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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.