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06:33
@Bohemianrelativist to know what a topological invariant is, you first need to get to the concept of a "homeomorphism"
@Bohemianrelativist I suppose you mean the metric completion; I don't remember the fastest way, but you can show that there is a "unique" ordered field, and then show that the metric completion of Q is such a field
 
2 hours later…
08:31
@RodrigodeAzevedo that's a scary thought but we don't have ads on math.SE right? only the community ads — Calvin Khor 6 hours ago
@CalvinKhor I don't see ads here, but I see ads on other SE sites where I have few rep points. I believe in making Math SE easy to use and useful for people for whom mathematics is utter torture. Having more tags to help people find questions similar to the ones in their homework would be a step in that direction. — Rodrigo de Azevedo 5 hours ago
@CalvinKhor I have never tried to find out all details, but at the very least I have seen screenshots showing ads on Mathematics in this question: Ads appearing before (on top of) questions?
When I search the help center it gives me this link: math.stackexchange.com/help/privileges/reduced-ads But the link just redirects to the list of privileges.
@MartinSleziak thank you for the intel. I'd bet ads are coming, just "when" rather than "if"
On some other sites, I see reduced ads as a privilege at 200 reputation points: meta.stackexchange.com/help/privileges/reduced-ads stackoverflow.com/help/privileges/reduced-ads
@CalvinKhor Well, judging by the linked post, the ads were on Mathematics in September 2020.
If the threshold is 200 reputation points (as on other sites), we're unlikely to notice them.
@MartinSleziak, yes, perhaps some scouting with a fresh account / no account might reveal something. odd that it doesn't appear in the list of privileges
There is also no option to see more ads in the Preferences (but I have this option in e.g. Meta.SE)
On the other hand this answer claims that there are no ads here. ("None of the SE 2.0 sites like Math earn any money at the moment.")
3
A: Where does math stackexchange get its money for hosting servers?

Jitendra SinghStack Exchange is earning money by ads on the three largest sites (Stackoverflow, Superuser and Serverfault) as well as their Careers platform. None of the SE 2.0 sites like Math earn any money at the moment.

I guess it might be reasonable to ask about this on Mathematics Meta. (Since in various places there is contradicting information.)
I see the option show/not show more advertisements if I go to my Meta Stack Exchange profile.
08:46
@MartinSleziak yes, that is exactly what I saw
I do not see such option in my Mathematics profile.
This list of sites doesn't include Mathematics and hasn't been updated since 2016
I would suggest to post a question about this on meta.
Should I post the question on meta, or do you want to post it?
@MartinSleziak What brach of math do you study?
@MartinSleziak you can post it if you're free :) something just occured irl for me
08:52
@CalvinKhor Ok, I'll do that.
Can you guys tell me the name of books so that I can study foundation of mathematics.
Instead of wasting my life I hope to contribute my time to mathematics
I need to restart everything again.
Build your own foundation. Start with a book you feel comfortable with and build your way up.
@MartinSleziak great, thanks very much and sorry for the inconvenience!
@eryceriousmatherfunker I haven't studied the 'foundations' from a book, and anything I say will just be things that google tells me
How did you study it then?
@CalvinKhor which question were you replying me?
09:00
@Bohemianrelativist you asked how to show that R is the completion of Q
You profile indicates me that you have good foundation.
@CalvinKhor is that related to metric?
@eryceriousmatherfunker thank you; but I learned it from lectures in a university, not from a book
@Bohemianrelativist well it depends what you mean by completion, but I would say yes
sorry, I forget Q represents rationals. I seldom encounter this symbol in physics.
@Bohemianrelativist :) my apologies for being lazy and typing Q
09:02
@user85795 I agree but I don't have experience in books and I don't remember anything after accident.
your school books are a good start
need to leave, leave a message if you want! bye
@user85795 I do think so.
09:04
When I read first few page I did my own discovery and made a conjecture.
Finding minima for two absolute value formula using triangle equation.
The book hard alternative approach.
|x-a|+|x-b|>=|2x-a-b|
@CalvinKhor I read "the real numbers are the completion of the rationals". from Wikipedia Metric Space - Types of metric spaces - Complete spaces. I actually I have read this kind of statement many times, but I keep not knowing why is so.
if you graph you will find |x-a|+|x-b| and |2x-a-b| you will see that minima of |2x-a-b| can location is location of |x-a|+|x-b|
Above was mine approach. I don't remember how to prove things and I am baffled by looking at the notebook where I write mathematical proof.
@Bohemianrelativist What can be said about this depends on what you already know about metric spaces and completions and what is your definition of real numbers.
In fact, this can be taken as the definition of real numbers.
If you wish, we can briefly discuss this - but I would suggest to go to another room: chat.stackexchange.com/transcript/19167/2021/3/13 (To avoid having two discussions in the same room at the same time.)
09:23
@eryceriousmatherfunker Since $(2x-a-b) = (x-a) + (x-b)$, triangle inequality indeed seems like the most natural approach here.
I certainly agree than when studying various inequalities with absolute value, graphs can help a lot.
I want to be more rigorous about it so I am going to study first logic again.
After car accident it is really hard to recall most of the stuff I have learn.
I have tried to mention graphic approaches (and also the interpretation as the distance) when posting about similar topics, such as here: Determine all solutions to $|x+12|+|x-5|=15$ and How to solve equations involving modulus function of the type $|x+1| - |1-x|=2$ and $|x-1|=|x|+a$?
I don't even understand the note I have written in for example: Numerical analysis
For determine all the solution to |x+12|+|x-5|=15=|2x+7|
finding |2x+7|=15 you can find the solution
but remember (x+12)(x-5)>=0
Because equality holds only when above inequality holds.
Exactly sa you say, |x+12|+|x-5|=15=|2x+7| is not true for all real numbers.
I think my approach is not presented in that post but it seems your approach is like mine but is kinda modified.
I have counter example for finding minima which is |x|+|1| and |x+1| is not going to give you minima at x location
minima at 2x-a-b=0

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