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Self-studying Gelfand pre-calculus (Functions and graphs, pg.92). Are there any more ways to prove that there is an integral point at a distance of less than $1/1000$ from the straight line $ y=\sqrt{3}x $ given the value of $\sqrt{ 3}$ is provided.(Wanted to understand elegant ways to prove this)

One possible approach: Assuming $\sqrt3 = 1.73205080757$, for a distance of $1/10000 < 1/1000$ (given distance in the question) from the line, suppose we calculate a point $(x,y)$ as $(y/x) = \sqrt3 - (1/10000) =17320/10000$ (approx.) which is the integral point $(10000,17320)$, then the value of $17320/ \sqrt3 = 9999.70666236$. On considering the value $9999.707$ (nearby round off in the x-axis, since $9999.70666236 * \sqrt3 - 17320 = 0$ ), $(\sqrt3*9999.707- 17320)$ turns out to be $0.00058480215$ which is less than a distance of $(1/1000)$. Hence there exists such integral points like $(10000,17320)$.

studyingsicp
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  • I take it the origin doesn't count? – Mike Jan 28 '24 at 09:17
  • Yes @Mike the required point cannot lie on this straight line. – studyingsicp Jan 28 '24 at 09:18
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    Hint: Take a sufficiency good rational approximation $\frac x y$ of $\sqrt 3$. – Gary Jan 28 '24 at 09:23
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    ... which can be done using the continued fraction representation of $\sqrt{3}$ that you will find for example here with the "greek ladder" method : https://maa.org/press/periodicals/convergence/a-disquisition-on-the-square-root-of-three-the-classical-greek-ladder – Jean Marie Jan 28 '24 at 09:27
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    Sorry, it should be $\frac y x$. – Gary Jan 28 '24 at 10:06
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    @JeanMarie Or you could take its decimal expansion and truncate it, which needs lees theory to be developed in order to prove such rational numbers exists. – jjagmath Jan 28 '24 at 10:16
  • @Gary, Doesn't it take trial and error to find the coordinates if we are to calculate their ratio $y/x$ – studyingsicp Jan 28 '24 at 10:27
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    Quick beginner guide for asking a well-received question + please avoid "no clue" questions. – Anne Bauval Jan 28 '24 at 10:32
  • @studyingsicp Not really, you look at your fraction and the numerator will be $y$ and the denomination will be $x$. – Gary Jan 28 '24 at 12:11
  • @Gary suppose we calculate a point (x,y) as (y/x) = sqrt(3) - (1/10000) =17320/10000 (approx.) which is an integral point (10000,17320), then the value of '(sqrt(3)*9999.707- 17320' turns out to be 0.00058480215 which is indeed less than (1/1000). Now even though we did manage to find an integral point, is this the way to prove the statement. – studyingsicp Jan 28 '24 at 20:41
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    This is a possible way provided you are allowed to know the numerical value of $\sqrt 3$. Note that $17320/10000$ is just $1732/1000$ or even $433/250$. Another method is to invoke the fact that the rationals are dense among the reals. Why did you use $9999.707$? It isn't an integer. – Gary Jan 28 '24 at 22:22
  • Assuming sqrt(3) = 1.73205080757, for the distance 1/10000 < 1/1000 (provided distance in the question) suppose we calculate a point (x,y) as (y/x) = sqrt(3) - (1/10000) =17320/10000 (approx.) which is the integral point (10000,17320), then the value of 17320/ sqrt(3) = 9999.70666236. On considering the approximation 9999.707 (next round off in the x-axis, since 9999.70666236 * sqrt(3) - 17320 = 0 ), '(sqrt(3)*9999.707- 17320' turns out to be 0.00058480215 which is less than (1/1000). – studyingsicp Jan 28 '24 at 23:11

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