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17h
comment interior and closure of (0,1]
$(0,1)$ is the union of all the basic opens $[1/n,1)$ for $n\geq 2$, and therefore perfectly open.
17h
comment Proving that a normal, abelian subgroup of G is in the center of G if |G/N| and |Aut(N)| are relatively prime.
Woah, look that formatting! Calm down on the dollar signs. Please use > at the start of a line for a block quote instead (while still having dollar signs around the math, of course). (I'd do it for you if I weren't on mobile.)
1d
comment Find two functions $f$ and $g$ which are integrable, but whose composition $f \circ g$ is not
A function is Riemann integrable iff it is bounded and continuous except at a set of measure zero. $f$ is discontinuous only at the rationals, and $g$ is discontinuous only at $0$, but $g\circ f$ is discontinuous everywhere.
1d
revised Find two functions $f$ and $g$ which are integrable, but whose composition $f \circ g$ is not
added 76 characters in body
1d
answered Find two functions $f$ and $g$ which are integrable, but whose composition $f \circ g$ is not
1d
answered Multiplying two logarithms (Solved)
1d
comment $k[x,y]/(xy-1)$ isomorphic to $k[x,\frac{1}{x}]$
What Jendrik means is that the most common way to define the relation between the elements $x$ and $\frac1x$ is exactly by this quotient, i.e. that $x\cdot\frac1x-1=0$.
1d
comment solve the Following Differential equation: $y''=2yy'$
@aribaldi The usual substitution rule for integrals tells you that $\int f(y)y'dx=\int f(y)dy$. So integrating $2yy'$ with respect to $x$ gives the same result as integrating $2y$ with respect to $y$.
1d
comment solve the Following Differential equation: $y''=2yy'$
What is non-rigourous with "We may rewrite the equation as $y''=(y^2)'$, and integration yields $y'=y^2+C$"?
1d
comment solve the Following Differential equation: $y''=2yy'$
It is so because you can differentiate and check. The derivative of $y^2$ is $2yy'$, while the derivative of $y^3$ is $3y^2y'$.
1d
comment solve the Following Differential equation: $y''=2yy'$
The right-hand side is $(y^2)'$.
1d
comment How many closed surfaces (up to homeomorphism) are there with Euler characteristic -2?
@thinker I think of it as common knowledge, although it clearly isn't for the uninitiated. The main result is that Euler characteristic and orientability together determines a unique closed surface up to homeomorphism. After that it's just a matter of using connected sums of the torus and the projective plane to construct the rest. By the main result, for instance, the connected sum of two projective planes must give a Klein bottle, and when you have something that is unorientable, summing with the torus is the same as summing with the Klein bottle.
1d
answered How many closed surfaces (up to homeomorphism) are there with Euler characteristic -2?
1d
comment So many logs with different bases
You can do a lot of simplifications. For instance, $5^{\log_6 6x} = 5^{\log_6 6 + \log_6 x} = 5\cdot 5^{\log_6x}$.
1d
comment Prove that $\Bbb Z[x, y, z,w]/(xz-yw)$ is not a UFD.
You should perhaps be aware that writing $xy$ as an element of $R$ is abuse of notation. It should really be written as $\bar x \bar y$ or $xy + (xw-yz)$ or something along those lines. Most people don't bother, though.
1d
comment How to change complex numbers into polar form?
Have you tried googling it? There are plenty of primers out there that are much better than what we can put into an answer.
1d
comment Prove neither $F_4$ nor $F_5$ is isomorphic to $\mathbb{Z}^4$?
The free group on $n>1$ elements is non-abelian by definition. The free abelian group on $n$ elements is abelian, however.
1d
comment $n(p(p(p(p(b))))$ Start from left to right?
When there are parentheses involved, it's usually a good idea to start furthest in. This includes functions like the power set.
1d
answered Laurent Series about $z=0$ of $f(z) = \frac{1}{z^3 - iz}$
1d
comment Does $\displaystyle \lim_{x \rightarrow 0^{+}}f(\ln x+ x)=-\infty$ imply $\displaystyle \lim_{x \rightarrow -\infty}f(x)=-\infty$?
It's true for any value of $\lim f(x+\ln x)$, not just $-\infty$.