1. ## Integer identity

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

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

2. ## Integer proof

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

$\displaystyle \Rightarrow \exists\, p \in \mathbb{N} : p <nx<p+1$ (1)

$\displaystyle \Rightarrow p = \lfloor nx\rfloor$

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

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

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

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

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

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

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

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

3. hello,

Thanks Grandad, very nice proof, so for ii) no one can help me ?

4. Hello linda2005
Originally Posted by linda2005
ii) Prove there is an only real(?) $\displaystyle a$ such that:$\displaystyle \forall n \in \mathbb{N}^{*}, \left\lfloor a\lfloor na\rfloor\right\rfloor - \lfloor na\rfloor = n - 1$, (you can use i) )
I must confess I haven't given this much thought. I'm not really sure what the question means. Have you got the wording right?