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4

You say $X$ is a finite set, not a group. Thus "on group $X$" (quoting your question as it now stands) is wrong. One may say $G$ is a transitive permutation group on the set $X$, but not on the group $X$ unless $X$ is a group. It makes sense to speak of a transitive permutation group on a set $X$, but one does not say "$G$ is a transitive permutation". ...


3

Natural language is hard to interpret (for everyone) without a context (and may be hard even then). Math does not have a clear connection to natural language, and thus we do not necessarily have a clear interpretation just because we are using math. So in general you can not translate something, just because it is a mathematical context. Thus we may not give ...


2

Check the equation $\dot x = x^2$. All the nontrivial solutions have "blowup in finite time" (i.e., a vertical asymptote).


2

They are equivalent. Suppose your first statement holds, then $\forall y_1,y_2 \in S$, $$ d(y_1,y_2) \leq d(x,y_1)+d(x,y_2) < 2r. $$ Conversely, suppose your second statement holds, then $\forall x \in S$, $S\subset B_x(2r)$. Your second definition is more commonly seen in literature. There is, however, a different definition of boundedness in general ...


2

You are exploiting both the commutitative law $a + b = b + a$ and the associative law $a + (b + c) = (a + b) + c$


2

\begin{align} 97+198&=\\(100-3)+(200-2)&=\\ 100+(-3+200)-2&=\quad\quad&\text{Associativity.}\\ 100+(200-3)-2&=\quad\quad&\text{Commutativity.}\\ (100+200)-3-2&=\quad\quad\quad\quad\quad&\text{Associativity.}\\ \vdots \end{align}


1

After correcting a mistake, I agree with the formula : $$ F''(x)= −\frac{B(D−A)(\frac{x}{C})^B\Big((B+1)(\frac{x}{C})^B−B+1\Big)}{x^2\Big((\frac{x}{C})^B+1\Big)^3} $$ Anyways $F''(C)\neq0$ This is obvious for low values of $B$. Moreover, there is no inflexion of the curve if $B\leq 1$ : in the formula below, $x_{inflexion}$ is not real. In fact, saying ...


1

There is not much to say about this relation and thus it does not deserve a special name: If $\mathbb F$ has characteristic $2$, it just the equality-relation, since $x^2=y^2 \Longleftrightarrow x=y$. Otherwise we have $x^2=y^2 \Longleftrightarrow x=y \text{ or } x=-y$, hence the relation pairs any element of the field with its additive inverse.


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Yes. This follows from a stronger version of Jordan Curve Theorem, Jordan–Schoenflies theorem.


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Yes, an interpolating function should pass exactly through all the data points. You need enough adjustable parameters to make this happen. It can also be useful to find an approximating function that has fewer parameters and does not pass exactly through the points. If there is noise in your data, some forms of interpolating function, like a high degree ...


1

I would again means the same in maths as in English. I would say the answer to your example as 30. Similarly, Finally also means the same in math as in English. Eg: I would toss a coin. I then would toss another coin. I will do this experiment again. means that I am gonna toss 4 coins.



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