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Math Help - Surd Explanation

  1. #1
    Member GAdams's Avatar
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    Surd Explanation

    I have a rectangle sides of sq.rt 2 and 1

    I cut it in half to show that the ratio of the new rectangles is teh same as the original.

    Here's what I did:

    New rectangle is sq.rt 2 / 2 : 1

    That's as far as I got.

    The book says:

    sq.rt 2 / 2 : 1 = 1 / sq.rt 2 : 1 = sq.rt 2 : 1

    -I don't understand how they do this last line.

    Thanks.
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  2. #2
    Member GAdams's Avatar
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    Another problem:

    Do these equate, if so how:

    1 - 1 / sq.rt 2 = (sq.rt 2 - 1) / sq.rt 2
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  3. #3
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    The first one.

    So you cut the sqrt(2) by 1 rectangle into to equal parts. Each half is
    (1/2)sqrt(2) by 1
    Or, [sqrt(2)]/2 by 1
    So the ratio of the new dimensions is
    [sqrt(2)]/2 : 1

    You want to check if that is the same as the ratio in the original, sqrt(2) : 1.

    2 is sqrt(2)*sqrt(2), so,
    [sqrt(2)]/2 : 1
    = [sqrt(2)]/[sqrt(2) *sqrt(2)] : 1
    = 1/[sqrt(2)] : 1
    Multiply both by sqrt(2),
    = 1 : sqrt(2) ---------------------same as the original.

    --------------------------------------
    The second one.

    1 - 1 / sq.rt 2 = (sq.rt 2 - 1) / sq.rt 2

    Clear the fractions, multiply both sides by sqrt(2),
    sqrt(2) -1 = sqrt(2) -1

    Hey, one step only.

    ------------------
    Another way.

    The RHS is (sqrt(2) -1) / sqrt(2)
    So divide,
    = sqrt(2)/sqrt(2) -1/sqrt(2)
    = 1 - 1/sqrt(2)

    That's it.
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  4. #4
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    Hello, GAdams!

    I have a rectangle sides of \sqrt{2} and 1.

    I cut it in half to show that the ratio of the new rectangles is the same as the original.

    Here's what I did:

    New rectangle is \frac{\sqrt{2}}{2}\,:\,1 . . . . not quite correct

    This is the original rectangle:
    Code:
          * - - - - - - - - - *
          |                   |
          |                   |
        1 |                   | 1
          |                   |
          |                   |
          * - - - - - - - - - *
                   √2
    The ratio of length-to-width is: . {\color{blue}\sqrt{2}\,:\,1}


    Bisect it:
    Code:
          * - - - - * - - - - *
          |         |         |
          |         |         |
        1 |         |         | 1
          |         |         |
          |         |         |
          * - - - - * - - - - *
              √2       √2


    Each small rectangle looks like this:
    Code:
          * - - - - - *
          |           |
          |           | √2
          |           |
          * - - - - - *
                1
    The ratio of length-to-width is: . {\color{blue}1\,:\,\frac{\sqrt{2}}{2}}

    And we must show that the two ratios are equal.


    The first ratio is: . \frac{\sqrt{2}}{1} \;=\;{\color{red}\sqrt{2}}


    The second ratio is: . \frac{1}{\frac{\sqrt{2}}{2}} \:=\:\frac{2}{\sqrt{2}}

    . . Rationalize: . \frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} \;=\;\frac{2\sqrt{2}}{2} \;=\;{\color{red}\sqrt{2}}


    Therefore, the ratios are equal.

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  5. #5
    Member GAdams's Avatar
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    1 - 1 / sq.rt 2 = (sq.rt 2 - 1) / sq.rt 2

    Sorry, what I meant to ask is how do I get from

    1 - 1 / sq.rt 2

    to

    (sq.rt 2 - 1) / sq.rt 2
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  6. #6
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    Quote Originally Posted by Soroban View Post


    And we must show that the two ratios are equal.


    The first ratio is: . \frac{\sqrt{2}}{1} \;=\;{\color{red}\sqrt{2}}


    The second ratio is: . \frac{1}{\frac{\sqrt{2}}{2}} \:=\:\frac{2}{\sqrt{2}}

    . . Rationalize: . \frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} \;=\;\frac{2\sqrt{2}}{2} \;=\;{\color{red}\sqrt{2}}


    Therefore, the ratios are equal.



    I don't understand why in the first ratio you have used the 1 as the denominator and then as the numerator for the second ratio.
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  7. #7
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    Hello, GAdams!

    I don't understand why in the first ratio you have used the 1 as the denominator
    and then as the numerator for the second ratio.
    We can't order a proportion randomly.


    In both ratios, it is Length-to-Width.

    . . That is, .the longer side : the shorter side.


    If you don't "line them up" correctly, you're asking for trouble.
    Code:
          * - - - - - - - - - *
          |                   |       * - - *
          |                   |       |     |
       40 |                   |     3 |     |
          |                   |       |     | 
          * - - - - - - - - - *       * - - *
                  60                     2

    These two rectangles are similar; the side are proportional.

    But not because: . \frac{60}{40} \:=\:\frac{2}{3} . . . . which is not true.


    Get the idea?

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  8. #8
    Member GAdams's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, GAdams!

    We can't order a proportion randomly.


    In both ratios, it is Length-to-Width.

    . . That is, .the longer side : the shorter side.


    If you don't "line them up" correctly, you're asking for trouble.
    Code:
          * - - - - - - - - - *
          |                   |       * - - *
          |                   |       |     |
       40 |                   |     3 |     |
          |                   |       |     | 
          * - - - - - - - - - *       * - - *
                  60                     2

    These two rectangles are similar; the side are proportional.

    But not because: . \frac{60}{40} \:=\:\frac{2}{3} . . . . which is not true.


    Get the idea?

    Yes. So in your example it should be 60/40 : 3/2

    In the original example it's 1 over the second time because that's the longer side.


    Thank you!
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  9. #9
    Member GAdams's Avatar
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    Quote Originally Posted by GAdams View Post
    1 - 1 / sq.rt 2 = (sq.rt 2 - 1) / sq.rt 2

    Sorry, what I meant to ask is how do I get from

    1 - 1 / sq.rt 2

    to

    (sq.rt 2 - 1) / sq.rt 2

    .........
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  10. #10
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    Quote Originally Posted by GAdams View Post
    1 - 1 / sq.rt 2 = (sq.rt 2 - 1) / sq.rt 2

    Sorry, what I meant to ask is how do I get from

    1 - 1 / sq.rt 2

    to

    (sq.rt 2 - 1) / sq.rt 2
    1 - 1/sqrt(2)

    = 1/1 -1/sqrt(2)

    Combine them. The common denominator is sqrt(2),
    = (1*sqrt(2) -1) / sqrt(2)
    = (sqrt(2) -1) / sqrt(2)
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  11. #11
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    Quote Originally Posted by ticbol View Post
    1 - 1/sqrt(2)

    = 1/1 -1/sqrt(2)

    Combine them. The common denominator is sqrt(2),
    = (1*sqrt(2) -1) / sqrt(2)
    = (sqrt(2) -1) / sqrt(2)
    I still don't get it.

    (1 ) - 1 / sq.rt 2 is what I start with

    Which is teh same as:

    (1 / 1) - (1/ sq.rt 2) I got that.

    Combining them would give: (1 - 1) / (1 - sq.rt 2)

    You seem to be multiplying the (1 - 1) by sq.rt 2.?
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  12. #12
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    Quote Originally Posted by GAdams View Post
    I still don't get it.

    (1 ) - 1 / sq.rt 2 is what I start with

    Which is teh same as:

    (1 / 1) - (1/ sq.rt 2) I got that.

    Combining them would give: (1 - 1) / (1 - sq.rt 2) <----this is not correct.

    You seem to be multiplying the (1 - 1) by sq.rt 2.?
    You seem to be multiplying the (1 - 1) by sq.rt 2.?

    No. I multiplied the numerator 1 of the 1/1 by sqrt(2).

    It is subtraction of fractions.
    Example, 1/2 -3/4
    Common denominator is 4.
    4 divided by 2 ...[equals 2], times 1, equals 2
    4 divided by 4 ...[equals 1] , times 3, equals 3
    So,
    1/2 -3/4
    = (2 -3)/4
    = -1/4

    Now compare that to your surd above.
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  13. #13
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    Quote Originally Posted by ticbol View Post
    You seem to be multiplying the (1 - 1) by sq.rt 2.?

    No. I multiplied the numerator 1 of the 1/1 by sqrt(2).

    It is subtraction of fractions.
    Example, 1/2 -3/4
    Common denominator is 4.
    4 divided by 2 ...[equals 2], times 1, equals 2
    4 divided by 4 ...[equals 1] , times 3, equals 3
    So,
    1/2 -3/4
    = (2 -3)/4
    = -1/4

    Now compare that to your surd above.
    Ok:

    (1 / 1) - (1/ sq.rt 2)

    common denomintaor is sq.rt 2

    So,

    sq.rt 2 divide by 1 = sqrt2/1 times 1 is sq.rt2/1
    sq.rt 2 divide by sq.rt 2 = 1 times 1 is 1

    sq.rt 2 - 1 / sq.rt 2

    Voila!!

    I didn't know that subtraction method for fractions! Thanks. I know it now.
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