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 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. Feb 6 revised Is $0$ a natural number? style Feb 6 comment Is $0$ a natural number? @goblin Hm, what’s wrong with calling them “finite numbers”? Feb 6 revised Is $0$ a natural number? deleted 88 characters in body Feb 6 revised Is $0$ a natural number? deleted 1 character in body Feb 6 answered Is $0$ a natural number? Feb 6 comment Is $0$ a natural number? Judging from other posts by you, I like your style of thinking about mathematics. I’m using $ℕ$ and $ℕ_0$ to denote natural numbers without and with zero respectively. Naturally, I’m now wondering why you do include zero in $ℕ$. I’m with Noldorin on this issue, thinking of natural numbers as the counting numbers with which we all grow up (that’s what makes them so natural). Is your use of $ℕ$ sheer pragmatics or do you think of zero as natural in some sense? Feb 2 comment What's the name for the property for which $x + x = 0 \Longleftrightarrow x = 0$? AndreasCaranti. Ah okay. I still think your answer is phrased confusingly. Perhaps it’s better to say “You might say that a group with such a property …, so $(\mathbb S, +)$ does not …” or so. Whatever, though, I didn’t see the title question. Feb 2 comment What's the name for the property for which $x + x = 0 \Longleftrightarrow x = 0$? Although strictly speaking, he has found that $(\mathbb S, +)$ does have $2$-torsion, or is lacking $2$-torsion-freeness. Feb 2 comment Looking for a non-combinatorial proof that $a! \cdot b! \mid (a+b)!$ Marlu once suggested using the obvious injective group map $S_a × S_b → S_{a+b}$. Do you consider this combinatorial? Jan 26 revised Are the polynomial functions on $S^1$ dense in $C(S^1,ℂ)$? rolled back to a previous revision Jan 25 comment 35 + 6 = 30 + 11 Okay, taking you seriously: You don’t add quantity, but stuff, but as if it were quantity – and that’s why you fail. If you’re talking quantities, then 31 + 10 = 41 = 35 + 6, if you’re talking apples and oranges, then 31 oranges + 10 apples ≠ 35 oranges + 6 apples. That’s why people have a saying “You can’t add apples and oranges.”, which mathematicians jokingly extend by saying “… unless, of course, within the free abelian group generated by an apple and an orange”. It turns out, mathematicians have invented structures where we can add apples and oranges! Jan 23 comment Monomorphisms of monoids are stable under coproducts @EricWotsey One point is, though, that it strongly depends on the notion of a set and that it has to be done for every pair or triple of $M$, $N$ or $M$, $N$, $K$. Whereas, if you could do the proof only using a really specific monoid, the proof can be seen in a rather self-contained axiomatic framework of the category of monoids (which is always nicer and hopefully even simpler). Jan 23 comment Monomorphisms of monoids are stable under coproducts Ah, so you want to avoid a set-theoretic construction of the monoid coproduct, is that it? Jan 23 comment Monomorphisms of monoids are stable under coproducts If a proof only relied on the universal property of the coproduct, wouldn’t that proof be reusable to prove that all coproducts preserve monomorphisms? Jan 12 comment Is $\operatorname{End}(V)\simeq M_n(D)$? Well, actually. Come to think of it. It probably goes like $\operatorname{End} R^n \cong M_n (R^\mathrm{op})$. Doesn’t matter much, though, because the opposite ring of a division ring is again a division ring and the definition of prime rings seems to be symmetric.