Pouring water from bottles

This proof uses the same basic idea of having $N$ bottles with $\frac1N$ of the volume as @mathworker21’s proof and merely provides a more elementary proof of the existence of suitable fractional parts that doesn’t require ergodic theory or linear independence over $\mathbb Q$.

Without loss of generality, rescale the bucket sizes so that $x_1+x_2+x_3=1$. Consider $N$ bottles with volume $\frac1N$ each. The fractional parts of $Nx_i$ can add up to $0$, $1$ or $2$. If we can choose $N$ such that they add up to $2$, it will be necessary to split $2$ bottles (since one bottle can only fill fractional parts that add up to $1$). They add up to $2$ if and only if two of them add up to more than $1$. So we want to show that e.g. $(\{Nx_1\},\{Nx_2\})$ lies in the upper right half of $[0,1]^2$ for some $N$. It would be slightly surprising if it required advanced concepts to show that we can hit an entire half of the square.

If all $x_i$ are rational with common denominator $d$, choose $N=d-1$. Since $\{dx_i\}=0$, we have $\{Nx_i\}=\{dx_i-x_i\}=\{-x_i\}=1-x_i$, and thus $\sum_i\{Nx_i\}=3-\sum_ix_i=2$.

Else, at least two of the $x_i$ must be irrational; assume without loss of generality that $x_1$ and $x_2$ are and that $x_1\lt x_2$. If $x_2\lt\frac12$, some multiple $kx_2$ lies in $\left[\frac12,1\right]$, and either $\sum_i\{k x_i\}=2$ and we are done, or $\sum_i\{k x_i\}=1$ and we can replace the $x_i$ by $\{k x_i\}$; so we can assume $x_2\gt\frac12$.

Since $x_1+x_2\lt1$, by Dirichlet’s approximation theorem (which can be proved by an elementary application of the pigeonhole principle), there is $M\in\mathbb N$ such that $\{M x_1\}\gt x_1+x_2$ and thus $\{(M-1)x_1\}\gt x_2$. At least one of $\{Mx_2\}$ and $\{(M-1)x_2\}$ is at least $1-x_2$. (This is where $x_2\gt\frac12$ is needed.) Thus, for at least one of $N=M$ and $N=M-1$ we have $\{Nx_1\}+\{Nx_2\}\gt x_2+1-x_2=1$.

Yes. Take $N$ as in Lemma $1$ with $\alpha = \frac{x_1}{x_1+x_2+x_3}$ and $\beta = \frac{x_2}{x_1+x_2+x_3}$. Let $b_1,\dots,b_N = \frac{x_1+x_2+x_3}{N}$. Suppose we could only split one bottle, giving $\delta_1,\delta_2$ of it to the buckets with $x_1,x_2$ liters (initially), respectively and possibly some to bucket $x_3$. Then there are $m_1,m_2 \in \mathbb{N}$ with $m_1\frac{x_1+x_2+x_3}{N} = x_1-\delta_1$ and $m_2\frac{x_1+x_2+x_3}{N} = x_2-\delta_2$. Then $\frac{Nx_1}{x_1+x_2+x_3} = m_1+\frac{N\delta_1}{x_1+x_2+x_3}$ and $\frac{Nx_2}{x_1+x_2+x_3} = m_2+\frac{N\delta_2}{x_1+x_2+x_3}$. Since $\delta_1,\delta_2 < \frac{x_1+x_2+x_3}{N}$ (if one of them were equal to $\frac{x_1+x_2+x_3}{N}$, then one of $\{\frac{Nx_1}{x_1+x_2+x_3}\},\{\frac{Nx_2}{x_1+x_2+x_3}\}$ would be $0$, contradicting that their sum is more than $1$), we see $\frac{N\delta_1}{x_1+x_2+x_3}+\frac{N\delta_2}{x_1+x_2+x_3} = \{\frac{Nx_1}{x_1+x_2+x_3}\}+\{\frac{Nx_2}{x_1+x_2+x_3}\} > 1$, meaning $\delta_1+\delta_2 > \frac{x_1+x_2+x_3}{N}$, a contradiction to having split a bottle properly.

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Lemma 1: Given any $\alpha,\beta > 0$ with $\alpha+\beta < 1$, there is some $N \in \mathbb{N}$ with $\{\alpha N\}+\{\beta N\} > 1$.

Proof: If $\alpha,\beta$ are $\mathbb{Q}$-linearly independent, then $\{(\{\alpha N\},\{\beta N\}) : N \ge 1\}$ is dense in $\mathbb{T}^2$, so clearly a desired $N$ exists. Otherwise, $\beta = \frac{c}{d}\alpha+\frac{p}{q}$ for some $c,d,p,q \in \mathbb{Z}^{\ge 0}$. Then $\{\alpha N\}+\{\beta N\} = \{\alpha N\}+\{\alpha\frac{c}{d} N+\frac{Np}{q}\}$. First suppose $\alpha$ is irrational. Then since $\{\{\alpha N'dq\} : N' \ge 1\}$ is dense in $\mathbb{T}$, we get a desired $N$ by taking $N = N'dq$ with $\{\alpha N'dq\} > 1-\frac{1}{c^2d^2}$, since then $\alpha cqN' = \frac{(k+1)c}{d}-\frac{c}{d}\epsilon$ for some $k \in \mathbb{Z}$ and $0 < \epsilon < \frac{1}{c^2d^2}$, meaning $\{\alpha cqN'\}$ is either at least $1-\frac{c}{d}\epsilon$ or at least $\frac{1}{d}-\frac{c}{d}\epsilon$, both large enough. Now suppose $\alpha = \frac{m}{n}$ is rational, with $\gcd(m,n) = 1$. Then write $\frac{m}{n}\frac{c}{d}+\frac{p}{q} = \frac{m'}{n'}$ with $\gcd(m',n') = 1$. We wish to show $\{\frac{m}{n}N\}+\{\frac{m'}{n'}N\} > 1$ for some $N \in \mathbb{N}$. WLOG suppose $n \ge n'$. We can take $N$ so that $Nm \equiv -1 \pmod{n}$ and $n' \nmid N$; indeed, if $n' \mid n$, then clearly $Nm \equiv -1 \pmod{n'}$ as well, and otherwise, $N = kn+m^*$ for an appropriate $k$ works, where $m^*m \equiv -1 \pmod{n}$. For this $N$, we have $\{\frac{m}{n}N\}+\{\frac{m'}{n'}N\} \ge \frac{n-1}{n}+\frac{1}{n'} \ge 1$, with equality only if $n' = n$ and $mN \equiv -1 \pmod{n}$ and $m'N \equiv 1 \pmod{n}$. But if we had equality, then $(m+m')N \equiv 0 \pmod{n}$, meaning $\alpha+\beta = \frac{m}{n}+\frac{m'}{n} = 1$, which is false.

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Fact: If $\alpha,\beta$ are $\mathbb{Q}$-linearly independent, then $\{(\{\alpha N\},\{\beta N\}) : N \ge 1\}$ is dense in $\mathbb{T}^2$.

Proof: Define $T: \mathbb{T}^2 \to \mathbb{T}^2$ by $T(x,y) = (x+\alpha,y+\beta)$. It suffices to show that $T$ is ergodic w.r.t. the Lebesgue measure. Suppose $f \in L^2(\mathbb{T}^2)$ is $T$-invariant. By basic fourier analysis, $f(x_1,x_2) = \sum_{k_1,k_2 \in \mathbb{Z}} c_{k_1,k_2}e^{2\pi i (k_1x_1+k_2x_2)}$. Then $\sum_{k_1,k_2} c_{k_1,k_2} e^{2\pi i (k_1x_1+k_2x_2)} = f(x_1,x_2) = f\circ T(x_1,x_2)= \sum_{k_1,k_2} c_{k_1,k_2} e^{2\pi i (k_1(x_1+\alpha)+k_2(x_2+\beta))} = \sum_{k_1,k_2} e^{2\pi i(k_1\alpha+k_2\beta)}c_{k_1,k_2}e^{2\pi i (k_1x_1+k_2x_2)}$ Therefore, $c_{k_1,k_2} = c_{k_1,k_2}e^{2\pi i (k_1\alpha+k_2\beta)}$ for each $k_1,k_2 \in \mathbb{Z}$. Since $\alpha,\beta$ are $\mathbb{Q}$-linearly independent, we have $c_{k_1,k_2} = 0$ for all $(k_1,k_2) \not = (0,0)$. It follows that $f$ is a.e. constant, as desired.