8,782 reputation
11741
bio website
location
age 43
visits member for 2 years, 5 months
seen Dec 14 at 18:42

Oct
19
answered What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
Oct
18
comment Are there countably many infinities?
Ah, now I got it. Thank you!
Oct
18
comment Are there countably many infinities?
Well, I didn't yet agree that $\aleph_\alpha$ exists for all $\alpha<\epsilon_0$ exist. For that I first need an argument that all countable ordinals are reachable through the addition/limit process.
Oct
18
comment Are there countably many infinities?
Let's start with $\aleph_0$. Now with the $\alpha+1$ rule, I can generate $\aleph_1$, $\aleph_2$ and so on. Basically, everything $\aleph_n$, $n\in\mathbb N$. With the limit rule I then get $\aleph_\omega$. Then I get $\aleph_{\omega+1}$ etc., then again with the limit rule $\aleph_{2\omega}$, and indeed $\aleph_{m\omega+n}$. I even see how you can extend that to arbitrary polynomials of $\omega$. But those are still countably many ordinals, and thus countably many cardinals associated to them. So, how do you get beyond countability?
Oct
18
comment Are there countably many infinities?
But I don't see how you can get more than countably many cardinalities by that procedure.
Oct
18
comment Are there countably many infinities?
Is there an easy way to see that there is an $\aleph_\alpha$ for each ordinal $\alpha$?
Oct
18
comment Does $1.0000000000\cdots 1$ with an infinite number of $0$ in it exist?
@DavidRicherby: Maybe the cardinality of proper classes?
Oct
18
comment Does $1.0000000000\cdots 1$ with an infinite number of $0$ in it exist?
@AsafKaragila: The banner at the top of the site contains, besides a set of images, "StackExchange" and some numbers, as well as the words "review", "help" and "search", but not the word "Mathematics". ;-)
Oct
18
comment Does $1.0000000000\cdots 1$ with an infinite number of $0$ in it exist?
Actually, you don't need $10^{-\omega}$ to be well defined, since that doesn't occur in the answer. What you need is a well defined limit.
Oct
10
answered Addition of Vectors?
Oct
9
answered Is it possible to not have irrational numbers?
Oct
9
comment How is greater than defined for real numbers?
@Lano: Good point, see my edit.
Oct
9
revised How is greater than defined for real numbers?
Added how to define the topology without comparison operators
Oct
8
comment The “Empty Tuple” or “0-Tuple”: Its Definition and Properties
@SteveKass, Mathemanic: Thank you for explaining why my idea doesn't work.
Oct
8
comment The “Empty Tuple” or “0-Tuple”: Its Definition and Properties
@Mathemanic: The Wikipedia definition mirrors the way lists are defined in Lisp (or, more generally, in functional languages). Also, it allows to distinguish e.g. the triple $(a,b,c)$ from both the tuple $((a,b),c)$ and the tuple $(a,(b,c))$.
Oct
8
comment The “Empty Tuple” or “0-Tuple”: Its Definition and Properties
A quite obvious definition of tuples I cannot find on the Wikipedia page: $()=\emptyset$, $(a)=\{\{a\}\}$, $(a,b)=\{\{a\},\{a,b\}\}$, $(a,b,c)=\{\{a\},\{a,b\},\{a,b,c\}\}$, …
Oct
8
comment How is greater than defined for real numbers?
Well, transitivity holds only if the sum of any two squares is again a square; the rational numbers fail this property. Antisymmetry holds only if $x^2+y^2=0$ implies $x=0$ and $y=0$; the complex numbers fail this property.
Oct
8
comment How is greater than defined for real numbers?
@AsafKaragila: It works for $\mathbb C$, it just doesn't give anything useful there (especially no order relation).
Oct
8
answered How is greater than defined for real numbers?
Oct
6
comment Unitary and Self-adjoint complex operator .
Err ... with $v=x+y$, $U(v)=x+y$ means $U(v)=v$; given that this decomposition is possible for all vectors, $U$ is clearly the identity, which is both unitary and self-adjoint. But somehow I think that is not the problem you meant ...