Exercise 4.1.8 (If $x0, y0$, then $\lfloor x - y \rfloor \leq \lfloor x \rfloor - \lfloor y \rfloor \leq \lfloor x -y \rfloor + 1.$)

Question

For any positive real numbers $x$ and $y$ prove that
$$\lfloor x - y \rfloor \leq \lfloor x \rfloor - \lfloor y \rfloor \leq \lfloor x -y \rfloor + 1.$$

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

\begin{proof} We write as usual
$$\begin{array}{lll}
x = n + 
u,& n \in \Z, &0 \leq 
u < 1,\
y = m + \mu,& m \in \Z, &0 \leq \mu < 1,
\end{array}
$$
so that $\lfloor x floor = n, \lfloor y floor = m$.

Then
$$-1 < -\mu \leq 
u - \mu \leq 
u<1.$$
Put $\eta = 
u - \mu$. Then $\lfloor \eta floor = 0$ or $\lfloor \eta floor = -1$, and 
$$x - y  = n- m + 
u - \mu = n +m + \eta,$$
thus by Theorem 4.1(3),
$$\lfloor x- y floor = n- m + \lfloor \eta floor , \qquad \lfloor \eta floor \in \{0,-1\}.$$
Then $-1 \leq  \lfloor \eta floor \leq 0$, thus
$$n - m -1 \leq \lfloor x- y floor \leq n-m,$$
that is
$$\lfloor x floor - \lfloor y floor - 1 \leq \lfloor x- y floor \leq \lfloor x floor - \lfloor y floor,$$
which is equivalent to
$$\lfloor x - y floor \leq \lfloor x floor - \lfloor y floor \leq \lfloor x -y floor + 1.$$
\end{proof}

Exercise 4.1.8 (If $x>0, y>0$, then $\lfloor x - y \rfloor \leq \lfloor x \rfloor - \lfloor y \rfloor \leq \lfloor x -y \rfloor + 1.$)

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Exercise 4.1.8 (If $x>0, y>0$, then $\lfloor x - y \rfloor \leq \lfloor x \rfloor - \lfloor y \rfloor \leq \lfloor x -y \rfloor + 1.$)

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