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 Jul 13 comment When a function is a dimension? @JonathanHebert: So what would you then call a criterion in the everyday-language sense that is not necessary (for example, a sufficient criterion)? The "being seven" criterion sounds ridiculous because it's ridiculously narrow. More similar to my dimension sentence would be "an obvious criterion for the term "numbers" making sense for elements of an algebraic structure is if that structure includes the integers." Jul 12 comment Logically equivalent formulas and contradiction @Sushil: $A\implies(\lnot B\land B)$ is not a contradiction, it is the statement that assuming $A$ would imply a contradiction. As such, it is equivalent to $\lnot A$. Jul 12 comment When a function is a dimension? I didn't claim it's always true. I said it's one obvious criterion. Which implies that there are others as well. Jul 12 comment When a function is a dimension? One obvious criterion would be that if you can apply the definition to mathematical objects where you already have the term "dimension" defined, and it turns out that for those objects it gives exactly the previously defined dimension, then it probably makes sense to call that newly defined quantity a dimension as well. Jul 12 comment Can a vector field be conservative if its domain is not a star domain? Another way to get a conservative function on a non-star shaped domain is to take a conservative function on a star-shaped domain and restrict it to a non-star shaped subset of that domain. Jul 12 answered Vector spaces - $\mathbb{R}^n$ and $\mathbb{R}^m$ Jul 12 comment Order in writing composed morphisms @darijgrinberg: But that's exactly what we do with operator notation on linear spaces: $Av$ is the application of the linear operator $M$ to the vector $v$ (in finite dimension: multiplication of the matrix M with the vector v), while $AB$ is the composition of linear functions (matrix multiplication). The point is, you have to know what sort of object you are dealing with (another example: if $\alpha$ and $\beta$ are scalars and $v$ is a vector, then $(\alpha\beta)v$ also involves two different operations; one field multiplication, and one scalar multiplication of a vector). Jul 12 comment Order in writing composed morphisms Another argument for the "reverse" (compared to the traditional) notation is that it gives a more natural composition rule for relations: With the usual rules, you get $a(R_1\circ R_2)b \iff \exists c: cR_1b\land aR_2c$. It would be much more natural to define it as $a(R_1\circ R_2)b \iff \exists c: aR_1c\land cR_2b$. But with the usual composition rule, that would not reduce to the usual function composition. It would, instead, reduce to the reverse function composition. Jul 12 comment Limit of ratio of sequences Your first limit after "I proceeded" does not need to exist. For example, take $a_n=n^2$, $b_n=n^3$, $s=0$. Jul 12 comment Does Pi contain all possible number combinations? How do you know that the set of non-normal numbers is Lebesgue measurable? Jul 12 comment Probability of winning a tournament by winning all matches in tournament Your terminology is very confusing: You are using "tournament" for two different things, a combination of three matches, and a series of tournaments in the first sense. In particular, your players can not win the tournament, and yet win the tournament … because the "tournaments" in both parts of the sentence are different things. Jul 12 revised Probability of winning a tournament by winning all matches in tournament added 7 characters in body Jul 12 answered Probability of winning a tournament by winning all matches in tournament Jul 12 reviewed Approve Prove $\{a(x,y,z)=(ax,y,z)\}$ is a vector space Jul 12 comment Computing the volume of this weird object, @mathcounterexamples.net: in the interval $[-1,1]$ the sphere of radius $2$ has cylindric radius $\ge\sqrt{3} > 1$. Since the function is restricted to be $\le 1$, the surface clearly is completely inside the sphere. Jul 12 comment Show that $\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}$ for all $n\in\mathbb{N}$. If you really feel that you have to prove that $2^k$ is positive, you can of course do that using induction as well. It definitely is not impossible. Jul 11 comment Zero to the zero power - Is $0^0=1$? $0^0=999$ would be a contradiction to the power laws, because then $(0^0)^2 = 999^2 \ne 0^{0\cdot2} = 999$. The only two values for $0^0$ consistent with the power laws are $0$ and $1$. Jul 11 revised Zero to the zero power - Is $0^0=1$? improved formatting Jul 11 comment Derivative that doesn't care about countable subsets? Both answers would have been worthy of an accept, but I can accept only one. Jul 11 accepted Derivative that doesn't care about countable subsets?