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Math Help - Writing a Fraction in Simplest Form

  1. #1
    Junior Member Euclid Alexandria's Avatar
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    Question Writing a Fraction in Simplest Form

    Say we are dividing out the common factors in \frac{7 \cdot a \cdot a \cdot a}{7 \cdot 8 \cdot a \cdot a}

    Is an answer of \frac{1a}{8} just as valid and correct as \frac{a}{8}?
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    Hello, Euclid Alexandria!

    Say we are dividing out the common factors in \frac{7 \cdot a \cdot a \cdot a}{7 \cdot 8 \cdot a \cdot a}

    Is an answer of \frac{1a}{8} just as valid and correct as \frac{a}{8}?
    Yes, it is . . . but a coefficient of 1 is usually omitted.


    Similarly: . \frac{7\cdot a\cdot a\cdot a}{a\cdot a} is equal to \frac{7a}{1},
    . . but a denominator of 1 is usually omitted.
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  3. #3
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Euclid Alexandria
    Say we are dividing out the common factors in \frac{7 \cdot a \cdot a \cdot a}{7 \cdot 8 \cdot a \cdot a}

    Is an answer of \frac{1a}{8} just as valid and correct as \frac{a}{8}?
    I've had some teachers that marked me down for doing something like that so if you're giving the answers to a test, it is better to get rid of the 1.
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  4. #4
    Junior Member Euclid Alexandria's Avatar
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    Thumbs up

    Thanks guys, I'll get in the habit of omitting the 1 just in case.

    Oh hey look, I just became a senior member.
    Last edited by Euclid Alexandria; June 23rd 2006 at 05:54 PM. Reason: Bragging about senior member status
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  5. #5
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    Quote Originally Posted by Euclid Alexandria
    Thanks guys, I'll get in the habit of omitting the 1 just in case.

    Oh hey look, I just became a senior member.
    That means you're old now. You're entitled to some discounts now. You can be given incomplete answers now.
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  6. #6
    Junior Member Euclid Alexandria's Avatar
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    Smile

    Quote Originally Posted by ticbol
    That means you're old now. You're entitled to some discounts now. You can be given incomplete answers now.
    I'm glad to hear \frac{3 \cdot a \cdot a}{9 \cdot a \cdot a} of that!
    Last edited by Euclid Alexandria; June 24th 2006 at 01:05 PM. Reason: Snarkiness
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  7. #7
    Super Member malaygoel's Avatar
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    Post Writing fraction in simplest form

    When we write a fraction in the simplest form, we cancel out what is common in the numerator and denominator. This is what a teacher tells us in a class. After this the practice problem was given:
    Simplify \frac{16}{64}
    One student cancelled out the 6(common in numerator and denominator) and got the answer \frac{1}{4}, which was suprisingly the right answer. How many more fractions are there like this?

    KeepSmiling
    Malay
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    Hello, Malay!

    Besides \frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}, we have:

    . . \frac{2\!\!\!\not{6}}{\not{6}5} = \frac{2}{5}

    . . \frac{1\!\!\!\not{9}}{\not{9}5} = \frac{1}{5}

    . . \frac{4\!\!\!\not{9}}{\not{9}8} = \frac{4}{8} = \frac{1}{2}
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  9. #9
    Junior Member Euclid Alexandria's Avatar
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    Cool

    Quote Originally Posted by Soroban
    Hello, Malay!

    Besides \frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}, we have:

    . . \frac{2\!\!\!\not{6}}{\not{6}5} = \frac{2}{5}

    . . \frac{1\!\!\!\not{9}}{\not{9}5} = \frac{1}{5}

    . . \frac{4\!\!\!\not{9}}{\not{9}8} = \frac{4}{8} = \frac{1}{2}
    Interesting...are these all the two-digit fractions that are like this? Did you know this intuitively or did you use a method to figure it out?
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  10. #10
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    Hello, Euclid Alexandria!

    Are these all the two-digit fractions that are like this?
    Did you know this intuitively or did you use a method to figure it out?

    I had seen these listed a book many years ago.
    But you can derive your own formula for finding these "jokes".

    We want digits a,b,c so that: . \frac{10a + b}{10b + c} \:= \:\frac{a}{c}

    Solve for c:\;\;c\:=\:\frac{10ab}{9a + b}

    Now try all digits for a and b which make c a digit.

    You'll find that the list I gave is complete.

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

    One more joke:


    Reduce: . \frac{(1 + x)^2}{1 - x^2}

    Answer: . \frac{(1 + x)^{\not{2}}} {1 - x^{\not{2}}} \;=\;\frac{1 + x}{1 - x}

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  11. #11
    Junior Member Euclid Alexandria's Avatar
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    Exclamation

    Quote Originally Posted by Soroban
    I had seen these listed a book many years ago.
    But you can derive your own formula for finding these "jokes".

    We want digits a,b,c so that: . \frac{10a + b}{10b + c} \:= \:\frac{a}{c}

    Solve for c:\;\;c\:=\:\frac{10ab}{9a + b}

    Now try all digits for a and b which make c a digit.

    You'll find that the list I gave is complete.
    I am having difficulty following the example and reproducing the results. Here are the steps I took to reproduce the example:

    <br />
\frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}<br />

    Let a = 1, b = 6, and c = 4.

    \frac{10a + b}{10b + c} = \frac{a}{c}

    c = \frac{10ab}{9a + b}

    \frac{10 \cdot 1 \cdot 6}{10 \cdot 6 + 4}

    = \frac{16}{20}

    = \frac{1 \cdot \not2 \cdot 3}{5 \cdot \not2 \cdot 2}

    = \frac{3}{10}
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  12. #12
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    Hello, Euclid Alexandria!

    A few really gruesome errors . . .

    I am having difficulty following the example and reproducing the results. . . . No wonder!

    Here are the steps I took to reproduce the example:

    \frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}

    Let a = 1, b = 6, and c = 4.

    \frac{10a + b}{10b + c} = \frac{a}{c}

    c = \frac{10ab}{9a + b}

    = \frac{10 \cdot 1 \cdot 6}{10 \cdot 6 + 4}
    . . . . ↑? . .↑?
    .ten? . . b = 6
    . . . . a = 1


    . 10򈚖 = 60
    . . . ↓?
    = \frac{16}{20}
    . . . ↑?
    . 106 + 4 = 64


    . What is this? . 16 = 2򈭾2
    . . . . .
    = \frac{1 \cdot \not2 \cdot 3}{5 \cdot \not2 \cdot 2}

    = \frac{3}{10}
    Can you possibly make any more mistakes?
    Last edited by Soroban; June 28th 2006 at 05:16 AM.
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban
    Hello, Euclid Alexandria!

    A few really gruesome errors . . .


    Can you possibly make any more mistakes?
    It's mathematically incorrect, but I thought she had done it on purpose. If so, I found it rather humerous. Here's why:

     \frac{10 \cdot 1 \cdot 6}{10 \cdot 6 +4}
    Well, 10 x 1 = 10 and thus 10 x 1 x 6 = 10 x 6 = 1(0+6)=16.
    6 + 4 = 10 so 10 * 6 + 4 = 10 * 10 = (1+1)0 = 20.

    \frac{16}{20}
    Now, 16 = 1 (6) = 1 * (2 * 3)
    20 = 5 * 2 * 2

    \frac{1 \cdot 2 \cdot 3}{5 \cdot 2 \cdot 2}=\frac{3}{10}

    All of the Math mistakes are common (at least I've seen alot of them) mistakes similar to the cancelling joke that started this whole conversation.

    (At least I HOPE she was making a joke out of it!!)

    -Dan
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  14. #14
    Junior Member Euclid Alexandria's Avatar
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    Unhappy

    Yes, it is possible to make more mistakes in this instance.

    Thoughtful pause

    Here we are attempting to reproduce \frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4} as before, except with a newly revised method.

    Let a = 1, b = 6, and c = 4

    <br />
\frac{10a + b}{10b + c} = \frac{a}{c}<br />

    <br />
c = \frac{10ab}{9a + b}<br />

    = \frac{10 \cdot 1 \cdot 6}{9 \cdot 1 + 6}
    . . .. ...
    .nine! . b = 6
    . . . . a = 1


    = \frac{60}{15} = \frac{2 \cdot 2 \cdot \not3 \cdot \not5}{\not 3 \cdot \not5}

    = \frac{4}{1} = 4
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  15. #15
    Junior Member Euclid Alexandria's Avatar
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    Red face

    Here's another try at using that formula (if that's the correct terminology) to reproduce \frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}

    Let a = 1, b = 6, and c = 4

    \frac{10a + b}{10b + c} = \frac{a}{c}

    = \frac{10 \cdot 1 + 6}{10 \cdot 6 + 4} = \frac{16}{64}

    (Now it appears that a = 16 and c = 64)

    c = \frac{10ab}{9a+b}

    = \frac{10 \cdot 16 \cdot 6}{9 \cdot 16 + 6} = \frac{960}{150}

    = \frac{\not2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot \not3 \cdot \not5}{\not2 \cdot \not3 \cdot \not5 \cdot 5}

    = \frac{32}{5}
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