21,712 reputation
42866
bio website ie.linkedin.com/in/aymanh
location Dublin, Ireland
age
visits member for 3 years, 7 months
seen 1 hour ago

Senior software engineer at Google. B.Sc. and M.Sc. in Computer Science, Software Engineering. Studying for a M.Sc. in Mathematics.


1d
comment Let $I$ be an ideal generated by a polynomial in $\mathbb Q[x]$. When is $\mathbb Q[x] / I$ a field?
@BillDubuque Fixed.
1d
revised Let $I$ be an ideal generated by a polynomial in $\mathbb Q[x]$. When is $\mathbb Q[x] / I$ a field?
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1d
revised Resources for learning integral calculations
edited tags
Jul
10
revised $f(A) \cap f(B) = f(A \cap B)$ if $f$ is a bijection?
\cap is more appropriate in this context.
Jul
6
comment Calculate the distance from a point to a line
@IvoTerek Thank you!
Jul
6
comment Calculate the distance from a point to a line
Non-English posts are acceptable. Hopefully someone who knows Portuguese will translate it to English.
Jul
6
comment Show that $T\mathbb S^1$ is diffeomorphic to $\mathbb S^1\times\mathbb R$
Do you know that Lie groups are always parallelizable? $S^1$ is a Lie group via multiplication. Alternatively, try to find a global frame for $S^1$.
Jul
6
comment Is $\cup_{k=1}^\infty (r_k-\frac{1}{k}, r_k+\frac{1}{k}) = \mathbb{R}$?
@Shine The condition that $\Bbb N \to \Bbb Q$ is a bijection (note the codomain) is already satisfied by my answer. This is the definition of an enumeration of $\Bbb Q$. Different enumerations lead to different results indeed. I provide an example in which (2) fails. See Rene's answer for an example in which (2) is true.
Jul
6
comment Is $\cup_{k=1}^\infty (r_k-\frac{1}{k}, r_k+\frac{1}{k}) = \mathbb{R}$?
@ReneSchipperus Nice observation!
Jul
6
answered Is $\cup_{k=1}^\infty (r_k-\frac{1}{k}, r_k+\frac{1}{k}) = \mathbb{R}$?
Jul
5
awarded  analysis
Jul
4
awarded  Excavator
Jul
4
revised $\sin x$ does not satisfy this quadratic equation
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Jul
4
comment $\sin x$ does not satisfy this quadratic equation
@Michael You're right. We need to clear the denominators first.
Jul
4
revised $\sin x$ does not satisfy this quadratic equation
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Jul
1
comment Deleting a subset from a topological manifold
It can be anything $\le n$. You can remove a subset of the sphere to get a circle or single point.
Jun
30
comment Complex analysis - existence of field $\mathbb{C}$
@user160738 No problem at all. I've seen your comment above. I think this is what the author had in mind, but it does require some background in ring theory. Particularly, you need to study quotients of polynomial rings and the first isomorphism theorem.
Jun
30
comment Complex analysis - existence of field $\mathbb{C}$
@user160738 What's the context of this question? Did you study quotients of polynomial rings? What about field extensions?
Jun
30
answered Complex analysis - existence of field $\mathbb{C}$
Jun
26
revised Funky Fundamental Group Question
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