Integer identity

**linda2005** · April 17th 2009, 02:45 AM

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) )

**Grandad** · April 17th 2009, 05:26 AM

Hello linda2005

Originally Posted by linda2005

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

$x \in \mathbb{R} - \mathbb{Q} \Rightarrow nx\in \mathbb{R} - \mathbb{Q}$

$\Rightarrow \exists\, p \in \mathbb{N} : p <nx<p+1$ (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) $n-1 \le \left\lfloor{\frac{\lfloor{nx}\rfloor}{x}}\right\r floor < n$

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

Grandad

**linda2005** · April 18th 2009, 01:31 PM

hello,

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

**Grandad** · April 19th 2009, 12:21 AM

Hello linda2005

Originally Posted by linda2005

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) )

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?

Grandad

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