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Oct
28
revised Which of the following subsets of $R[x]$ are subrings of $R[x]$?
Updated to match updated question.
Oct
28
answered Show that the integers have infinite index in the additive group of rational numbers.
Oct
28
answered Which of the following subsets of $R[x]$ are subrings of $R[x]$?
Oct
28
comment “Problem of points (POP)” explanation
No need to apologize: a good many native speakers make exactly the same mistake. My comment was purely for information, not a criticism.
Oct
28
revised Show a subring contains certain elements.
Improved LaTeX.
Oct
28
revised how to solve $x^2 \equiv -1 \pmod{13}$
Fixed LaTeX.
Oct
28
comment “Problem of points (POP)” explanation
Note: The word that you want is the noun rationale ‘reason, justification’, not the adjective rational.
Oct
28
comment how to solve $x^2 \equiv -1 \pmod{13}$
How did you get $721$ from André’s suggestion? You know that $720$ is one solution to your congruence; he’s suggesting that you reduce it mod $13$ to get a nicer form for it. Once you’ve done that, getting the other solution, $-x$, will be easy.
Oct
28
answered Sum of $k$ consecutive integers
Oct
28
revised Sum of $k$ consecutive integers
added 2 characters in body
Oct
28
comment Sum of $k$ consecutive integers
The reasoning is correct; the OP omitted a lot of parentheses that would have made it much clearer. I’ve taken the liberty of inserting them.
Oct
28
revised Sum of $k$ consecutive integers
LaTeX, formatting.
Oct
28
answered What is $R(\omega)$ (and where can I find definitions for similar common notation)?
Oct
28
comment Is the intersection of a sequence of nested subspaces nonempty?
Properly speaking the finite intersection property is a property of families of sets, not of spaces, compact or otherwise. What you mean is the characterization of compactness in terms of families of closed sets with the FIP.
Oct
28
answered Can someone check my answer for this area between 2 polar curves question?
Oct
28
answered Non-trivial open dense subset of $\mathbb{R}$.
Oct
28
revised $f : X \to Y $ continuous and surjective. $A $ dense in $X$ $\implies$ $f(A)$ dense in $Y$
added 210 characters in body
Oct
28
answered $f : X \to Y $ continuous and surjective. $A $ dense in $X$ $\implies$ $f(A)$ dense in $Y$
Oct
28
answered Limit point compactness implies sequential compactness
Oct
28
comment Hard floor function problem
Very nice. ${}$