The floor function [closed]

Question

How do I prove or disprove the following statements: $$(a) ∀n ∈ \mathbb N, ∃k ∈ \mathbb N, ∀x ∈ \mathbb R , \lfloor nx \rfloor − n \lfloor x \rfloor ≤ k$$ and $$ (b)\exists k \in \mathbb N, \forall n \in \mathbb N, \forall x \in \mathbb R, \lfloor nx \rfloor − n \lfloor x \rfloor ≤ k $$

I've also been given the following three properties to use: $ (i) \forall x \in \mathbb R, \exists \epsilon \in R, 0 ≤ \epsilon < 1 ∧ x = \lfloor x \rfloor + \epsilon \\ (ii) \forall x \in \mathbb Z, \forall y \in \mathbb R, \lfloor x + y \rfloor = x + \lfloor y \rfloor \\ (iii) \forall x \in \mathbb Z, \lfloor x \rfloor = x $

I know that we must negate the statement$(a)$ first in order to pave the way for a disproof(if it false) by proving: $$ \exists n \in \mathbb N, \forall k \in \mathbb N, \exists x \in \mathbb R, \lfloor nx \rfloor - n \lfloor x \rfloor > k $$ Other than that I have no idea which property to use and in which order. The help would be really appreciated.

Answer 1

You have been given that $$0\le\epsilon<1$$ multiplying by $-1$ \begin{equation} \therefore 0\ge-\epsilon>-1\tag{1} \end{equation} adding $x$ $$\therefore x\ge x-\epsilon> x-1$$ \begin{equation} \therefore x\ge \lfloor x \rfloor >x-1\tag{2} \end{equation} We can use this inequality $(2)$ to show that \begin{equation} \therefore nx\ge n\lfloor x\rfloor > nx - n\tag{3} \end{equation} adding $nx$ to inequality $(1)$ we get $$nx\ge nx-\epsilon > nx -1$$ \begin{equation} nx\ge \lfloor nx \rfloor > nx -1 \tag{4} \end{equation} Using $(3)$ and $(4)$ we can write $$nx-1 - nx<\lfloor nx\rfloor - n \lfloor x\rfloor<nx - (nx - n)$$ $$\therefore -1<\lfloor nx\rfloor - n \lfloor x\rfloor<n $$

Answer 2

The first step is to understand for each $n,k$ the following property holds: $$ P(n,k) \Leftrightarrow \forall x \in \mathbb{R}, \lfloor nx \rfloor - n\lfloor x \rfloor \leq k. $$ If $nx = na + b + \epsilon$, where $a,b \in \mathbb{Z}$, $0 \leq b < n$, and $0 \leq \epsilon < 1$, then $$ \lfloor nx \rfloor = na + b, \quad n\lfloor x \rfloor = na \quad \Longrightarrow \quad \lfloor nx \rfloor - n \lfloor x \rfloor = b. $$ This means that $$\{\lfloor nx \rfloor - n \lfloor x \rfloor : x \in \mathbb{R}\} = \{0,1,\ldots,n-1\}.$$ In particular, $$ P(n,k) \Leftrightarrow k \geq n-1. $$ Your two statements are thus equivalent to $$ \forall n \in \mathbb{N} \exists k \in \mathbb{N}, k \geq n-1, \\ \exists n \in \mathbb{N} \forall k \in \mathbb{N}, k \geq n-1. $$ You take it from here.