1. ## Integer identity

i)Prove that if $x > 1$ and $x\in (\mathbb{R}{ - }\mathbb{Q})$ then : $\left\lfloor \frac {\lfloor nx\rfloor}{x}\right\rfloor = n - 1$

ii) Prove there is an only real $a$ such that: $\forall n \in \mathbb{N}^{*}, \left\lfloor a\lfloor na\rfloor\right\rfloor - \lfloor na\rfloor = n - 1$, (you can us i) )

2. ## Integer proof

Hello linda2005
Originally Posted by linda2005
i)Prove that if $x > 1$ and $x\in (\mathbb{R}{ - }\mathbb{Q})$ then : $\left\lfloor \frac {\lfloor nx\rfloor}{x}\right\rfloor = n - 1$
$x \in \mathbb{R} - \mathbb{Q} \Rightarrow nx\in \mathbb{R} - \mathbb{Q}$

$\Rightarrow \exists\, p \in \mathbb{N} : p (1)

$\Rightarrow p = \lfloor nx\rfloor$

$\Rightarrow \frac{p}{x} = \frac{\lfloor{nx}\rfloor}{x}< n$, from (1)

$\Rightarrow \left\lfloor{\frac{\lfloor{nx}\rfloor}{x}}\right\r floor < n$ (2)

and from (1) $\frac{p+1}{x} > n$

$\Rightarrow \frac{p}{x}+\frac{1}{x}>n$

$\Rightarrow \frac{\lfloor{nx}\rfloor}{x} > n - \frac{1}{x}> n -1$, since $x > 1$

$\Rightarrow \left\lfloor{\frac{\lfloor{nx}\rfloor}{x}}\right\r floor \ge n - 1$

So from (2) $n-1 \le \left\lfloor{\frac{\lfloor{nx}\rfloor}{x}}\right\r floor < n$

$\Rightarrow \left\lfloor{\frac{\lfloor{nx}\rfloor}{x}}\right\r floor = n - 1$

ii) Prove there is an only real(?) $a$ such that: $\forall n \in \mathbb{N}^{*}, \left\lfloor a\lfloor na\rfloor\right\rfloor - \lfloor na\rfloor = n - 1$, (you can use i) )