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If you solve this I'd award you with 500rep


4h
revised Evaluate the limit: $\lim_{x\to \infty}$
formatting
1d
revised Is the Axiom of Choice necessary to prove $\mathbb R \approx \mathcal P(\omega)$?
added 9 characters in body
2d
comment Proof for '$AB = I$ then $BA = I$' without Motivation?
This proof only use matrix multiplication.
2d
revised If $AB = I$ then $BA = I$
added 2 characters in body
Jul
14
revised How does one prove that if $f$ and $g$ are linear functionals on $V$ such that $h=fg$ is also a linear functional, then either $f=0$ or $g=0?$
Formatting
Jul
12
revised Existence of rational sequence such that a polynomial is split over $\Bbb{Q}$
Avoiding trivial solutions
Jul
12
comment Existence of rational sequence such that a polynomial is split over $\Bbb{Q}$
Yes, here the roots are not necessarily distinct. There should be some literature about this. Both seems like very natural questions, I wonder where we can find some.
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