# Can every terminating rational number be approximated with a square root?

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?

• 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. Sep 18 at 17:10
• 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. Sep 18 at 17:15
• 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 $\bar 0$) Sep 18 at 17:18
• 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. Sep 19 at 8:35
• 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. Sep 19 at 14:21

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].\$$

• So how should I understand the fractional part of the expression $(\{n\sqrt2\}\colon n\in\mathbb{N})$...? Sep 19 at 8:30
• @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: $\{\ 0.4142\ldots,\ 0.8284\ldots,\ 0.2426\ldots,\ \ldots\ \}.$ Sep 19 at 8:38
• 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? Sep 19 at 8:42
• $\{\cdot\}$ 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. Sep 19 at 9:11

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.

• Adam's answer essentially says 'here is a link to a theorem that says it's true' whereas you actually prove the theorem. Sep 19 at 7:24
• Your answer is just as nice/"clever" in my opinion. Sep 19 at 16:41

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.

• 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?
• $m=2b^2$, of course. Sep 19 at 9:32
• I am probably being stupid but isn't $\sqrt{2b^2} = b\sqrt{2}$ rather than $-b\sqrt{2}$?
• @OscarLanzi can't you just replace $20$ by $21$ (or any odd exponent)? Sep 19 at 13:37