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May
26
comment Bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$
@egreg: OK, I think I now fixed all problems. And it turned out that my $h$ would have been wrong if the previously claimed properties for $g$ had held, but is exactly right for the actual properties of $g$. :-) Please have a look if I overlooked anything else.
May
26
revised Bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$
Corrected errors
May
26
comment Bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$
Oops, I've just noticed another problem ... but I think that can be fixed as well. Stay tuned :-)
May
26
comment Bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$
I've just noticed one problem, namely that you need to include $y=1$ for $X=Y=\mathbb{N}$. But that is easily fixed by another intermediate mapping step. But other than that, the mapping is indeed bijective (note that I explicitly avoided the period 0/period 2 ambiguities in the ternary representation). I'll fix the "endpoint issue" and add a short bijectivity proof for $g$.
May
26
comment Bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$
@egreg: If it's impossible, where's the error in my post?
May
26
answered Bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$
May
26
comment Bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$
I'd guess he means $(X,Y)$ is the pair of the sets $X$ and $Y$.
May
26
comment Find integer in the form: $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$
When $\gcd(a+b,b+c,c+a)=\lambda>1$, then $x:=(a+b)/\lambda$, $y:=(b+c)/\lambda$, $z:=(c+a)/\lambda$ are integers. Thus $x + y - z = 2b/\lambda$ also is an integer, same for $a$ and $c$. If we write $a = \lambda a'/2$ etc, you can see that all $\lambda/2$ cancel out, and we have a new set of integers $(a',b',c')$ which gives the same number $I$, with $\gcd(a'+b',b'+c',c'+a')=2$. So it is enough to consider numbers where this gcd is 2.
May
26
comment How to prove $G_2$ is a normal subgroup of $G_1\times G_2$?
I've taken the liberty to make the image more readable.
May
26
revised How to prove $G_2$ is a normal subgroup of $G_1\times G_2$?
I've taken the liberty to make the inage more readable
May
26
accepted Does there exist a prime number decomposition for ordinal numbers?
May
26
comment Are there infinitely many “super-palindromes”?
Actually the question becomes whether products of a known super-palindrome with a prime larger than the largest prime factor of that super-palindrome are palindromes.
May
26
comment Quantum Information: Tensor with Outer Product
@user12290: Sorry, I didn't see your comment until now (as you might have noticed on my activity, I've not been here for quite some time). The point is that for tensor products, $(A\otimes B)(C \otimes D) = (AC) \otimes (BD)$. Thus we actually get $\left<i\middle|0\right>\left<0\middle|k\right> \otimes \left<i\middle|I\middle|l\right>$. But both factors are just complex numbers, therefore the tensor product reduces to the ordinary product.
May
26
answered What is $-i$ exactly?
Feb
19
awarded  Nice Question
Nov
20
awarded  Popular Question
Oct
15
awarded  Nice Question
Oct
6
comment What are imaginary numbers?
@MakotoKato: That depends very much on what you mean with "the Schrödinger equation". If you mean the equation exactly as normally formulated, then yes, because there's an explicit i in it. But that doesn't tell us much because nature doesn't care about which of many equivalent ways to formulate a theory we use.
Oct
5
answered How to solve this conditional probability problem?
Oct
5
comment An element within a set within a set from a power set
@alexthebake: Exactly. That's what I wrote in the second paragraph and proved in the third.