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Math Help - Canceling Down Confusion

  1. #1
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    Canceling Down Confusion

    Thanks loads to Chop Suey for help with latex


    I know this is going to be so simple, I just can't see it yet.

    I am working on canceling down fractions to make them easier to work with.

    For example

    <br /> <br />
\frac{6}{8} = \frac{6/2}{8/2} = \frac{3}{4}<br /> <br />

    I understand how to do it and thought I was fine. Then the book I'm using throws the following at me and I can't understand how they got to their answer.

    <br /> <br />
\frac{3}{7} * \frac{5}{6} = \frac{1}{7} * \frac{5}{2} = \frac{1*5}{7*2} = \frac{5}{14}<br /> <br />


    IN the book they say canceling is possible because 3 divides into both the 3 and the 6, which is true but I thought the numerator and the denominator were divided by their highest common factor to cancel down.

    This has confused me.

    I think I may have it figured out, however I don't want to get carried away with the wrong idea.

    I can see the they are doing 6/3 = 2 then 3/3 = 1, obviously 7/5 doesn't work as it produces a decimal so this is not canceled down. I found out their method by looking at some other examples in the book, however when they explained canceling down they said to find the highest common factor of the numerator and denominator then divide the numerator and denominator by it.

    e.g

    <br /> <br />
\frac{24}{60} = \frac{24/6}{30/6} = \frac{4}{5}<br /> <br />

    Many thanks in advance.
    Last edited by tkdvipers; August 11th 2008 at 01:02 PM. Reason: Math was messed up
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  2. #2
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    I can't understand what you wrote, but here's a quick list for what need to write it in latex.

    For every expression you want to write, begin by writing the math tags: [ math][/tex] (without that extra space)

    Write your code between [ math] and [/tex].

    To form a fraction, use \frac{numerator}{denominator}. Example:

    \frac{22}{7} will become \frac{22}{7}

    You can use the symbols + - * for arithmetic operations.
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  3. #3
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    What you are referring to is called reducing to lowest terms. Now you said this confused you:

    <br /> <br />
\frac{3}{7} * \frac{5}{6} = \frac{1}{7} * \frac{5}{2} = \frac{1*5}{7*2} = \frac{5}{14}<br /> <br />

    Look at it differently:

    \frac{3}{7} \cdot \frac{5}{6} = 3 \cdot \frac{1}{7} \cdot 5 \cdot \frac{1}{6}

    It is legal to rearrange them. With this fraction, it's more easier if you notice that:

    \frac{3}{6}

    can be reduced to:

    \frac{1}{2}

    Back to the original fraction:

    \frac{3}{6} \cdot \frac{5}{7} = \frac{1}{2} \cdot \frac{5}{7} = \frac{5}{14}

    Another way of doing this is simply multiply both fractions and then reduce:

    \frac{3}{7} * \frac{5}{6} = \frac{15}{42}

    You should realize that you can divide both by 3:

    \frac{15/3}{42/3} = \frac{5}{4}
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  4. #4
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    Cheers

    I am getting there,

    What confuses me is the way that.

    <br /> <br />
\frac{3}{7} * \frac{5}{6} <br /> <br />

    becomes

    <br /> <br />
 \frac{1}{7} * \frac{5}{2}<br /> <br />

    I can see that 6/3 = 2 and 3/3 = 1, however I thought the fractions were unrelated and you could only work within the same fraction.

    for example 3/7 and 5/6

    I can do the sums now but want to understand why it is legal.
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  5. #5
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    Let's look at the fraction you gave again:

    \frac{3}{7} \cdot \frac{5}{6}

    Do you agree that this can be written like this?

    3 \cdot \frac{1}{7} \cdot 5 \cdot \frac{1}{6}

    Consider these two fractions:

    \frac{a}{b} and \frac{c}{d} where b and d are not equal to zero.

    Let's multiply them:

    \frac{a}{b} \cdot \frac{c}{d} = a \cdot \frac{1}{b} \cdot c \cdot \frac{1}{d} = \frac{c}{b} \cdot \frac{a}{d}

    Since you are multiplying the two fractions, you can manipulate them.
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  6. #6
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    Thank you very much.

    I do get it, I am just going to have to practice.

    You have been very very helpful.
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  7. #7
    Super Member Matt Westwood's Avatar
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    Even better, you can look at it like this:

    \frac a b \cdot \frac c d = \frac {a \cdot c} {b \cdot d} = \frac {a \cdot c} {d \cdot b} = \frac {c \cdot a} {b \cdot d}

    and so on.

    Any factor that appears in both the top and the bottom can be cancelled out.

    For example:

    \frac 3 5 \cdot \frac 5 6 = \frac {3 \cdot 5} {5 \cdot 6} = \frac {5 \cdot 3} {5 \cdot 6} = \frac 5 5 \cdot \frac 3 6  = 1 \cdot \frac 3 {2 \cdot 3} = 1 \cdot \frac 3 3 \cdot \frac 1 2 = 1 \cdot 1 \cdot \frac 1 2 = \frac 1 2
    Last edited by Matt Westwood; August 11th 2008 at 01:36 PM. Reason: typos
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  8. #8
    A riddle wrapped in an enigma
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    Quote Originally Posted by tkdvipers View Post
    Cheers

    I am getting there,

    What confuses me is the way that.

    <br /> <br />
\frac{3}{7} * \frac{5}{6} <br /> <br />

    becomes

    <br /> <br />
 \frac{1}{7} * \frac{5}{2}<br /> <br />

    I can see that 6/3 = 2 and 3/3 = 1, however I thought the fractions were unrelated and you could only work within the same fraction.

    for example 3/7 and 5/6

    I can do the sums now but want to understand why it is legal.
    Look at the cancelation that took place in red. The 3 and the 6 have a common factor of 3.

    <br /> <br />
 \frac{3}{7} * \frac{5}{6} = \color{red}\frac{\rlap{---}{3}^1}{7} * \color{red}\frac{5}{\rlap{---}{6}^2} =\color{black}\frac{1}{7}\cdot\frac{5}{2}= \frac{1\cdot5}{7\cdot2}=\frac{5}{14}<br /> <br />

    Another approach:

    \frac{3}{7} \cdot \frac{5}{6}=\frac{3}{6} \cdot \frac{5}{7}=\color{red}<br />
\frac{\rlap{---}{3}^1}{\rlap{---}{6}^2}\cdot\frac{5}{7}=<br />
\color{black}\frac{1\cdot5}{2\cdot7}=\frac{5}{14}<br />
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