9,672 reputation
828
bio website danshved.wordpress.com
location Moscow, Russia
age 27
visits member for 1 year, 5 months
seen 11 hours ago

I'm a graduate student at MIPT, Russia. My thesis is in the theory of groups, but I like lots of other things too )


Apr
6
comment Showing two things are homotopic to each other
About the last paragraph: haw can we even talk about a homotopy between $1_X$ and $1_{S^1}$? You can have a homotopy between two maps $f, g: A \to B$. But in your case $1_X$ is $X \to X$, and $1_{S^1}$ is $S^1 \to S^1$. It doesn't make sense to call these two maps homotopic.
Apr
6
comment Question about singular homology
$C_0(X)$ is what is called a free module freely generated by set $X$. You can have a module generated freely by absolutely any set.
Apr
6
comment Showing two things are homotopic to each other
Please be more precise. (i) When you say "unit square", do you mean the border of a unit square? (ii) When you say "homotopic", do you actually mean "homotopic maps", or do you mean "homotopy equivalent spaces"?
Apr
6
comment Groups - abstract algebra
Also, this statement is false: "$G$ is cyclic only if the order of $G$ is a prime". A counterexample to that is the cyclic group $\mathbb{Z}/4\mathbb{Z}$ of order $4$.
Apr
6
comment Groups - abstract algebra
I strongly suspect that instead of "associative" you want to say "commutative" (equivalently, "abelian"). Your statement isn't true for non-abelian groups anyway, as witnessed by the symmetric group $S_3$ which has order $2\cdot3$, has elements of orders $2$ and $3$, but is not cyclic.
Apr
6
comment Proving that if $E, F$ are equivalence relation on $A$ and $E \subseteq F$, then there is a surjection from $A\setminus E$ to $A\setminus F$
It looks like in the second paragraph you would want to talk about individual elements from $A$, not pairs of them.
Apr
6
awarded  Generalist
Apr
5
comment Are these subgroups of G only subgroups if G is abelian?
You did implicitly use that $G$ is abelian. For instance, how did you get from $x_1^2 x_2^2 \in H$ to $(x_1 x_2)^2 \in H$?
Apr
5
comment If $a,b,c(a,b,c\in\mathbb{R} )$ satisfy $b^2-4ac<0$ then equation $f(x)=0$ has complex root
If you want a proof that doesn't use the fundamental theorem of algebra, say so. Otherwise people will not try to do it.
Apr
5
revised How to find the multiplication of $pq \times abc$ such that the result is producing the same digits from the original problem?
added 96 characters in body
Apr
5
revised If $a,b,c(a,b,c\in\mathbb{R} )$ satisfy $b^2-4ac<0$ then equation $f(x)=0$ has complex root
added 24 characters in body
Apr
5
answered If $a,b,c(a,b,c\in\mathbb{R} )$ satisfy $b^2-4ac<0$ then equation $f(x)=0$ has complex root
Apr
5
answered How to find the multiplication of $pq \times abc$ such that the result is producing the same digits from the original problem?
Apr
5
revised Irreducibility of $X^n-a$
edited body
Apr
5
answered Irreducibility of $X^n-a$
Apr
3
comment Difficult volume computation inside an ellipsoid and above a plane
As a simple check of this answer, take some point on the original plane. For instance, $A = (x_0, y_0, z_0) = (0,0,b)$. Then find its image under $f$: $f(A) = f(x_0, y_0,z_0) = (0,0,b/c)$. Then check that $f(A)$ satisfies the equation that we have found: $(b/c)c = b - 0 \cdot b$. The equality is correct. You can also try some other points to be sure. For instance, take $B = (0,b,0)$. Clearly, $B \in P$. Now take $f(B) = (0,1,0)$. Does it belong to the new plane that we found? Yes, because $0 \cdot c = b - 1 \cdot b$ is true.
Apr
3
comment Difficult volume computation inside an ellipsoid and above a plane
@josh It is quite simple. The transformation $f: \mathbb{R}^3 \to \mathbb{R}^3$ that sends the original ellipsoid into a unit sphere is $f(x,y,z) = (x/a, y/b, z/c)$. The original plane is $P = \{(x,y,z) \mid z = b - y \}$. To find the transformed plane $f(P)$, note that $(x,y,z) \in f(P) \Leftrightarrow f^{-1}(x,y,z) \in P \Leftrightarrow (xa, yb, zc) \in P \Leftrightarrow zc = b - yb$. So the equation of the new plane is $zc = b - yb$.
Apr
2
comment What about the index of this subgroup?
This more or less answers your question.
Apr
2
comment Why is this the method to getting transpositions from disjoint cycles?
@Frumpy then what is the trouble? Just verify that both expressions are the same permutation, by definition of right-to-left multiplication.
Apr
1
revised Show all the harmonic functions over $\mathbb{R}^N\setminus\{0\}$ such that $u(x)=f(|x|)$.
edited body