Problem Let $f:I\rightarrow \mathbb{R}$ be differentiable in an arbitrary interval I. If $f'(x)=0 $ for all $x\in I$, then $f$ is constant. My idea If $x \in \operatorname{int}(I)$, we know that $\exists$ $n_0\ge0$ such that $x\in [x-\frac{1}{n_0},x+\frac{1}{n_0}]$ then by the Intermediate...

This can be done more easily. We have $f'(x)=\frac{df(x)}{dx}=0\implies df(x)=0dx.$ Now, integrating both sides, we have, $\int f(x)=\int 0dx\implies f(x)=C,$ where, $C$ is an arbitrary constant. This seems a one-liner.

Round the following number to $4$ significant figures: $794834.$ For the number $794834$, I simply observed that all the digits in here are significant. The $4th$ significant digit in this case, is $8$ and the $5th$ digit is $3$. Now, as $3\lt 5$ so, we can represent the given number as $7948\tim...

This question was asked here, where the answer uses this description. The last line reads: "The special case of $0$ does not have a unique representation in scientific notation, i.e., $0=0×10^0=0×10^1=..."$ My question is the value of $a$ cannot be $0$ since, as they state: $a$ is a $\color{blue}...

Here's a proof without using complex analysis, differential geometry or the projective plane, and only using results from general topology. Consider the equivalence relation on $\mathbb{C}^n$ given by $x\sim y$ iff $y = (x_{s(1)}, ..., x_{s(n)})$ for some $s\in S_n$. Give $\mathbb{C}^n/\sim$ the ...

The Question: Suppose we are given the definition that a torus $T$ of a linear algebraic group over $k$ is a subgroup isomorphic to $\Bbb D_n$, show that $$T\cong(k^*)^m$$ for some $m\in\Bbb N$. Context: I have looked in Springer, Borel, and Humphreys; a proof is nowhere obvious in them. The text