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Math Help - [SOLVED] Find the values of p and q

  1. #1
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    [SOLVED] Find the values of p and q

    (2√3 - 1) = p + q√3

    Does anyone know how to solve this?
    So far I've only got:

    13 - 4√3 = p + q√3

    However, as with a similar problem I have done, I am pretty sure there are a range of answers, not just p = 13 and q = -4.

    Can someone help me with this?
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  2. #2
    Moo
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    Hello,

    It may depend on how you define p and q. Do they belong to \mathbb{N}, or \mathbb{R} ?
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    Quote Originally Posted by Moo View Post
    Hello,

    It may depend on how you define p and q. Do they belong to \mathbb{N}, or \mathbb{R} ?
    Or \mathbb{Q}, ya know, the rationals
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  4. #4
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    If it belongs to \mathbb{Q} there can't be several solutions, just one


    By the way, I meant \mathbb{Z}, certainly not \mathbb{N}
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    The question did not contain any further detail, also, I do not know what \mathbb{N} or \mathbb{R} are...we haven't learned it, so I don't think they belong to....those things.


    Erm....could someone explain what those large, nice looking letters mean?

    Thanks.
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  6. #6
    Moo
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    \mathbb{N} : set of natural integers (means the positive ones)

    \mathbb{Z} : set of all integers (positive and negative)

    \mathbb{Q} : rational numbers, which means \frac pq with p and q in \mathbb{Z} (and q different from 0)

    \mathbb{R} : set of all numbers, including irrational ones.

    \mathbb{N} is included in \mathbb{Z}, which is included in \mathbb{Q}, which is included in \mathbb{R}

    -----------------------------

    Why did I ask ? Because if we're working on \mathbb{Z}, there is only one solution.

    If we're working on \mathbb{R}, then p=10-5\sqrt{3} and q=1+\sqrt{3} are solutions, for example, but there is an infinity of possibilities. (note that \sqrt{3} is an irrational and only belongs to \mathbb{R}).
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    Lord of certain Rings
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    Quote Originally Posted by Moo View Post
    If it belongs to \mathbb{Q} there can't be several solutions, just one


    By the way, I meant \mathbb{Z}, certainly not \mathbb{N}
    In the most general sense, shouldnt you be saying \mathbb{Q}, since they include \mathbb{Z}(this is where uniqueness of the representation holds)?
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  8. #8
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    Well the question did not state anything, so I'd assume the solution would belong to \mathbb{R}.
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    Moo
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    So the solutions will be

    p=a+b\sqrt{3} and q=c+d\sqrt{3}

    p+q\sqrt{3}=a+b\sqrt{3}+c\sqrt{3}+3d

    So we want a+3d=13 and b+c=-4

    Hence a=13-3d and b=-4-c.


    -------> general solutions are p=13-3d-(4+c)\sqrt{3} and q=c+d \sqrt{3}, this time, c and d belonging to \mathbb{Q} (yep Isomorphism, it can be done this way indeed )

    This is only for the ones involving \sqrt{3} the basic way. Otherwise, there is an infinity of them.
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  10. #10
    Moo
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    Since they ask you the "values of p and q", I'd follow the idea that the solutions are simply p=13 and q=-4
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  11. #11
    Lord of certain Rings
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    Quote Originally Posted by anne. View Post
    Well the question did not state anything, so I'd assume the solution would belong to \mathbb{R}.
    Then you have \mathbb{R} solutions

    Pick any real number. Let it be p. Substitute in your equation and solve for q, another real number!

    From my experience, I think you are asked to find rational values(that is the numbers from \mathbb{Q} and not \mathbb{R}).

    Doing it for \mathbb{R} looks uneducational
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  12. #12
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    O_O;;

    wow. I didn't get a word of the second-last post that you posted (the one with all the equations...)

    I'll be sticking with the simple solution of p = 13 and q = -4 then.

    Thanks for your help

    Edit: oops, yes I did mean Q. all the letters are getting to my head.
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  13. #13
    Moo
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    I just tried to show you another possibility of solution, but yep, it's complicated and not well-explained Sorry about that
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