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seen Dec 14 at 18:42

Apr
27
comment In a group of 26 people, is it possible for each person to shake hands with exactly 3 other people?
And it's also easy to prove that for an odd number of people, it's not possible for everyone to shake hands with exactly three other people (because the total number of shaken hands would have to be odd, but in each handshake two hands get shaken), therefore that condition is both necessary and sufficient.
Apr
26
comment How can I write the numbers 5 and 7 as some sequence of operations on three 9s?
@Awesome: Care to explain? (For me, it just gives an overview of the windows on the current desktop ...)
Apr
22
comment Is it possible that “A counter-example exists but it cannot be found”
I forgot the @TrevorWilson in the previous comment.
Apr
22
comment Average Line of a Set of Lines
I'd say that completely depends on what you need the average line for (that is, what you want the average line to tell you), and/or where the original lines come from (indeed, this may decide whether the concept of an average line is actually meaningful in your context).
Apr
22
comment Prove that if $G$ is a group and $H$ is a subgroup of $G$ generated by all elements of order $N$ in $G$, then $H$ is a normal subgroup of $G$.
But $H$ is generated by the elements of order $N$. So for each $h\in H$, $h=h_1h_2h_3\dots$ where each $h_i$ is of order $N$. And $ghg^{-1} = (gh_1g^{-1}) (gh_2g^{-1}) (gh_3g^{-1}) \dots$, with each factor being of order $N$.
Apr
22
comment Prove that if $G$ is a group and $H$ is a subgroup of $G$ generated by all elements of order $N$ in $G$, then $H$ is a normal subgroup of $G$.
You're welcome.
Apr
22
comment Prove that if $G$ is a group and $H$ is a subgroup of $G$ generated by all elements of order $N$ in $G$, then $H$ is a normal subgroup of $G$.
And if $h$ is an element of order $N$, then this is ...
Apr
22
comment How to prove that the velocity field $u$ is always $\nabla\cdot u=o$ except at the origin?
What about simply calculating $\nabla\cdot \mathbf u$?
Apr
22
comment Prove that if $G$ is a group and $H$ is a subgroup of $G$ generated by all elements of order $N$ in $G$, then $H$ is a normal subgroup of $G$.
What is $(ghg^{-1})^N$?
Apr
22
comment Proof that $e^x$ is the eigenvector or the derivative operator
"You can (illustratively and not very well defined) understand a function as an infinitely dense vectors with values at $y(x)$. But then it makes no sense to talk about vectors at all." Of course it does, and it is actually very well defined. It's probably not very useful, and you'll certainly not be able to write down a basis, but it is well defined (indeed, it's exactly the definition of a function!) and a valid vector space. The only problematic term in your description is "infinitely dense" because that requires a topology; I don't know if a reasonable one can be defined in that case.
Apr
22
comment 2.71828. And then another 1828.
Here's another thing to "worry" about. Look at the first 30 digits of the decimal expansion of $\pi$. And search for the pattern "aba". What do you find? $3.\color{red}{\,141\,} 5926 \color{red}{\,535\,} 8 \color{red}{\,979\,323\,} 84 \color{red}{\,626\,} 43 \color{red}{\,383\,} 279$. What are four digits repeated once compared with that? :-)
Apr
22
awarded  Nice Question
Apr
22
comment Reflexive for belonging ($\in$)
@Sibi: Because one could imagine that there are sets which contain themselves as elements, for example $A=\{A\}$. For that set, if it existed, $A\in A$ would be true.
Apr
22
comment Reflexive for belonging ($\in$)
@Sibi: $\in$ is "is element of". The set $\{0,1\}$ has two elements, namely $0$ and $1$. That is, $x\in A$ if and only if either $x=0$ or $x=1$.
Apr
22
comment Reflexive for belonging ($\in$)
Actually, in the standard construction, $0$ and $1$ are sets, namely the empty set, and the set containing (only) the empty set. Indeed in that construction, $A$ is also a number, namely the number $2$.
Apr
21
comment How many different messages can be transmitted in n microseconds using three different signals…
You can also start at $0$ by noting there's exactly one message you can send in $0$ microseconds, namely the empty message (no signal). Then the recursion naturally gives you $a_2=a_1+2a_0=1+2\cdot 1=3$.
Apr
21
comment Knot theory: Braids
Did you try drawing both braids?
Apr
21
comment If $A + B = \frac{\pi}{3} (A,B>0),$ Then the minimum value of sec A + sec B is?
Hint: $B=\pi/3-A$
Apr
21
revised Plotting a region in $3D$ space
added 74 characters in body
Apr
21
revised Plotting a region in $3D$ space
Added labels and image example