Timeline for Disproving surjectivity of $f : \Bbb Z \times \Bbb Z \rightarrow \Bbb Z$, $f(u,v) = 3u + 6v$
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| when toggle format | what | by | license | comment | |
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| May 28, 2024 at 10:09 | audit | Reopen votes | |||
| May 28, 2024 at 10:10 | |||||
| May 14, 2024 at 13:23 | audit | Close votes | |||
| May 14, 2024 at 16:38 | |||||
| May 5, 2024 at 1:44 | comment | added | Arturo Magidin | @Trebor whether it is prefix or suffix is irrelevant here. You should not both say "Let $y$" and "there exists $y$". And you cannot both have $y=u+2=$ with $u$ and $v$ integers and $y\in\mathbb{Z}\times\mathbb{Z}$. | |
| May 5, 2024 at 1:02 | comment | added | Trebor | @ArturoMagidin There are certain conventions where $\exists$ is a postfix. That's just an arbitrary convention though. | |
| May 1, 2024 at 19:18 | history | edited | Viktor Vaughn | CC BY-SA 4.0 |
edited tags; edited title
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| May 1, 2024 at 14:33 | answer | added | Jo Wehler | timeline score: 1 | |
| May 1, 2024 at 12:52 | answer | added | egreg | timeline score: 1 | |
| May 1, 2024 at 9:20 | history | became hot network question | |||
| May 1, 2024 at 6:56 | answer | added | ultralegend5385 | timeline score: 7 | |
| May 1, 2024 at 2:08 | comment | added | Rrasco88 | Thank you @ArturoMagidin that was very helpful thank you for the constructive criticism! | |
| May 1, 2024 at 1:54 | comment | added | copper.hat | I would say $3 \mid f(u,v)$ for all $u,v \in \mathbb{Z}$. Since $3\not\mid 2$, $f$ cannot be surjective. | |
| May 1, 2024 at 1:53 | comment | added | copper.hat | There is some weirdness above starting with $\exists$. | |
| May 1, 2024 at 1:51 | comment | added | Karl | Dwelling on 2 obscures the main point too, which is that $f(u,v)$ is always a multiple of 3, so all non-multiples of 3 (i.e. 1,2,4,5,7,8,10,11,...) are missing from the range of $f$. | |
| May 1, 2024 at 1:25 | answer | added | csch2 | timeline score: 14 | |
| May 1, 2024 at 1:24 | comment | added | Arturo Magidin | I get what you tried to do, but what you wrote doesn't work. You are trying to show that $2$ is not in the image. You do so by noting that $f(u,v)$ is a multiple of $3$, which means you would need some pair of integers $(u,v)$ such that $u+2v=\frac{2}{3}$. Which is impossible. But that is, unfortunately, not what you actually wrote. For that matter, it is much easier to note that $f(x,y)$ is always an integer multiple of $3$, so it cannot be equal to $2$, which is not a multiple of $3$. | |
| May 1, 2024 at 1:22 | comment | added | Arturo Magidin | "Let $y=u+2v$ $\exists y\in \mathbb{Z}\times\mathbb{Z}$" is nonsense. You can let $y=u+2v$, but then it is an integer, not an element of $\mathbb{Z}\times\mathbb{Z}$. And having the "$\exists y\in\mathbb{Z}\times\mathbb{Z}$" tacked on at the end gives a formula that is not a well-formed formula. "This is a contradiction because $\frac{2}{3}\notin\mathbb{Z}\times\mathbb{Z}$"... yeah, well, that contradiction is to your false assertion that the integer $y$ was in $\mathbb{Z}\times\mathbb{Z}$, so the only contradiction you found was one you introduced. I would have marked this wrong as well. | |
| May 1, 2024 at 1:22 | comment | added | Randall | I think your argument is good, but a little roundabout and not written correctly. You can just note that such an integer $u+2v$ cannot exist. No need to call it $y$. | |
| S May 1, 2024 at 1:17 | review | First questions | |||
| May 1, 2024 at 1:21 | |||||
| S May 1, 2024 at 1:17 | history | asked | Rrasco88 | CC BY-SA 4.0 |