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Find four rational numbers $~\frac{p}{q}~$ with $~\left|\sqrt 2 - \frac{p}{q}\right| \le \frac{1}{q^2}~$

Is there a way to find such numbers?

The only one I can think of is $~p = 1, ~q = 1~$ and I thought maybe if I could convert $~\sqrt 2~$ to it's nearest rational that would help but it didn't.

nmasanta
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Gooby
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2 Answers2

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If $p,q\in \Bbb N$ then $$|p^2-2q^2|=1\implies$$ $$\implies |\frac {p}{q}-\sqrt 2\,|\cdot |\frac {p}{q}+\sqrt 2\,|=|\frac {p^2}{q^2}-2|=\frac {|p^2-2q^2|}{q^2}=\frac {1}{q^2}\implies$$ $$\implies |\frac {p}{q}-\sqrt 2\,|=\frac {1}{q^2}\cdot \frac {1}{|\frac {p}{q}+\sqrt 2\,|}\le\frac {1}{q^2}\frac {1}{\sqrt 2}<\frac {1}{q^2}.$$

E.g. $(p,q)\in \{(1,1),(3,2),(7,5),(17,12)\}.$

DanielWainfleet
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  • There is a theorem that if $x$ is an irrational real then there are infinitely many pairs $(p,q)$ of integers satisfying $|x-p/q|<1/(q^2\sqrt 5,)$. – DanielWainfleet Sep 10 '19 at 17:09
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HINT:

Calculate some values $q\cdot \sqrt{2}$ and choose those close to an integer. For instance, the first $100$ multiples with $10$ decimals are $$ 1.414213562, 2.828427125, 4.242640687, 5.656854249, 7.071067812, 8.485281374, 9.899494937, 11.31370850, 12.72792206, 14.14213562, 15.55634919, 16.97056275, 18.38477631, 19.79898987, 21.21320344, 22.62741700, 24.04163056, 25.45584412, 26.87005769, 28.28427125, 29.69848481, 31.11269837, 32.52691193, 33.94112550, 35.35533906, 36.76955262, 38.18376618, 39.59797975, 41.01219331, 42.42640687, 43.84062043, 45.25483400, 46.66904756, 48.08326112, 49.49747468, 50.91168825, 52.32590181, 53.74011537, 55.15432893, 56.56854249, 57.98275606, 59.39696962, 60.81118318, 62.22539674, 63.63961031, 65.05382387, 66.46803743, 67.88225099, 69.29646456, 70.71067812, 72.12489168, 73.53910524, 74.95331881, 76.36753237, 77.78174593, 79.19595949, 80.61017306, 82.02438662, 83.43860018, 84.85281374, 86.26702730, 87.68124087, 89.09545443, 90.50966799, 91.92388155, 93.33809512, 94.75230868, 96.16652224, 97.58073580, 98.99494937, 100.4091629, 101.8233765, 103.2375901, 104.6518036, 106.0660172, 107.4802307, 108.8944443, 110.3086579, 111.7228714, 113.1370850, 114.5512986, 115.9655121, 117.3797257, 118.7939392, 120.2081528, 121.6223664, 123.0365799, 124.4507935, 125.8650071, 127.2792206, 128.6934342, 130.1076477, 131.5218613, 132.9360749, 134.3502884, 135.7645020, 137.1787156, 138.5929291, 140.0071427, 141.4213562$$

(Dirichlet's theorem states that among the first $N$ integers of any number $\alpha$ there exists at least one $q\alpha$ that is close to an integer by less than $\frac{1}{N+1}$, so by less than $\frac{1}{q}$ )

So one can take $q=7$ or $99$, or others.

orangeskid
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