Thread: math: why do we divide the way we do?

  1. #1
    Set Apart -- jrahhali's Avatar
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    math: why do we divide the way we do?

    Code:
        2.8
       ___
    5| 14
       -10
       ---
        4 0
    i'm a bit confused behind the logic of this "long hand" way to division.

    ok, so
    5 goes into 14 2 times, so you write a 2. you then subtract the amount of times 5 can go into 14 from 14, and you get 4. Up to this point, i understand it.

    why, now, do you bring down a 0, and see how many times 5 can go into 40? why does this give you your fractional part, logically?
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  2. #2
    Mayor of Awesometown Govtcheez's Avatar
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    Technically it's how many times it goes into 4.0, but it's a lot easier to look at it as 40 instead. The decimal point does actually come down, but we never show it because it's a lot easier to learn that way.

  3. #3
    Registered User VirtualAce's Avatar
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    A whole number is not just the integer part that you see.

    4/100 = 25

    But it's 4.00... /100.00...

    The zeroes are not made up, they are really there since there is 0 amount of tenths, hundreths, etc. That's why you bring the zero down because it is really there and it allows you to continue the division.

  4. #4
    aoeuhtns
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    The algorithm uses this relationship:

    For all values of k, x/y = k/y + (x-k)/y

    For example:
    Code:
    To divide 84732/7, let:
    
    84732/7 = 10000 * (8/7)     + 4732/7  (We lop off all but the last
            = 10000 * (1 + 1/7) + 4732/7   four digits.)
            = 10000 + 10000/7   + 4732/7
            = 10000 + 14732/7             (At this point, we write down '1'.)
    
    14732/7 =  1000 * (14/7)    + 732/7   (We lop off all but the last
            =  1000 * (2 + 0/7) + 732/7    three digits.)
            =  2000 + 0/7       + 732/7
            =  2000 + 732/7               (At this point, we write down '2'.)
    
      732/7 =   100 * (7/7)     + 32/7    (We lop off all but the last
            =   100 * (1 + 0/7) + 32/7     two digits.)
            =   100 + 0/7       + 32/7
            =   100 + 32/7                (At this point, we write down '1'.)
    
       32/7 =    10 * (3/7)     + 2/7     (We lop off all but the last
            =    10 * (0 + 3/7) + 2/7      one digit.)
            =     0 + 30/7      + 2/7
            =     0 + 32/7                (At this point, we write down '0'.)
    
       32/7 =     1 * (32/7)    + 0/7     (We lop off all the digits.)
            =     1 * (4 + 4/7) + 0/7
            =     4 + 4/7       + 0/7
            =     4 + 4/7                 (At this point, we write down '1'.)
    
        4/7 =   0.1 * (40/7)    + 0/7     (We 'lop off' past the decimal
            =   0.1 * (5 + 5/7) + 0/7      place.)
            =   0.5 + 0.5/7     + 0/7
            =   0.5 + 0.5/7               (We write a decimal point and a
                                           '5'.)
    
      0.5/7 =  0.01 * (50/7)    + 0/7
            =  0.01 * (7 + 1/7) + 0/7
            =  0.07 + 0.01/7    + 0/7
            =  0.07 + 0.01/7              (We write down a '7'.)
    
    
    And so on and so forth.
    Substituting expressions back up, we get the answer:
    Code:
    10000 + 2000 + 100 + 00 + 4 + 0.5 + 0.7 + 0.01/7
    = 12104.57 + 0.01/7

  5. #5
    Registered User VirtualAce's Avatar
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    Ok.

    Our numbers are base 10. So:

    12500 is

    1 * (10^4)+
    2 * (10^3)+
    5* (10^2)+
    0 *(10^1)+
    0 *(10^0)

    for decimals

    .01 is
    0 * (10^-1)+
    1 * (10^-2)

    I'm not sure where the other stuff from above came from, but this is why we drop zeros. If the value was not zero, you would drop that value.

  6. #6
    Registered User white's Avatar
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    did anyone knows that there is a similar way of finding the square root of any number?
    http://www.nist.gov/dads/HTML/squareRoot.html
    ----------------

  7. #7
    Set Apart -- jrahhali's Avatar
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    >>Technically it's how many times it goes into 4.0, but it's a lot easier to look at it as 40 instead.

    ok. since there are 5 8's in 40, then there should 5 0.8's in 4.
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  8. #8
    Rad gcn_zelda's Avatar
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    Quote Originally Posted by white
    did anyone knows that there is a similar way of finding the square root of any number?
    http://www.nist.gov/dads/HTML/squareRoot.html
    I think I'll stick to local linear approximation.

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