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Math Help - surds fraction question

  1. #1
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    surds fraction question

    ok bit of a silly question but i really dont understand how to rearange

    \frac{\sqrt{2}}{2\sqrt{y}}
    to get
    \frac{1}{\sqrt{2y}}
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by renlok View Post
    ok bit of a silly question but i really dont understand how to rearange

    \frac{\sqrt{2}}{2\sqrt{y}}
    to get
    \frac{1}{\sqrt{2y}}

    \frac{\sqrt{2}}{(\sqrt{2})^2}. It's of the form \frac{a}{a^2} = \frac{1}{a}
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  3. #3
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    but would you not have to square the \sqrt{y} aswell?
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by renlok View Post
    but would you not have to square the \sqrt{y} aswell?
    No, you can treat this sum as \frac{\sqrt2}{2} \times \frac{1}{\sqrt{y}}

    As \frac{\sqrt2}{2} = \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}} = \frac{1}{\sqrt2} then it becomes \frac{1}{\sqrt2 \sqrt{y}}

    As \sqrt{a} \sqrt{b} = \sqrt{ab} \: , \: a,b \geq 0 we can combine the square roots in the denominator:

    \frac{1}{\sqrt{2y}}
    Last edited by e^(i*pi); October 30th 2009 at 07:45 AM. Reason: adding a step to make it clearer
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  5. #5
    Member rowe's Avatar
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    Nice and quick, but little motivation:

    \frac{\sqrt{2}}{2\sqrt{y}} \cdot \frac{\sqrt{2}}{\sqrt{2}}

    =\frac{2}{2\sqrt{y}\sqrt{2}}

    =\frac{1}{\sqrt{y}\sqrt{2}}

    =\frac{1}{\sqrt{2y}}
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