did you try to do it using a wall to get up? that's what I tried at first but I had the same problem
you just gotta kick yourself up from the ground and strike the right balance, your head won't ache so much since you'll probably fall before the blood rushes
Imagine the graph with vertices $\pm2,\pm1,\pm i$, connecting $\pm2$ with $\pm1$ (for each of the two choices of $\pm$) and having a cycle from $1$ to $i$ to $-1$ to $-i$
My proposed graph isn't $2$-pathed, since there's only one path from $-2$ to $-1$, but letting $u=-2$ and $v=2$ shows that the statement $a_i\ne b_j$ doesn't need to hold
For the circle with the arc, we have the following paths:
Clockwise along the circle, Counterclockwise along the circle, Clockwise until arc, follow the arc over, take the shortest way to the destination Counterclockwise until arc, follow the arc over, take the shortest way to the destination
Lets flatten the arc to a diameter-length bisector. This gives us more descriptive power.
If they're on the same semi-circle, there are 3, because traveling one direction around the circle will reach the destination point before reaching the base of the arc/separater
Adding a point (to which this arced cycle is an allegory for) immediately transitions from a $2$-pathed graph to a strictly more-than-$3$-pathed graph.
To connect $a$ and $b$, we can either go along the unique path on the spanning tree, or we can do modifications of that (occasionally using additional edges)
I have to prove that if a group $G$ is of even order then it contains one element of order 2
Proof 1: We know that the identity $e$ does not have any inverse (it has itself, I mean to say it cannot be paired) that is $e=e^{-1}$.Assume there is no element of order 2(i.e , for some $a\in G, a^2=e \implies a=a^{-1}$) But every other element has an inverse . So we can makes pairs for each of the elements that is the element and its inverse. All such elements add up to an even number, but since there is the identity which is till left , the total sum will be an odd number. This is a contradicti…
@TheKindCat @Axoren Subdivision does not work. Take $K_4$ with vertices labeled $1,2,3,4$, and add a vertex $x$; remove the edge $12$ and add edges $1x$ and $x2$.
Then there are $5$ paths between $1$ and $2$, as usual, but
Why does $\mathbb{Z}^{\times}_{20}\cong\mathbb{Z}^{\times}_{4}\times\mathbb{Z}^{\times}_{5}$ imply that $\mathbb{Z}^{\times}_{20}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{4}$? I know the first bit is the Chinese remainder theorem, but why does the implication follow?
@robjohn, Polya's Problems and Theorems in Analysis says, 'In mathematical induction the result to be obtained and the means available for its proof are proportional, they stand in the ratio of n + 1 to n. Hence, strengthening the statement to be proved may also be advantageous, for we strengthen at the same time the means available for its proof.' How is that ration n+1/n? I do not get this.
@Silent That is a terrible formulation, which you can probably ignore. The important thing is that strengthening the thing to be proved also strengthens the means available for proving it
(this does not always mean that one can just strengthen the statement, but it does mean that smetimes it makes things easier)
is there any name for this group: For a prime $p$ $$\Bbb{Z}(p^{\infty})=\{\overline{\frac{a}{b}}\in \Bbb{Q}/\Bbb{Z}| a,b\in \Bbb{Z}, b=p^i, i\geq 0 \}$$
where the operation is usual addition as it is for the rationals modulo integers
though, wikipedia's page on prufer also mentions at the bottom that p-adic integers can be defined as the inverse limit of the finite subgroups of the Prufer group
i don't really know pontryagin duality either, alas, aside from it somehow encompassing the usual notion of position space - reciprocal space duality in Fourier analysis
though when i use the word reciprocal space i'm really thinking in terms of applications to solid state
