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I am interested in proving the following theorem. Let $q$ be a terminating rational number. Then there exist integers $m$ and $n$ such that the decimal expansion of $\sqrt{m}+n$ begins with $q$. To give an example given a rational number $0.123$ the decimal expansion of $\sqrt{17}-4$ begins with $0.123$.

I have managed to prove this statement under the assumption that $\sqrt{2}$ is a normal number. Can this theorem be proven unconditionally?

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    $\begingroup$ Whether $\sqrt{2}$ is normal , cannot be decided with the current known tools. But that every rational number can be approximated in the desired way , should relatively easy be provable. $\endgroup$
    – Peter
    Commented Sep 18, 2023 at 17:10
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    $\begingroup$ I think you want to assume that $q$ is a terminating decimal (non-repeating still leaves open the possibility that the decimal is infinite). This is indeed provable: we're aiming for a window of size $10^{-k}$ where $k$ is the length of $q$, but the difference between $\sqrt m$ and $\sqrt{m+1}$ becomes less than $10^{-k}$ when $m$ is large enough. $\endgroup$
    – Greg Martin
    Commented Sep 18, 2023 at 17:15
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    $\begingroup$ The author only wants an approximation that begins with a given finite digit string. And the $\bar 0$ ambiguity is no problem here. In this context it is clear that we have only to deal with finite digitstrings, so terminating number (being written without the artificial $\bar 0$) $\endgroup$
    – Peter
    Commented Sep 18, 2023 at 17:18
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    $\begingroup$ It's silly to phrase questions like this in terms of decimal expansion. Much more to the point to directly ask about $\varepsilon$-close approximation. "Terminating" is irrelevant to the maths of interest here. $\endgroup$
    – leftaroundabout
    Commented Sep 19, 2023 at 8:35
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    $\begingroup$ The answers are all fine (+1). I just want to remark that I think the claim follows already from the density of the numbers of the form $$m\sqrt2+n=\sqrt{2m^2}+n.$$ See this oldie. Here $m,n$ obviously range over the integers. $\endgroup$
    – Jyrki Lahtonen
    Commented Sep 19, 2023 at 14:21

3 Answers 3

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Yes. Let $ \phi(x)$ denote the fractional part of $x.$ Since $\sqrt{2}$ is irrational, $ \left\{ \phi\left(n\sqrt{2}\right) : n\in\mathbb{N} \right\} $ is a dense subset of $[0,1],\ $ i.e. $ \left\{ \phi\left(\sqrt{2n^2}\right) : n\in\mathbb{N} \right\} $ is a dense subset of $[0,1].\ $

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    $\begingroup$ So how should I understand the fractional part of the expression $(\{n\sqrt2\}\colon n\in\mathbb{N})$...? $\endgroup$
    – leftaroundabout
    Commented Sep 19, 2023 at 8:30
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    $\begingroup$ @leftaroundabout I can see how the notation might be confusing. $ \left\{ \left\{n\sqrt{2}\right\} : n\in\mathbb{N} \right\} $ is the set of fractional parts of numbers of the form $n\sqrt{2}\ $ for positive integers $n.$ So, the fractional part of $\sqrt{2}\ $ is one member of the set, the fractional part of $2\sqrt{2}\ $ is another member of the set, the fractional part of $3\sqrt{2},\ $ is another member of the set, etc. So the set is: $\{\ 0.4142\ldots,\ 0.8284\ldots,\ 0.2426\ldots,\ \ldots\ \}.$ $\endgroup$
    – Adam Rubinson
    Commented Sep 19, 2023 at 8:38
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    $\begingroup$ It would be a lot less confusing if you hadn't chosen $\{\cdot\}$, of all notations. What's wrong with $\operatorname{fr}(x)$ or $\phi(x)$ or whatever else that doesn't look like a set bracket? $\endgroup$
    – leftaroundabout
    Commented Sep 19, 2023 at 8:42
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    $\begingroup$ $\{\cdot\}$ is the standard notation for fractional part of $x.$ Apparently at least $7$ other people did not find my answer confusing. However, I do actually agree with you that it might be confusing especially for those unfamiliar with the standard notation for fractional part, and I have edited my answer accordingly. $\endgroup$
    – Adam Rubinson
    Commented Sep 19, 2023 at 9:11
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Note that $(\sqrt{m} - \sqrt{m-1}) \downarrow 0$, so for $\varepsilon > 0$, there exists $M(\varepsilon)$ such that for every $ m \ge M$,

$$(\sqrt{m} - \sqrt{m-1}) < \varepsilon$$

Suppose that the rational number is $q = 0.a_1a_2...a_l$ with $a_i \in \{0,...,9\}$. Then, let $\varepsilon < 10^{-(l+1)}$ and WLOG letting $M(\varepsilon) = k^2$ be a square for convenience, consider $[\sqrt{m} ]$ for $n = k^2, ..., (k+1)^2-1$, where $[x]$ is taken to denote the fractional part of $x \in \mathbb{R}$. Further WLOG let $M$ be sufficiently large such that $[\sqrt{(k+1)^2-1}] > q+ \varepsilon$.

Then, $([\sqrt{m}])$ is an increasing sequence of numbers in $[0,1]$ which exceeds $q+\varepsilon$, but with the consecutive difference strictly upper-bounded by $\varepsilon$. Hence, we must have an element of the sequence in $[q, q+ \varepsilon]$, which will therefore have the same first $l$ digits as $q$ in its decimal expansion.

Thus, $\sqrt{m}$ as above and $n = -\lfloor \sqrt{m} \rfloor$ is sufficient.

Not as clever as Adam's answer, but to its credit, this approach is constructive.

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    $\begingroup$ Adam's answer essentially says 'here is a link to a theorem that says it's true' whereas you actually prove the theorem. $\endgroup$
    – quarague
    Commented Sep 19, 2023 at 7:24
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    $\begingroup$ Your answer is just as nice/"clever" in my opinion. $\endgroup$
    – Adam Rubinson
    Commented Sep 19, 2023 at 16:41
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Start with a quadratic surd that is less than $1$, such as $\sqrt2-1$. Raise it to a high power so as to get a very small number guaranteed to have the form $a-b\sqrt2$. Say we use $21$ as the exponent getting

$(\sqrt2-1)^{21}=-54608393+38613965\sqrt2\approx9.1561×10^{-9}$

Now multiply by a whole number that gets you close to the decimal you want to approximate, for instance

$0.123/(\sqrt2-1)^{21}\approx13433665$

$(\sqrt2-1)^{21}×13433665=-685229663750345+422004682131725\sqrt2\approx0.123000003$

The lower you dare to go with the power of the less than unit quadratic surd, the better you can approximate a given decimal.

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    $\begingroup$ I think I have spotted an issue with your method. If your method spits out the approximation $a-b\sqrt{2}$ where $b$ is positive, it may not be possible to express it as $\sqrt{m}+n$ for some integers $m$ and $n$. Or am I missing something? $\endgroup$
    – Adam
    Commented Sep 19, 2023 at 9:04
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    $\begingroup$ $m=2b^2$, of course. $\endgroup$
    – Oscar Lanzi
    Commented Sep 19, 2023 at 9:32
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    $\begingroup$ I am probably being stupid but isn't $\sqrt{2b^2} = b\sqrt{2}$ rather than $-b\sqrt{2}$? $\endgroup$
    – Adam
    Commented Sep 19, 2023 at 9:41
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    $\begingroup$ Failed to note that the positive square root was asked. Will need to try again $\endgroup$
    – Oscar Lanzi
    Commented Sep 19, 2023 at 9:44
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    $\begingroup$ @OscarLanzi can't you just replace $20$ by $21$ (or any odd exponent)? $\endgroup$
    – Especially Lime
    Commented Sep 19, 2023 at 13:37

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