@AfronPie No, it's definitely not. But if you assume $A$ and $B$ are independent, then you do get a contradiction. That's not saying the question is a contradiction.
@AdamL $\min(A)$ and $\max(A)$ are supposed to be elements of $A$. For example $(0,1)$ has no min or max the same as $\{\}$. Now $\inf(\{\})=\infty$ and $\sup(\{\})=-\infty$.
A while ago, I asked a similar question For any $k \gt 1$, if $n!+k$ is a square then will $n \le k$ always be true? where users mathworker21 and WE Tutorial School proved that for non-square $k$, $n\le k$ is always true when $n!+k$ is sqaure.
Recently I got the idea to check if the property is...
I will construct a sequence of functions on closed interval [a,b]. Let's say there is a countable infinity of rational numbers in this interval so there is a sequence in [a,b],which we denote by $x_1,x_2,...x_n$
@RealDumbfoggybrain yes. Any countable set (such as $\{x_1,x_2,x_3,\dots\}$) is a null set, and whether you are using Riemann or Lebesgue integrals, the integral will be $0$.
That is, $f(x)=0$ except on a null set, so the integral is $0$.
@RealDumbfoggybrain For a countable set, consider the partition $\bigcup\limits_{k=1}^\infty\left(x_k-\frac\epsilon{2^k},x_k+\frac\epsilon{2^k}\right)$
@skullpatrol Hey Pal! I want to talk to you about some topic, can you move me to some private chat (I know every chat is public but something like my room “Let’s do...” or yours “Corona update” )
@RealDumbfoggybrain Just an advice as a friend, some people here don’t like excessive joking (I do like jokes, just like you) so if you’re saying about spreading CORONA they might flag you
foggy brain: Depends on what level of metaphysics you are asking: An answer is a statement that address the needs of a question I am not sure if you can have integers without binary operators like addition, or more generally, an operation at all. Might think about such minimal notion of integers later although I currently don't need such deep metaphysical territory yet Leaky: That works too
I am not asking for the most minimal definition of an integer
I am asking for a definition of integer without reference to other number types, so Thorgott and Leaky have provided me the answers that address my question
But the basic theory is: For any solvable rubik cube configuration, it is done by applying a sequence of rotations and antirotations from the solved state to the current scrambled state. Since everything is invertible in a group, it guarentees there exists sequence of moves to solve it
Hmm... so you are following a more applied path to group theory. I don't know quantitative analysis well to comment, other than yes, there are some pretty high dimensional groups involved for complicated hedge fund portfolios
Having studied some group theory in my last term at university, I've found it to be quite interesting, although it's also something I want to improve on (mainly when it comes to proving statements), so I figured that it might be worth doing some group theory slightly beyond what I need for this y...
I think mathematician are not rare nowadays. Lots of people are studying in this field that are smarter than me. So it make me think I shall give up on math.
But why are some problems still remained unsolved?
This is my question.
I am sure there are millions of mathematician with phd but still why no one has been able to solve Reimann hypothesis? There are people who is learning math for their entire life.
https://oeis.org/A003418 "An assertion equivalent to the Riemann hypothesis is: | log(a(n)) - n | < sqrt(n) * log(n)^2. - Lekraj Beedassy, Aug 27 2006. " (due to Schoenfeld)
And continuing development seems dumb when there are smart people who can handle this
Math was my dream until I was 15 and start studying university level math and found too much people are smart thus I feel dumb to even tough math. But I kinda got addicted to it.
@geocalc33 Every kind of math. I am curious but depressed all the time. It feels like There will not be alot of change with my existence or without my existence.
@Astyx c'est vraiment "le set sature," desole. Le set c'est sature si le set c'est un intersection des sub-sets ouvert de X pour un espace topologuique $(X,\tau)$
@Astyx There is hurry since If I don't learn it fast and deeply then I am gonna be doomed forever. I must join university which is impossible because of my bad grade or just go get a job.
Must solve some open problems in math to get professor's attention lol.
A while ago, I asked a similar question For any $k \gt 1$, if $n!+k$ is a square then will $n \le k$ always be true? where users mathworker21 and WE Tutorial School proved that for non-square $k$, $n\le k$ is always true when $n!+k$ is sqaure.
Recently I got the idea to check if the property is...
