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Math Help - Equations

  1. #1
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    Equations

    Hello , help please I need to solve those equations :



    1) [x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345

    2) [x^2]=[x]^2

    3) [\frac{x^2-2x}{3}]=\frac{x}{2}
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  2. #2
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    Quote Originally Posted by linda2005 View Post
    Hello , help please I need to solve those equations :



    1) [x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345

    2) [x^2]=[x]^2

    3) [\frac{x^2-2x}{3}]=\frac{x}{2}
    What is being represented by the square brackets?
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  3. #3
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    Quote Originally Posted by Mush View Post
    What is being represented by the square brackets?
    whole part of number

    \left[ x \right]or\left\lfloor x \right\rfloor  = \max \left\{ {\left. {n \in \mathbb{Z}} \right|{\text{ }}n \leqslant x} \right\}.
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  4. #4
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    Quote Originally Posted by DeMath View Post
    whole part of number

    \left[ x \right]or\left\lfloor x \right\rfloor  = \max \left\{ {\left. {n \in \mathbb{Z}} \right|{\text{ }}n \leqslant x} \right\}.
    Ah, right. Are you familiar with the following property, for the first problem:

     \lfloor kx \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x+\frac{1}{k}\right\rfloor +\dots+\left\lfloor x+\frac{k-1}{k}\right\rfloor

    In other words:

     \lfloor 2x \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x+\frac{1}{2}\right\rfloor

     \lfloor 4x \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x+\frac{1}{4}\right\rfloor+ \left\lfloor x+\frac{2}{4}\right\rfloor + \left\lfloor x+\frac{3}{4}\right\rfloor

    Followed by the rule that:

     \lfloor x + n \rfloor =  \lfloor x \rfloor + n

    Hence:

     \lfloor 2x \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x\right\rfloor +\frac{1}{2} etc...
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  5. #5
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    For the 2nd one, think of it like this:

     x = k + p where k is an integer, and p is a non-integer between 0 and 1.

    In mathematical terms:  k \in Z \text{, } p \notin Z \text{ and } 0 < p<1

    Hence:

     \lfloor (k + p)^2 \rfloor = \lfloor k + p \rfloor^2

     \lfloor k^2+ 2pk+ p^2 \rfloor = \lfloor k + p \rfloor^2

    On the RHS, we know that k is the integer part of the number so:

     \lfloor k^2 +2pk+ p^2 \rfloor = k^2

    In other words, the NON-integer part of the LHS must be  2pk+p^2

    So now you need to come up with values of k and p such that:

     k \in Z \text{, } p \notin Z  \text{, } 0 < p < 1\text{ and } 0 < 2pk+p^2 < 1

    For example, p = 0.1, k = 1 is a solution. (x = 1.1, in other words).
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  6. #6
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    A fairly obvious solution to the last one is x = 0.
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  7. #7
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    Quote Originally Posted by Mush View Post
    A fairly obvious solution to the last one is x = 0.
    hello thanks Mush , but with x=4 it's ok? so you are wrong
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  8. #8
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    Quote Originally Posted by linda2005 View Post
    hello thanks Mush , but with x=4 it's ok? so you are wrong
    Eh? x = 4 is a solution to the equation, yes. But so is x = 0. I am not wrong, there are several solutions.
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  9. #9
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    i'm not sure about 1) but i guess there is no solution. my solutions to 2) and 3):

    Quote Originally Posted by linda2005 View Post

    2) [x^2]=[x]^2
    clearly any integer is a solution. so we'll assume that x \notin \mathbb{Z}. let [x]=n. then n < x < n+1 and [x^2 - n^2]=0, which is equivalent to 0 < x^2 - n^2 < 1. \ \ \ \ (1)

    case 1. x < 0: in this case, since x > n, we'll have: x^2 < n^2. but from (1) we have x^2 > n^2. this contradiction means there's no solution in this case.

    case 2. x > 0: then the conditions n < x < n+1 and (1) are equivalent to n < x < \sqrt{n^2 + 1}. conversely if n < x < \sqrt{n^2 + 1}, for some positive integer n, then

    n < x < n+1 and n^2 < x^2 < n^2 + 1. thus [x]=n and [x^2]=n^2=[x]^2.

    so the complete set of solutions of the equation is: \bigcup_{n \in \mathbb{N} \cup \{0 \}} (n , \sqrt{n^2 + 1}) \cup \mathbb{Z}.


    3) [\frac{x^2-2x}{3}]=\frac{x}{2}
    the equation is equivalent to these two conditions together: \frac{x}{2} \in \mathbb{Z} and \frac{x^2 - 2x}{3} - 1 < \frac{x}{2} \leq \frac{x^2 - 2x}{3}. \ \ \ \ (2)

    let x = 2n, where n \in \mathbb{Z}. solving the inequalities in (2) for n gives you: \left[\frac{7-\sqrt{97}}{8}, 0 \right] \cup \left[\frac{7}{4}, \frac{7 + \sqrt{97}}{8} \right], which gives us n = 0,2. thus x=2n=0,4.
    Last edited by NonCommAlg; April 7th 2009 at 12:57 PM. Reason: typo!
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  10. #10
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    Hello, Thanks you NonCommAlg Your solutions is , so you can help me with 1) thanks
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  11. #11
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    Quote Originally Posted by linda2005 View Post
    Hello, Thanks you NonCommAlg Your solutions is , so you can help me with 1) thanks
    as i said i'm not sure about 1) right now but i have some reasons to believe that the equation has no solution. so tell me, if you know the answer, am i right?
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  12. #12
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    There is no solution

    Using the following two properties:

     \lfloor kx \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x+\frac{1}{k}\right\rfloor +\dots+\left\lfloor x+\frac{k-1}{k}\right\rfloor

     \lfloor x + n \rfloor =  \lfloor x \rfloor + n

    It can be deduced that:

     \displaystyle \lfloor kx \rfloor = k \lfloor x \rfloor + \sum_{n = 1}^{k-1} \frac{n}{k}

    Hence:

      \displaystyle \lfloor 2x \rfloor =  2\lfloor x \rfloor + \sum_{n = 1}^{1} \frac{n}{2}  =2\lfloor x \rfloor + \frac{1}{2}

     \displaystyle \lfloor 4x \rfloor = 4 \lfloor x \rfloor + \sum_{n = 1}^{3} \frac{n}{4} = 4\lfloor x \rfloor + \frac{3}{2}

    And in general  \displaystyle \lfloor kx \rfloor =  k\lfloor x \rfloor  + \frac{k-1}{2}

    Continuing like this, the equation can be reduced to:

    63 \lfloor x \rfloor + \frac{1+3+7+15+31}{2} = 12345

     63\lfloor x \rfloor + \frac{57}{2} = 12345

     63\lfloor x \rfloor = \frac{24633}{2}

     \lfloor x \rfloor =  \frac{391}{2}

    It is non-integer, and so there are no solutions.
    Last edited by Mush; April 7th 2009 at 03:47 PM.
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  13. #13
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    Quote Originally Posted by Mush View Post

    It can be deduced that:

     \displaystyle \lfloor kx \rfloor = k \lfloor x \rfloor + \sum_{n = 1}^{k-1} \frac{n}{k}
    this is not true! why? see below for example:


    Hence:
     \displaystyle \lfloor 2x \rfloor = 2\lfloor x \rfloor + \sum_{n = 1}^{1} \frac{n}{2} =2\lfloor x \rfloor + \frac{1}{2}
    \lfloor 2x \rfloor - 2\lfloor x \rfloor is an integer but \frac{1}{2} is not. so they can never be equal. the function \lfloor 2x \rfloor - 2\lfloor x \rfloor is always either 0 or 1.
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  14. #14
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    Quote Originally Posted by NonCommAlg View Post
    this is not true! why? see below for example:



    \lfloor 2x \rfloor - 2\lfloor x \rfloor is an integer but \frac{1}{2} is not. so they can never be equal. the function \lfloor 2x \rfloor - 2\lfloor x \rfloor is always either 0 or 1.
    Oh very, true, my apologies.

    This is because the 2nd property I used only works if n is an integer. My mistake.
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  15. #15
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    look this
    When he plugs x = 196, the result is 12348.We're going to need to take off three. Now imagine we decrease that 196 slightly. Every term on the lefthand side will decrease by AT LEAST one (for example, [32\cdot 196] = 6272, [32\cdot 195.99] = 6271), therefore the sum will decrease by AT LEAST six, and that's too much.

    some one can more explain this way
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