v_Enhance wrote: Let $\sqrt 3 = 1.b_1b_2b_3 \dots _{(2)}$ be the binary representation of $\sqrt 3$. Prove that for any positive integer $n$, at least one of the digits $b_n$, $b_{n+1}$, $\dots$, $b_{2n}$ equals $1$. Suppose for some $n \ge 1$ we find $b_n=b_{n+1}=\dots=b_{2n}=0$. Now we observe that $$\lfloor 2^{2n}\sqrt{3} \rfloor=2^{2n}+2^{2n-1}b_1+\dots+2^{n+1}b_1$$hence $\lfloor 2^{2n}\sqrt{3} \rfloor=2^{n+1}k$ for some integer $k \ge 1$. Observe that $2^{n+1}k<2^{2n}\sqrt{3}<2^{n+1}k+1$ hence squaring gives $$2^{2n+2}k^2<3 \cdot 2^{4n}<2^{2n+2}k^2+2^{n+2}k+1.$$Now left-side shows that $k^2=3 \cdot 2^{2n+2}-r$ for some integer $r \ge 1$ (since $\sqrt{3} \not \in \mathbb{Q}$). Plug this value of $k^2$ on the right side hence $$3 \cdot 2^{4n}<3\cdot 2^{4n}+2^{n+2}(k-r \cdot 2^n)+1$$hence $k \ge r \cdot 2^n$; so $k^2 \ge r^2 \cdot 2^{2n}>3 \cdot 2^{2n-2}$ a contradiction!

My solution looks a bit too silly to be correct, so can someone check it ?

Wait Evan I think this problem was at mop.

Solved with jclash, mfang92, nukelauncher. Assume for contradiction $b_n$, $\ldots$, $b_{2n}$ are zero, and consider \[\frac pq=1.b_1\ldots b_{n-1}=1.b_1\ldots b_{2n},\]where $q=2^k$ for some $k\le n-1$. By hypothesis $0<\sqrt3-p/q<2^{-2n}$, from which \[0<q\sqrt3-p<\frac q{4^n}\le\frac1{4q}.\]Multiplying this by $0<q\sqrt3+p<2\sqrt3q$, we have \[0<3q^2-p^2<\frac{\sqrt3}2<1,\]absurd since $p$ and $q$ are integers.

Very silly problem. Suppose not, and there exists $n$ such that $b_n, b_{n + 1}, \ldots, b_{2n}$ are all $0$. Let $k$ be the integer $\overline{b_1b_2\ldots b_{n - 1}}_{(2)}$. We have \begin{align*} 2^{n - 1}\sqrt{3} &= k + 0.b_nb_{n + 1}\ldots b_{2n}\ldots_{(2)}. \end{align*}Note that all of $b_n, \ldots, b_{2n}$ are $0$. This implies that \begin{align*} k < 2^{n - 1}&\sqrt{3} < k + \frac{1}{2^{n + 2}} + \frac{1}{2^{n + 3}} + \cdots \\ k < 2^{n - 1}&\sqrt{3} < k + \frac{1}{2^{n + 1}}, \end{align*}where the second line is due to geometric series. Squaring, we find that \begin{align*} k^2 < &4^{n - 1} \cdot 3 < k^2 + \frac{k}{2^n} + \frac{1}{4^{n + 1}}. \end{align*}The left inequality implies that $k < 2^n$. Therefore, \begin{align*} \frac{k}{2^n} + \frac{1}{4^{n + 1}} &\leq \frac{2^n - 1}{2^n} + \frac{1}{4^{n + 1}} \\ &< 1. \end{align*}Therefore, \begin{align*} k^2 < 4^{n - 1} \cdot 3 < k^2 + 1, \end{align*}which is absurd as $3 \cdot 4^{n - 1}$ is an integer. This completes the proof.

Cute. Suppose otherwise. Then, for some integer $2^{n-1}<k<2^n$, the bound \[\frac{k}{2^{n-1}}<\sqrt{3}<\frac{k}{2^{n-1}}+\frac{1}{2^{2n}}\]holds. Equivalently, \[k<2^{n-1}\sqrt{3}<k+\frac{1}{2^{n+1}}\]holds. Then, we clearly have \[k^2<(2^{n-1}\sqrt{3})^2<k^2+\frac{k}{2^n}+\frac{1}{2^{2n+2}}<k^2+1.\]But this is absurd, so we are done.

My solution is different than everyone else so idk if it's correct. Assume for contradiction that for some $n$, $b_n, b_{n+1}, \ldots, b_{2n}$ are $0$. We will prove by induction, with the base case $n = 1$ being trivial. For the inductive step, let $x$ be the number formed by the digits up to $b_{2n}.$ We have $$3\cdot 2^{4n} > x^2\cdot 2^{4n} = k^2$$where $k^2$ is the biggest square less than the LHS. However, by the inductive step, we have that there exists a $k$ such that $$3\cdot 2^{4n-4} > k^2 > x'^2\cdot 2^{4n-4} = x^2\cdot 2^{4n-4}$$where $x'$ is the number formed by the digits up to $b_{2n-2}.$ Notice that $x' = x$ as the last $2$ digits of $x$ are $0$. Therefore, $$\implies 3\cdot 2^{4n} > (4k)^2 > x^2\cdot 2^{4n},$$so $x^2\cdot 2^{4n}$ is not the greatest square smaller than $3\cdot 2^{4n}$, a contradiction. Therefore, there doesn't exist an $n$ such that $b_n, b_{n+1}, \dots, b_{2n}$ are $0$.

This is hilarious.

dame dame

Does this work? Suppose towards contradiction that the digits $a_n$ through $a_{2n}$ are all $0$, then for some integer $m$ between $2^{n-1}$ and $2^n$ we must have $0 < \sqrt3 - \frac{m}{2^{n-1}} < \frac{1}{2^{2n}}$ $\implies 0 < 2^n\sqrt3 - 2m < \frac{1}{2^n}$ We will now multiply through by $2^n\sqrt3 + 2m$, upper bounding it weakly with $2^{n+2}$ on the RHS: $\implies 0 < 3\cdot 4^n - 4m^2 < 4$ $\implies 0 < 3\cdot 4^{n-1} - m^2 < 1$ But this is absurd as the middle expression is an integer, so we are done.

Relabel the $b_i$ sequence as $a_i$ for convenience, since my $b$ key is scuffed. FTSoC there is some $k$ and $a_k = a_{k+1} = \ldots = a_{2k} = 0$. Then, it follows that the floor of $2^{2k}\sqrt{3}$ is a multiple of $2^{k+1}$. Hence we can write\[2^{k+1}m < 2^{2k}\sqrt{3} < 2^{k+1}m+1 \implies 2^{2k+2}m^2 < 3 \cdot 2^{4k} < 2^{2k+2}m^2 + 2^{k+2}m + 1\]for some positive integer $m$. Write $m^2 = 3 \cdot 2^{2k - 2} - r$ where $r$ is minimal. Plugging this back yields $-1 < 2^{k+2}(m - 2^kr)$ so $m \geq 2^kr \implies m^2 \geq 2^{2k}r^2 > m^2 = 3 \cdot 2^{2k - 2} - r$ which is clearly impossible, the desired contradiction.

Let $n$ be arbitrary and suppose that the desired claim does not hold, i.e. $$\sqrt{3}<\frac{\lfloor 2^{n-1}\sqrt{3}\rfloor}{2^{n-1}}+\frac{1}{2^{2n}} \implies 2^{n-1}\sqrt{3}<\lfloor 2^{n-1}\sqrt{3}\rfloor+\frac{1}{2^{n+1}} \implies 2^{2n-2}\cdot 3<\lfloor 2^{n-1}\sqrt{3}\rfloor^2+\frac{1}{2^n}\lfloor 2^{n-1}\sqrt{3}\rfloor+\frac{1}{2^{2n+2}}$$by squaring. Since $\lfloor 2^{n-1}\sqrt{3}\rfloor<2^{n-1}\sqrt{3}<2^n$ as $\sqrt{3}<2$, it follows that $$\frac{1}{2^n}\lfloor 2^{n-1}\sqrt{3}\rfloor \leq \frac{2^n-1}{2^n} \implies \lfloor 2^{n-1}\sqrt{3}\rfloor+\frac{1}{2^{2n+2}}<1,$$so since $2^{2n-2}\cdot 3$ and $\lfloor 2^{n-1}\sqrt{3}\rfloor^2$ are both integers, $2^{2n-2}\cdot 3 \leq \lfloor 2^{n-1}\sqrt{3}\rfloor^2$. This in turn implies that $2^{n-1}\sqrt{3}\leq \lfloor 2^{n-1}\sqrt{3}\rfloor$, which is absurd as $2^{n-1}\sqrt{3}$ is not an integer: contradiction. Hence we are done. $\blacksquare$

Suppose otherwise. Then, for some positive integer $n$, the digits $b_n$ through $b_{2n}$ are all 0. Denote $d$ as the contribution of the digits prior to $b_{n}$ (including the 1 at the start), and denote $s$ as the contribution of the digits starting from $b_{2n+1}.$ Consider the value of $$(d+s)^2=d^2+2ds+s^2.$$Note that $d$ in base 2 terminates at the $n-1$th place after the decimal place, so the entire contribution of $d^2$ takes place in the integer part and up to the $2n-2$th decimal place, since $d^2$ is guaranteed to terminate at or before the $2n-2$th place after the decimal. Now note that $$d\leq 2-2^{-n+1},s\leq 2^{-2n}.$$This tells us that $$2ds+s^2\leq 2^{-2n+2}-2^{-3n+2}+2^{-4n}.$$This means that $$2ds+s^2$$is strictly less than $2^{-2n+2}.$ Therefore, the entire contribution of the $2ds+s^2$ term will lie strictly beyond the $2n-2$th digit after the decimal in base 2. We have established earlier that $d^2$ only contributes up to the $2n-2$th digit. This means it is impossible or $(d+s)^2$ to equal 3, since $d^2$ is of course not equal to 3 due to irrationality, but the additional $2ds+s^2$ cannot fix this since these terms only contribute beyond the $2n-2$th digit, but $d^2$ already terminates there, so everything up to the $2n-2$th digit will be the same as it is in $d^2$.

would love to see the motivation behind writing this problem, it seems like such an arbitrary thing to look at Assume otherwise. Let the number up to digit $b_{n-1}$ be $k$, so that $k < \sqrt 3$. Then since $b_n = b_{n+1} = \dots = b_{2n} = 0$, we must have \[\sqrt3 \,-\, k < \frac{1}{2^{2n}} \implies 3 < k^2+\frac{k}{2^{2n-1}}+\frac{1}{2^{4n}} < k^2+ \frac{\sqrt{3}+1/2^{2n+1}}{2^{2n-1}} < k^2+\frac{1}{2^{2n-2}}. \] However, $k^2 < 3-\frac{1}{2^{2n-2}}$ (since the error between $k^2$ and $3$ is at least the square of the value of $b_{n-1}$) hence $k^2+\frac{1}{2^{2n-2}} \le 3$, which is a contradiction.

The only good part of this problem is the problem statement. FTSOC suppose that $\sqrt{3} = 1 + c \cdot \frac{1}{2^{k}} + \varepsilon$ for $\varepsilon < \frac{1}{2^{2k+2}}, c < 2^k$. Then squaring both sides gives that $c^2 \cdot \frac{1}{2^{2k}} + 2c \cdot \frac{1}{2^k} + \delta = 2$ for $\delta = 2\varepsilon + 2c \cdot \varepsilon \cdot \frac{1}{2^k} + \varepsilon^2$. Bounding, we get that $\delta < 4\varepsilon$. It thus follows that \[ c^2 \cdot \frac{1}{2^{2k}} + 2c \cdot \frac{1}{2^k} < 2 < c^2 \cdot \frac{1}{2^{2k}} + 2c \cdot \frac{1}{2^k} + 4 \varepsilon \]Multiplying both sides by $2^{2k}$, we get \[ c^2 + c \cdot 2^{k+1} \le 2^{2k+1} < c^2 + c \cdot 2^{k+1} + 2^{2k+2} \cdot \varepsilon. \]However, this implies that the first equality is tight, giving a contradiction as then $c = (\sqrt{3} - 1) \cdot 2^k$ is an integer.

Lemma: Define $\{x\}=x-\lfloor x\rfloor$ as the fractional part function. For any non-zero integer $x$, we have \[\left\{x\sqrt{3}\right\}>\frac1{4x}\]Proof. Suppose for contradiction we have $\left\{x\sqrt{3}\right\}<\frac1{4x}$. Note that equality cannot hold as $x\sqrt3$ is irrational. Let $y=\left\lfloor x\sqrt{3}\right\rfloor$, and $k=3x^2-y^2$. Since $y<x\sqrt{3}$, we have $k>0$. Now, \[\sqrt{y^2+k}=x\sqrt3=y+\left\{x\sqrt3\right\}<y+\frac1{4x}\]\[\Longrightarrow 0<k<\frac y{2x}+\frac1{16x^2}<\frac{\sqrt{3}}2+\frac1{16}<2,\]which forces $k=1$. So, $3x^2-y^2=1\Longrightarrow y^2\equiv2\pmod{3}$, which is impossible. We have arrived at the contradiction. $\blacksquare$ We now return to the problem. Suppose for contradiction we have $b_n=b_{n+1}=\cdots=b_{2n}=0$. Then, the integer $\left\lfloor 2^{n+i}\sqrt3\right\rfloor$ is even for each $0\leq i\leq n$. We know that for any $x$, we have $\lfloor 2x\rfloor=2\lfloor x\rfloor$ iff $\{x\}<\frac12$ and $\lfloor 2x\rfloor=2\lfloor x\rfloor+1$ iff $\{x\}\geq\frac12$. Since $\left\lfloor 2^{n+i}\sqrt3\right\rfloor$ is even, it means $\left\{ 2^{n+i-1}\sqrt3\right\} < \frac12$ for every $0\leq i\leq n$. Finally, note that if $\{x\}<\frac12$, then $\{2x\}=2\{x\}$. Hence, if $\{x\},\{2x\}<\frac1{2^m}$ then $\{x\}<\frac1{2^{m+1}}$ for any $m>0$. A simple induction gives that $\left\{2^{n-1}\sqrt3\right\}<\frac1{2^{n+1}}$. Invoking our lemma, we see that this is impossible. The proof is complete. $\blacksquare$

Solved with Eka01. $$\sqrt{3}=1.b_1b_2b_3\dots_{(2)}=1+\frac{b_1}{2}+\frac{b_2}{2^2}+\frac{b_3}{2^3}+\dots$$For the sake of contradiction, assume there exists a positive integer $n$ such that $b_n=b_{n+1}=\cdots=b_{2n}=0$. We see that $$\sqrt{3}=1+\frac{b_1}{2}+\frac{b_2}{2^2}+\dots+\frac{b_{n-1}}{2^{n-1}}+\frac{b_{2n+1}}{2^{2n+1}}+\frac{b_{2n+2}}{2^{2n+2}}+\dots$$Multiplying both sides by $2^{2n}$, we get $$2^{2n}\sqrt{3}=2^{2n}+b_1\cdot2^{2n-1}+b_2\cdot2^{2n-2}+\dots+b_{n-1}\cdot2^{n+1}+\frac{b_{2n+1}}{2}+\frac{b_{2n+2}}{2^2}+\dots$$Notice that $\sum_{i=1}^{\infty} \frac{b_{2n+i}}{2^i} \le \sum_{i=1}^{\infty} \frac{1}{2^i}=1$ with equality only when $b_{2n+1}, b_{2n+2},\dots$ are all $1$. The equality case can easily be rejected because $\sqrt{3}$ is irrational. This suggests that $$\left\lfloor 2^{2n}\sqrt{3}\right\rfloor=2^{2n}+b_1\cdot2^{2n-1}+b_2\cdot2^{2n-2}+\dots+b_{n-1}\cdot2^{n+1} \implies 2^{n+1}\mid \left\lfloor 2^{2n}\sqrt{3} \right\rfloor.$$Suppose $\left\lfloor 2^{2n}\sqrt{3} \right\rfloor = 2^{n+1}m$ for a positive integer $m$. Then $$2^{n+1}m<2^{2n}\sqrt{3}<2^{n+1}m+1 \implies 2^{n-1}\sqrt{3}-\frac{1}{2^{n+1}}<m<2^{n-1}\sqrt{3}.$$Observe that $$3\cdot2^{2n-2}-\frac{\sqrt{3}}{2}+\frac{1}{2^{2n+2}}<m^2<3\cdot2^{2n-2} \implies -1<-\frac{\sqrt{3}}{2}+\frac{1}{2^{2n+2}}<m^2-3\cdot2^{2n-2}<0,$$which is not possible as $m^2-3\cdot2^{2n-2}$ is an integer. Hence, our assumption is false, and it follows that at least one of the digits $b_n, b_{n+1}, \dots, b_{2n}$ must be $1$ for any positive integer $n$, as desired. $\blacksquare$

FTSOC, assume that there exists an integer $n$ for which $b_n, b_{n+1}, \dots, b_{2n}$ are all $0$. Let $\sqrt{3} = a+b$, where $a$ is the block of digits before the drought of $1$'s (including the $1$ left of the decimal point) and $b$ is the block of digits after the drought. So, $a \leq 2-2^{1-n}$ and $b < 2^{-2n}$. Write $3$ as $(a+b)^2 = a^2 + (2ab + b^2)$. Since $\nu_2 (a) \geq 1-n$, the value $a^2$ is at least $2^{2-2n}$ less than $3$ (or any integer for that matter). But \[ 2ab + b^2 < 2^{-2n+2} - 2^{2-3n} + 2^{-4n} < 2^{-2n+2},\]so we have a contradiction.

Assume that there was an positive integer $n$ such that $b_n, \ldots b_{2n}$ are all $0$. The multiplying $\sqrt{3}$ by $2^n$ produces a number with integral part ending with $n + 1$ zeroes. Thus: \[ \lfloor 2^{2n} \sqrt{3} \rfloor = 2^{n + 1}k \]for some $k$. Additionally, let $x = \{ 2^{2n} \sqrt{3} \}$ so that $2^{n + 1}k + x = 2^{2n}\sqrt{3}$. Squaring both sides of this gives: \[ 3 \cdot 2^{4n} = 2^{2n + 2}k^2 + 2^{n + 2}kx + x^2 \]Now $2^{2n + 2}k^2 = 3 \cdot 2^{4n}$ is impossible because $3 \cdot 2^{4n}$ is not a square, so: \[ 2^{2n + 2}k^2 \le 3\cdot 2^{4n} - 2^{2n + 2} \]We know $x < 1$, so: \[ 2^{n + 2}kx + x^2 < 2^{2n + 2}k + 1 \]Adding these together and substituting into $3 \cdot 2^{4n} = 2^{2n + 2}k^2 + 2^{n + 2}kx + x^2$ produces: \[ 3 \cdot 2^{4n} < 2^{4n} - 2^{2n + 2} + 2^{2n + 2}k + 1 \]so: \[ 2^{2n + 2} < 2^{n + 2}k + 1 \]so $k > 2^n - \frac{1}{2^{n + 2}}$. Plugging this back into the first equation produces \[ \lfloor 2^{2n} \sqrt{3} \rfloor > 2^{2n + 1} - \frac{1}{2} \]We know $\sqrt{3} \le 2 - \frac{1}{2^{2n + 1}}$ for all $n$, so multiplying by $2^{2n}$ gives $ 2^{2n} \sqrt{3} \le 2^{2n + 1} - \frac{1}{2}$. But because $\lfloor 2^{2n} \sqrt{3} \rfloor \le 2^{2n} \sqrt{3} $, we have: \[ \lfloor 2^{2n} \sqrt{3} \rfloor \le 2^{2n + 1} - \frac{1}{2} \]a contradiction.

Here's the natural generalization about this problem: Call a non-dyadic (not of the form $\frac{n}{2^k}$) binary number $n$ with $n = a_1a_2 \dots a_k.b_1b_2b_3 \dots$ superone if there exists an integer $N$ such that for all $n > N$, all $b_{n+1}, \dots, b_{2n}$ equals $1$. Are all square roots $\sqrt{d}$ for $d$ that aren't perfect squares superone? Are all quadratic irrationals $\alpha$ superone? Are all irrational algebraic numbers $\alpha$ superone? (This seems related to Thue-Siegel-Roth and thus might not be feasible.)

bruh. Observe the problem is true for $n=1$. Now notice that \[ 3 \cdot 2^{2n-2} \ge \lfloor 2^{n-1} \sqrt{3} \rfloor^2 + 3 \]because the left hand side isn't a square, and is $0$ mod $4$ and also $3 \cdot 2^{2n-2} > \lfloor 2^{n-1} \sqrt 3 \rfloor^2$. It follows then because $3 > \sqrt{3} + 1/2$ that \[ 3 \cdot 2^{2n} > 2^n + 2^{n + 1} \cdot \lfloor 2^{n-1} \sqrt{3} \rfloor. \]whence multiplying the first inequality by $2^{2n}$, \[ 3 \cdot 2^{4n-2} > \left(1 + 2^n \cdot \lfloor 2^{n-1} \sqrt{3} \rfloor \right)^2. \]As such, \[ 2^{2n - 1} \sqrt{3} - 2^n \lfloor 2^{n-1} \sqrt{3} \rfloor > 1 \implies 2^{n-1} \sqrt{3} - \lfloor 2^{n-1} \sqrt{3} \rfloor > \frac{1}{2^n}, \]which is what we want.