Timeline for Rational points on an analytic curve
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 5 at 6:05 | comment | added | DavideLombardo | Right, sorry, I was thinking of a stronger property (that for every rational $x$ the corresponding $y$ are also rational). | |
| Aug 4 at 20:02 | comment | added | zxx | If you notice that rational points on the curve $x^2 + y^2 = 1$ are dense, then you can see that rational points on the curve $x^2 + (f(y))^2 = 1$ are also dense. | |
| Aug 4 at 19:41 | vote | accept | user534345 | ||
| Aug 4 at 18:32 | comment | added | DavideLombardo | I'm not sure if the previous suggestion works, but a slight variant should: let $A=[-1,1] \cap \mathbb{Q}$ and $B=\{ \pm \sqrt{1-r^2} : r \in A\}$. Using the same method (or applying an older result of Franklin, "Analytic transformations of everywhere dense point sets") it should be possible to construct $g : [-1,1] \to [-1,1]$ that gives a bijection from $A$ to $B$ (both are countable and dense, which are essentially the only properties used). For this $g$, $x^2 + g(y)^2 = 1$ should work (to ensure smoothness, we should also impose $g'(-1) \neq 0, g'(1) \neq 0$). | |
| Aug 4 at 18:13 | comment | added | zxx | Is the curve $x^2+(f(y))^2=1$ an example? $f$ is the function described in the answer. | |
| Aug 4 at 16:03 | comment | added | Dmitrii Korshunov | can these examples be made into a closed curve in $\mathbb R^2$? | |
| Aug 4 at 15:32 | comment | added | zxx | Interesting question and accurate answer! Some thoughts after reading: The construction mainly uses approximation by polynomials. Since the rational points are countable, we can find a polynomial that vanishes at the first n points, while maintaining good control over its growth. | |
| Aug 4 at 14:22 | history | edited | DavideLombardo | CC BY-SA 4.0 |
added 9 characters in body
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| Aug 4 at 14:11 | history | answered | DavideLombardo | CC BY-SA 4.0 |