal P(\mathcal P(\Bbb N))$ has no total order

@AkivaWeinberger Yup. Indeed, the statement "Any set can be well-ordered" is equivalent to the Axiom of Choice.

While consistently already $\Bbb R$ cannot be well ordered

@XanderHenderson I'm talking about a total order, not a well order

"Total order" is weaker

(A joke: The Axiom of Choice is obviously true, the Well-Ordering Theorem is clearly false, and no one knows what the hell is going on with Zorn's Lemma.)

@XanderHenderson but Akiva is talking about total orders, every set can be totally ordered is independent of ZF but weaker than choice

@AkivaWeinberger Oh, shoot, I missed that.

I thought we were talking about well-orders.

Nope. Literally any ol' order, (as long as it's not a sneaky partial order 'cause they don't count), you can't define one constructively on $\mathcal P(\Bbb R)$

Apparently the issue is the subset $\{a+\Bbb Q:a\in\Bbb R\}\subset\mathcal P(\Bbb R)$

@AkivaWeinberger That makes sense, but it is definitely outside of my wheelhouse.

Mine too but you can read more here

@AdilMohammed Well, a typical bad teacher answer. Show your teacher the order we gave you.

What the hell

I have no memory of writing that comment

(Andreas wrote a nicer argument in the comments, though it used a lemma I didn't know about)

@TedShifrin Maybe not "bad-teacher answer", but definitely a "bad answer from a teacher". It can be hard to give a good answer in the moment. But I would definitely go back to that teacher and (respectfully) ask about lexicographic order.

"Hey, after our conversation last week, I did some Googling and came across this 'lexicographic order' thingy. Can't we do that to the complex numbers? What's wrong with it?"

@AkivaWeinberger Ha!

@AkivaWeinberger to answer you other comment the fact that $E_0$-invariant sets have measure 0 or 1 is some standard 0-1 law for tail events whose name I forgot

There is also a topological version saying that tail sets with the Baire property are either meager or comeager

@XanderHenderson My string of words was ambiguous, yes.

@AlessandroCodenotti Kolmogorov 0-1 law

@AkivaWeinberger I thought about a little bit but I couldn't do it

@BalarkaSen I now think I have a substantially different solution to my first solution

(for the problem you have not yet solved)

Indeed I have, and dare I say, it's simpler than my previous one

I'm actually surprised I hadn't thought of this solution before because it's so much nicer