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Jul
12
revised Existence of rational sequence such that a polynomial is split over $\Bbb{Q}$
Avoiding trivial solutions
Jul
10
answered The annihilator of an intersection
Jul
10
revised The annihilator of an intersection
Formatting, english
Jul
6
revised Nilpotent Matrix
Fixed typo
Jul
6
comment Polynomial Linear transformation
As in 11.1, let $V$ be $n$-dimensional over... you are confusing that $V$ with the vector space of polynomials. That $V$ is any space, not necessarily that one.
Jul
6
revised If matrix $A$ is invertible, then there is a permutation of its rows leaving no-zeros on the diagonal
Formatting
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jul
2
comment If $f^{n_o}$ has a fixed point , then does $f$ also has a fixed point , where $f$ is continuous on $\mathbb R$?
True, but this function is not from $\mathbb R$ into $\mathbb R$ so this doesn't answer the question.
Jul
1
revised Proof that $a\equiv b \pmod n \iff a \pmod n = b\pmod n$
Formatting
Jun
30
revised Calculate the dimension of $U = \{(x_1,x_2,x_3,x_4,x_5) : x_1+x_3+x_5=x_2+x_4=0\}$
TeX
Jun
29
revised $V = \operatorname{Im} T + \ker T $ then $ \operatorname{Im} T \cap \ker T = \{0\}$
Formatting et al
Jun
25
comment Countable union of cartesian product.
It's never an interval. What you en up with looks like stairways.
Jun
24
revised Proof for Integral Inequality $|\int f| \le \int |f|$ - is it sufficient enough?
added 30 characters in body
Jun
23
comment proof by induction that every non-zero natural number has a predecessor
A bit technical answer.
Jun
23
comment Proof for Integral Inequality $|\int f| \le \int |f|$ - is it sufficient enough?
@user2637293 I edited my answer. If anything is unclear, let me know how I can help you.
Jun
23
revised Proof for Integral Inequality $|\int f| \le \int |f|$ - is it sufficient enough?
Expanding on some details on demand
Jun
21
comment showing $\int _a^b\left(f'\left(x\right)\right)dx\:=\:f\left(b\right)-f\left(a\right)$
@TonyPiccolo That's what he's asking
Jun
21
revised showing $\int _a^b\left(f'\left(x\right)\right)dx\:=\:f\left(b\right)-f\left(a\right)$
Fixing typos and formatting
Jun
12
comment Show: $M\subset\mathbb{R}^n$ Jordan-measurable, iff $vol^*(\partial A)=0$
Look at this