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Math Help - Integer equation

  1. #1
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    Integer equation

    Hello, I need your help, thanks

    - Solve in \mathbb{R} the equation :  \lfloor x^2 - 2x\rfloor + 2\lfloor x\rfloor = {\lfloor x\rfloor}^2 .
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  2. #2
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    x\in\mathbb{R}
    Try any number, this equation will obiously always end with x=x
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  3. #3
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    Quote Originally Posted by maria18 View Post
    Hello, I need your help, thanks

    - Solve in \mathbb{R} the equation :  \lfloor x^2 - 2x\rfloor + 2\lfloor x\rfloor = {\lfloor x\rfloor}^2 .
    clearly every integer is a solution. so i'll assume that x \notin \mathbb{Z}. let [x]=n and x=n+\theta, where 0 < \theta < 1. then x is a solution if and only if: 0 \leq \theta^2 + 2(n-1) \theta < 1.

    so \theta^2 + 2(n-1) \theta = \theta(\theta + 2n - 2) \geq 0. hence 0 \leq \theta + 2n - 2 < 2n - 1. thus we must have n \geq 1. the converse is obviously true, i.e. if n \geq 1, then \theta^2 + 2(n-1) \theta \geq 0.

    now, from quadratic formula it's clear that \theta^2 + 2(n-1)\theta - 1 < 0, if and only if 0 < \theta < 1 - n + \sqrt{n^2 - 2n + 2}. thus n < x < 1+ \sqrt{n^2 - 2n + 2}. so the general solution of

    your equation is: \mathbb{Z} \cup \bigcup_{n=1}^{\infty} (n, 1 + \sqrt{n^2 - 2n + 2}).

    by the way questions of this type have nothing to do with calculus and they should be posted in high school pre-algebra subforum.
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    clearly every integer is a solution. so i'll assume that x \notin \mathbb{Z}. let [x]=n and x=n+\theta, where 0 < \theta < 1. then x is a solution if and only if: 0 \leq \theta^2 + 2(n-1) \theta < 1.

    so \theta^2 + 2(n-1) \theta = \theta(\theta + 2n - 2) \geq 0. hence 0 \leq \theta + 2n - 2 < 2n - 1. thus we must have n \geq 1. the converse is obviously true, i.e. if n \geq 1, then \theta^2 + 2(n-1) \theta \geq 0.

    now, from quadratic formula it's clear that \theta^2 + 2(n-1)\theta - 1 < 0, if and only if 0 < \theta < 1 - n + \sqrt{n^2 - 2n + 2}. thus n < x < 1+ \sqrt{n^2 - 2n + 2}. so the general solution of

    your equation is: \mathbb{Z} \cup \bigcup_{n=1}^{\infty} (n, 1 + \sqrt{n^2 - 2n + 2}).

    by the way questions of this type have nothing to do with calculus and they should be posted in high school pre-algebra subforum.
    wow too advanced
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    clearly every integer is a solution. so i'll assume that x \notin \mathbb{Z}. let [x]=n and x=n+\theta, where 0 < \theta < 1. then x is a solution if and only if: 0 \leq \theta^2 + 2(n-1) \theta < 1.

    so \theta^2 + 2(n-1) \theta = \theta(\theta + 2n - 2) \geq 0. hence 0 \leq \theta + 2n - 2 < 2n - 1. thus we must have n \geq 1. the converse is obviously true, i.e. if n \geq 1, then \theta^2 + 2(n-1) \theta \geq 0.

    now, from quadratic formula it's clear that \theta^2 + 2(n-1)\theta - 1 < 0, if and only if 0 < \theta < 1 - n + \sqrt{n^2 - 2n + 2}. thus n < x < 1+ \sqrt{n^2 - 2n + 2}. so the general solution of

    your equation is: \mathbb{Z} \cup \bigcup_{n=1}^{\infty} (n, 1 + \sqrt{n^2 - 2n + 2}).



    by the way questions of this type have nothing to do with calculus and they should be posted in high school pre-algebra subforum.
    1) Hello NonCommAlg and thanks, so i dont understand why x is a solution if and only if: 0 \leq \theta^2 + 2(n-1) \theta < 1. ? Thanks

    2) I'm seeing that \lfloor x^2-2x\rfloor + 2\lfloor x\rfloor = {\lfloor x\rfloor}^2 \iff \lfloor (x - 1)^2 \rfloor = (\lfloor x - 1 \rfloor)^2 so Mr NonCommAlg we can use the same proof that \lfloor x^2 \rfloor = (\lfloor x \rfloor)^2 ? look here ==> http://www.mathhelpforum.com/math-he...equations.html
    Last edited by maria18; April 19th 2009 at 03:27 PM.
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