1,604 reputation
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bio website kmlinux.fjfi.cvut.cz/…
location Prague, Czech Republic
age 27
visits member for 2 years, 4 months
seen 32 mins ago

My former display name: tohecz

Math PhD student at Czech Technical University in Prague and at LIAFA in Paris. At the same time, I'm a typesetter (and partly the copy editor) of one scientific journal (done in LaTeX, of course).

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Dec
10
answered $1/i=i$. I must be wrong but why?
Nov
23
comment on two decompositions of a finite group $G$
+1 very nice, it just shows that people should be carefull about what they claim! However, it IMHO means that the OP likely wanted $AB=AC$ instead of $A\times B\cong A\times C$, which is what the other answer proves, and where your counterexample fails.
Nov
22
revised Cool simple solution to: $-A^2$ is not the identity matrix
markup corrected
Nov
22
suggested approved edit on Cool simple solution to: $-A^2$ is not the identity matrix
Nov
22
answered Cool simple solution to: $-A^2$ is not the identity matrix
Nov
22
awarded  Civic Duty
Nov
22
comment Notation: Why do we learn to write the higher powers in an equation first?
@AsafKaragila I don't call that a series. That is a set of functions (most of which are not possible to express as series)
Nov
22
comment Notation: Why do we learn to write the higher powers in an equation first?
@AsafKaragila The convention for series is opposite because you can't write them in the same direction as polynomials.
Nov
22
comment on two decompositions of a finite group $G$
Do I miss something? People really use $b^\gamma$ instead of $\gamma(b)$, $\gamma b$ or $(b\gamma)$?
Nov
22
comment Integral value of z
Well, your solution works for the current version of the question (which is hopefully the one OP wanted to ask). You just might want to explain that $x,y$ need to be integers (which you explained in the comments I think). Sorry, I seem to be getting lost in it myself :-/
Nov
22
comment Integral value of z
Well, you just got very close to getting a downvote from me. It's true that it's the same, however after dividing by $z$ you get $x^2/z+y^2/z=1$ or (after substitution) $zx^2+zy^2=1$. Now, you have to say that $z$ divides $1$ therefore $z=\pm1$ and we can ignore it. I don't think it's a good idea to omit such details, they seem trivial to us maybe, but probably are not trivial to OP. Still, they are quite crucial in understanding the ideas behind.
Nov
22
comment Integral value of z
However, he asks for $x,y\in\mathbb Q$ and $x^2+y^2\in\mathbb Z$. You might want to explain in detail that this is the same as $x,y\in\mathbb Z$ and $x^2+y^2\in\mathbb Z$.
Nov
22
comment Integral value of z
Dietrich, if you checked the edit history, you would see that ulead86 made an invalid edit, changing rationals to $\mathbb R$.
Nov
22
revised Integral value of z
the previous edit was wrong! Exchanged rationals and reals!
Nov
22
suggested approved edit on Integral value of z
Nov
19
awarded  Informed
Nov
19
awarded  Nice Answer
Nov
18
comment SageMath: Embed all roots of a polynomial
You can identify the roots in another way. You may do qRoot=L.polynomial().complex_roots()[0] and redefine L as L.<gg> = NumberField(q, embedding=qRoot). Then f.roots(L) give 3 roots, and when you .N() them, you see that they are gamma, gammabar and the real root in some order. They will permute depending on which qRoot you choose.
Nov
18
answered Show that $x \mapsto \frac{f(x)}{x}$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0
Nov
18
answered Is it true that $E[X^2]-E[Y^2] = 0?$