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Math Help - Surds problem

  1. #1
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    Surds problem

    Hey can you guys help me out with this problem:

    It is given that \frac{a}{b+\sqrt{c}} + \frac{d}{\sqrt{c}} is rational, where a,b,c and d are positive integers and c is not a square. Show that as a consequence, db^2 = c(a+d). Use this result to show that \frac{a}{1+\sqrt{c}} + \frac{d}{\sqrt{c}} is not rational.

    Thanks!
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  2. #2
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    Hello, eskimogenius!

    I have a truly "clunky" approach to the first part . . .



    \text{It is given that: }\:\dfrac{a}{b+\sqrt{c}} + \dfrac{d}{\sqrt{c}}\,\text{ is rational,}
    . . \text{where }a,b,c,d\text{ are positive integers and }c\text{ is not a square.}

    \text{Show that: }\:db^2 \,=\, c(a+d).

    We have: . \dfrac{a}{b +\sqrt{c}} + \dfrac{d}{\sqrt{c}} \;=\;\dfrac{(a+d)\sqrt{c} + db}{b\sqrt{c} + c}



    Long division:
    . . \begin{array}{ccccccc}<br />
&&&& \frac{a+d}{b} \\<br />
&& ----- & - & ----- \\<br />
b\sqrt{c} + c & \bigg) & (a+d)\sqrt{c} & + & db \\<br />
&& (a+d)\sqrt{c} & + & \frac{c(a+d)}{b} \\<br />
&& ----- & - & ----- \\<br />
&&&& db - \frac{c(a+d)}{b}\end{array}


    Since the division produces a rational answer, the remainder must be zero.

    . . db - \dfrac{c(a+d)}{b} \;=\;\dfrac{db^2 - c(a+d)}{b} \;=\;0


    Therefore: . db^2 - c(a+d) \:=\:0

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, eskimogenius!

    I have a truly "clunky" approach to the first part . . .




    We have: . \dfrac{a}{b +\sqrt{c}} + \dfrac{d}{\sqrt{c}} \;=\;\dfrac{(a+d)\sqrt{c} + db}{b\sqrt{c} + c}



    Long division:
    . . \begin{array}{ccccccc}<br />
&&&& \frac{a+d}{b} \\<br />
&& ----- & - & ----- \\<br />
b\sqrt{c} + c & \bigg) & (a+d)\sqrt{c} & + & db \\<br />
&& (a+d)\sqrt{c} & + & \frac{c(a+d)}{b} \\<br />
&& ----- & - & ----- \\<br />
&&&& db - \frac{c(a+d)}{b}\end{array}


    Since the division produces a rational answer, the remainder must be zero.

    . . db - \dfrac{c(a+d)}{b} \;=\;\dfrac{db^2 - c(a+d)}{b} \;=\;0


    Therefore: . db^2 - c(a+d) \:=\:0

    Wow, truly impressive. I have a question but, how did you know \dfrac{a+d}{b} was the quotient?

    Now, there's only the second part of the question
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  4. #4
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    Never mind about that second part, my math teacher has solved it. Thank you very much Soroban.

    For those who want to know the solution:

    The first fraction that was given when compared to the second fraction: it can be seen that b = 1. Now for it to be true, the above proven equation has to hold.

    When you sub b=1 into the proven equation above, it can be seen that  c = \dfrac{d}{a+d}

    However, as a, c and d are positive integers, that fraction has to be less than one, and hence this is impossible because c is a positive integer.

    Therefore, the second fraction is never rational.

    PS. By fraction I mean everything that is need to be proved irrational and everything that is need to be proved irrational.
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