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Math Help - Integer identity

  1. #1
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    Thumbs up 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) )
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  2. #2
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    Integer proof

    Hello linda2005
    Quote Originally Posted by linda2005 View Post
    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
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  3. #3
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    hello,

    Thanks Grandad, very nice proof, so for ii) no one can help me ?
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  4. #4
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    Hello linda2005
    Quote Originally Posted by linda2005 View Post
    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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