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Math Help - Rational Number

  1. #1
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    Rational Number

    Hi,

    A rational number is described as being a fraction where both the numerator and denominator are integers. If this is so, why does my book say that (\surd3)/3 is a rational number?
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  2. #2
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    Quote Originally Posted by webguy View Post
    Hi,

    A rational number is described as being a fraction where both the numerator and denominator are integers. If this is so, why does my book say that (\surd3)/3 is a rational number?
    \frac{\sqrt{3}}{3} is not a rational number ... are you sure that is what the book says? a direct quote would be helpful.
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    Book says: To rationalise the fraction 1/(\surd3), multiply both the numerator and denominator by \surd3 giving: (\surd3)/3.
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    Quote Originally Posted by webguy View Post
    Book says: To rationalise the fraction 1/(\surd3), multiply both the numerator and denominator by \surd3 giving: (\surd3)/3.
    There is a failure to communicate there. At least a failure to translate correctly.
    The word rationalize does not mean strictly to make rational in mathematics-speak.
    It is considered poor form to have a radical in the denominator.

    Therefore instead of \dfrac{2}{\sqrt{3}} we rationalize that number writing it as \dfrac{2\sqrt{3}}{3} by multiplying by \dfrac{\sqrt{3}}{\sqrt{3}}.

    In other words we make the denominator rational.
    Last edited by Plato; August 27th 2010 at 03:54 PM.
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    Understood, thanks . However, why is it better to have an irrational number as the numerator instead of the denominator?
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    Quote Originally Posted by webguy View Post
    Understood, thanks . However, why is it better to have an irrational number as the numerator instead of the denominator?
    goes back to the days before calculators ... making decimal approximations was easier with a whole number in the denominator.
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    Quote Originally Posted by webguy View Post
    Understood, thanks . However, why is it better to have an irrational number as the numerator instead of the denominator?
    Yours is actually a very good question.
    Before good calculators and/or CAS it was very difficult to divide by an irrational number (actually it still is in practice).
    Consider \dfrac{1}{\sqrt3} over against \dfrac{\sqrt3}{3} which is easier to do by ‘hand’?
    They are both equal. But the second is easier to work with.
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  8. #8
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    Hello, webguy!


    Edit: Plato beat me to it . . . *sigh*
    . . . .But I refuse to delete all this typing . . . so there!


    Why is it better to have an irrational number in the numerator
    instead of the denominator?

    Probably the same reason we prefer positive denominators.

    We know that \frac{2}{3} is read "two-thirds".

    And \frac{\text{-}2}{3} is read "negative-two thirds".

    And \text{-}\frac{2}{3} is read "negative two-thirds".

    But how do we read \frac{2}{\text{-}3} ? . . . Maybe "two negative-thirds" ?


    Interesting . . . This problem never came up before. .Why?

    Because "they" always use positive denominators at the start.
    (And didn't bother to tell us that it's a Rule.)



    A bit of trivia . . .
    A fraction has three signs: . \begin{Bmatrix}\text{one in the numerator} \\ \text{one in the denominator} \\ \text{one "out in front"} \end{Bmatrix}
    So that \dfrac{2}{3} is actually: +\dfrac{+2}{+3}

    We can change any two signs without changing the value of the fraction.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Back to the question: Why do we "rationalize" denominators?

    The main reason arose back in 1000 B.C. (before calculators).


    If we need the value of \frac{1}{\sqrt{2}}, we have a lot of work to do!

    First, look up \sqrt{2} on a square-root table and get, say, 1.4142

    Then we must divide: . 1 \div 1.4142 . . . a very unplesant task!



    It starts like this:

    . . \begin{array}{cccccccc}<br />
&& --&--&--&--&-- \\ 1\;.4\;1\;4\;2 & | & 1 & .0 & 0 & 0 & 0  \end{array}



    We "move the decimal points" and divide:

    . . \begin{array}{cccccccccccccc}<br />
&&&&&&0 & .7 & 0 & 7 & 1 & 1 \\ && --&--&--&--&--&--&--&--&--&-- \\<br />
1.4142 & | & 1 & 0 & 0 & 0 & 0 & .0 & 0 & 0 & 0 & 0 \\<br />
&&& 9 & 8 & 9 & 9 & 4 \\ <br />
&& --&--&--&--&--&-- \\<br />
&&&& 1 & 0 & 0 & 6 & 0 & 0 \\<br />
&&&&& 9 & 8 & 9 & 9 & 4 \\<br />
&&&& --&--&--&--&--&-- \\<br />
&&&&&& 1 & 6 & 0 & 6 & 0 \\<br />
&&&&&& 1 & 4 & 1 & 4 & 2 \\<br />
&&&&&&--&--&--&--&-- \\<br />
&&&&&&& 1 & 9 & 1 & 8 & 0 \\<br />
&&&&&&& 1 & 4 & 1 & 4 & 2 \\<br />
&&&&&&& --&--&--&--&-- \\<br />
&&&&&&&&5&0&3&8<br />
\end{array}

    So, to four decimal places, we have: . \dfrac{1}{\sqrt{2}} \:=\:0.7071



    If rationalize first, we have: . \dfrac{1}{\sqrt{2}} \:=\:\dfrac{\sqrt{2}}{2}


    And the division looks like this:

    . . \begin{array}{ccccccc}<br />
&& 0 & .7 & 0 & 7 & 1 \\<br />
&&--&--&--&--&-- \\<br />
2 & | & 1 & .4 & 1 & 4 & 2 \\<br />
&& 1 & 4 \\<br />
&& --&-- \\<br />
&&&& 1 & 4 \\<br />
&&&& 1 & 4 \\<br />
&&&& --&-- \\<br />
&&&&&&2 \\<br />
&&&&&& 2 \\<br />
&&&&&& -- <br />
\end{array}


    Which way do you prefer?

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  9. #9
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    Thank you for all of your answers! I think I can declare this thread as solved
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