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Feb
21
comment $\mathbb{Z}$ is a vector space over $\mathbb{Q}$
@user1952009 Yeah, but the addition on $ℤ$ gets really weird. I’d have no intuition for such an implementation of $ℚ^n$.
Feb
21
revised $\mathbb{Z}$ is a vector space over $\mathbb{Q}$
added 63 characters in body
Feb
21
comment $\mathbb{Z}$ is a vector space over $\mathbb{Q}$
Oh, actually: Do you want $ℤ$ to be equipped with the usual addition?
Feb
21
answered $\mathbb{Z}$ is a vector space over $\mathbb{Q}$
Feb
20
revised Why are matrices indexed by row first?
added 490 characters in body
Feb
20
revised Why are matrices indexed by row first?
deleted 1 character in body
Feb
20
answered Why are matrices indexed by row first?
Feb
16
comment Can any uncountable dimensional real vector space be made into a Banach space?
@SaunDev I took the liberty of changing the title into a somewhat more concise question (for the sake of readability). Oh yeah, forgot uncountable.
Feb
16
revised Can any uncountable dimensional real vector space be made into a Banach space?
shortened the title
Feb
16
comment Can any uncountable dimensional real vector space be made into a Banach space?
Why are the questions posed in title and body different?
Feb
16
answered Finding limit of a 2 variable function (or show a lack of)
Feb
16
answered Determinant equality issue
Feb
16
comment Determinant equality issue
… and the coefficient $2,3$ of the first one might be $c$?
Feb
16
comment normal closures and generating sets of a group
Okay, I just deleted an answer of mine where I mentioned that symmetric groups yield counterexamples to the claim. This probably isn’t of much interest to you (as symmetric groups are not residually nilpotent for $n ≥ 3$ as far as I know).
Feb
6
comment How to find a onto homomorphism between two groups?
The upper right entry of $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ doesn’t belong to $8ℤ$, you meant to write the identity matrix, right?
Feb
6
comment motivation for the direct limit
@ZhenLin Can’t one use some sort of choice to find a subdiagram of finite fields of characteristic $p$ that has all objects and is commuting?
Feb
6
comment motivation for the direct limit
I’d say that the algebraic clouse of a field of prime order $p$ is mereley isomorphic to a direct limit of all finite fields of order of a power of $p$ (or one could say, the direct limit is an algebraic closure). I really think one should stress this because due to the nontriviality of the absolute Galois group, the algebraic closure of a field does not exhibit any universal property (except of course when the base field was algebraically closed to begin with).
Feb
6
comment Is $0$ a natural number?
By the way: Are people downvoting because they don’t like the proposal or because they genuinely think it’s a bad answer?
Feb
6
revised Is $0$ a natural number?
added 50 characters in body
Feb
6
comment Is $0$ a natural number?
@goblin None? I don’t follow you … Oh, actually, I get it: We use “of finite measure” a lot. So, convinced. Finite ordinals/cardinals it is.