Timeline for A translation of the Cantor set contained in the irrationals
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 17, 2017 at 21:42 | history | edited | Anthony Quas | CC BY-SA 3.0 |
added 62 characters in body
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| Oct 17, 2017 at 15:47 | history | edited | Did | CC BY-SA 3.0 |
added 1 character in body
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| Oct 17, 2017 at 14:52 | comment | added | user21820 | @Did,Anthony: If I didn't make a mistake myself, there is another simple way; see my answer! =) | |
| Oct 17, 2017 at 2:21 | history | edited | Anthony Quas | CC BY-SA 3.0 |
added 114 characters in body
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| Oct 16, 2017 at 21:12 | comment | added | Did | Anthony: Add a mention to that effect at the beginning of the post? | |
| Oct 16, 2017 at 17:47 | comment | added | Anthony Quas | @KajetanJaniak: Thanks for the comment. It seems I was too hasty in my last but 2 paragraph. I feel hopeful there is some way to modify the example and make it work. | |
| S Jan 19, 2016 at 18:44 | comment | added | Kajetan Janiak | More precisely, for each subsequent pair of digits we use the following two formulas: $$ 0.(0)^n00 + 0.(0)^n20 = 0.(0)^n20 $$ and $$ 0.(0)^n11 + 0.(0)^n02 = 0.(0)^n20. $$ | |
| S Jan 19, 2016 at 18:44 | comment | added | Kajetan Janiak | Anthony, I have some doubts about your explicit example of $t$. Let's write it like that: $$ t = 0.1(00)^{2^0}(11)^{2^1}(00)^{2^2}(11)^{2^3}\ldots (00)^{2^{2n}}(11)^{2^{2n+1}}\ldots $$ Then take $x \in C$: $$ x = 0.0(20)^{2^0}(02)^{2^1}(20)^{2^2}(02)^{2^3}\ldots (20)^{2^{2n}}(02)^{2^{2n+1}}\ldots $$ As a result we get $$ x + t = 0.1(20)^\infty \in \mathbb Q, $$ because within each block of length $2^n$ we can calculate the sum separately. (continued) | |
| Feb 23, 2014 at 23:05 | comment | added | Andrés E. Caicedo | Thank you. This is the kind of answer I was expecting, and it is nice that the calculations end up not being as messy as I feared. I myself lost patience trying to keep track of what you call the 1-blocks of $x+t$ (for the $t$ I used myself, which I see is different from yours) when I first thought about this. | |
| Feb 22, 2014 at 21:04 | history | edited | Anthony Quas | CC BY-SA 3.0 |
fix appearance
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| Feb 22, 2014 at 20:58 | history | answered | Anthony Quas | CC BY-SA 3.0 |