# Proof regarding the floor function

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: $(i) \space \forall x \in \mathbb R, \exists \epsilon \in R, 0 ≤ \epsilon < 1 ∧ x = \lfloor x \rfloor + \epsilon \\ (ii) \ \forall x \in \mathbb Z, \forall y \in \mathbb R, \lfloor x + y \rfloor = x + \lfloor y \rfloor \\ (iii) \ \forall x \in \mathbb Z, \lfloor x \rfloor = x$

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.

For (a) : TRUE

Just apply the definition $\lfloor x\rfloor\le x\lt\lfloor x\rfloor+1$

Note that (i) is just an alternate definition.

multiply by n then $n\lfloor x\rfloor\le nx\lt n\lfloor x\rfloor+n$

And since the bounds are integers $n\lfloor x\rfloor\le \lfloor nx\rfloor\lt n\lfloor x\rfloor+n$

Now substract to get $0\le \lfloor nx\rfloor - n\lfloor x\rfloor < n$ this is $(a)$ for $k=n-1$ since it does not depends on $x$. (if an integer is $<n$ then it is $\le n-1$)

For (b) : FALSE

A counter example is $x=\frac12$ and $n=2k+2$ then

$\lfloor nx\rfloor - n\lfloor x\rfloor=\lfloor \frac{2k+2}{2}\rfloor - n\lfloor \frac 12\rfloor=k+1-n*0=k+1>k$

For both we have merely applied only (ii) and (iii).