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Math Help - Working in base 12...

  1. #1
    Junior Member Bartimaeus's Avatar
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    Question Working in base 12...

    To work in base 12, we must invent new symbols for digits 10 and 11. Use t for ten and e for eleven. For example, 2x5=t 12
    a Write out 3, 5 and 8 times table in base 12, up to e times.
    b In the times tables in a, some have answers with the last digit 0. Which tables in base 10 [up to 9 times] have answers with the last digit 0?
    c Explain why some tables have answers with last digit 0 and others do not, in base 10 [up to 9 times] and in base 10 [up to e times]
    d Which tables in base 12 [up to e times] will have no answers with last digit 0?
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  2. #2
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    Hello, Bartimaeus!

    I assume you are somewhat familiar with base-12 notation.


    To work in base-12, we must invent new symbols for digits 10 and 11.
    Use t for ten and e for eleven. For example, 2 \times 5 = t_{12}

    (a) Write out 3-, 5- and 8-times table in base-12, up to e-times.
    Code:
      x |  1 |  2 |  3 |  4 |  5 |  6 |  7 |  8 |  9 |  t |  e |
     -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- +
      3 |  3 |  6 |  9 | 10 | 13 | 16 | 19 | 20 | 23 | 26 | 29 |
     -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- +
      5 |  5 |  t | 13 | 18 | 21 | 26 | 2e | 34 | 39 | 42 | 47 |
     -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- +
      8 |  8 | 14 | 20 | 28 | 34 | 40 | 48 | 54 | 60 | 68 | 74 |
     -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- +
    (b) In the times tables in (a), some have answers with the last digit 0.
    Which tables in base-10 [up to 9-times] have answers with the last digit 0?

    In base-10, the rows that have answers ending in 0 are:
    . . 2-times, 4-times, 6-times, 8-times, and 5-times.



    (c) Explain why some tables have answers with last digit 0 and others do not
    in base-10 [up to 9-times] and in base-12 [up to e times]

    For base-ten: 10 = 2\cdot5
    Any row which shares a common factor with 10 has answers ending in 0.
    Any row which is relative prime to 10 has no answers ending in 0.

    For base=twelve: 12 = 2^2\cdot3
    Any row which shares a common factor with 12 has answers ending is 0.
    . . They are: 2-times, 3-times, 4-times, 6-times, 8-times, 9-times, and t-times



    (d) Which tables in base-12 [up to e-times] will have no answers with last digit 0?

    Any row which is relatively prime to 12 has no answers ending in 0.
    . . They are: 5-times, 7-times, and e-times.

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  3. #3
    Junior Member Bartimaeus's Avatar
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    Thankyou Soroban. You are a Super member.
    I am a little familiar with writing in bases, but never in any higher than base ten. Is there, I suppose you could call it, formula, table or a trick in working in converting on base into another?
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    Hello again, Bartimaeus!

    There is a procedure for changing a base-ten number to another base.


    Example: write 389 in base-5.

    Step 1: Divide 389 by 5 and note the remainder:
    . . . . . . 389 \div 5\:=\:77\quad rem.\boxed{4}

    Step 2: Divide the quotient by 5 and note the remainder:
    . . . . . . . 77 \div 5 \,=\,15\quad rem.\boxed{2}

    Repeat Step 2 until a zero quotient each reached.
    . . . . . . . 15 \div 5 \,=\,3\quad rem.\boxed{0}
    . . . . . . . . 3 \div 5\,=\,0\quad rem.\boxed{3}\;\;\uparrow

    Now read up the remainders: . 3024_5


    The division can be written like this:
    Code:
        5 ) 3 8 9
          -------
          5 ) 7 7   4
            -----
          5 ) 1 5   2
            -----
            5 ) 3   0
              ---
                0   3 ↑
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  5. #5
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    Now, suppose you wanted to change this base 5 to base 10.

    3024_{5}

    Starting from the far right multiply each digit by successive powers of 5(starting with 0):

    4*5^{0}=4
    2*5^{1}=10
    0*5^{2}=0
    3*5^{3}=375

    Now add them up: 375+0+10+4=389
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    Nice explanation, Cody!

    But has anyone ever seen this method?
    [The explanation/derivation is long, but the method is very fast.]


    Example: .Evaluate f(x) = 2x^3 - 5x^2 + 3x - 1 at x = 3

    We have: . f(x)\:=\:2x^3 - 5x^2 + 2x - 1

    Factor x from the first three terms: . 2x^2 - 5x + 3)\cdot x - 1" alt="f(x)\:=\2x^2 - 5x + 3)\cdot x - 1" />

    Factor x from the first two terms: . [2x-5]\cdot x + 3)\cdot x - 1" alt="f(x)\:=\[2x-5]\cdot x + 3)\cdot x - 1" />

    Then: . [2\cdot3 - 5]\cdot3 + 3)\cdot3 - 1" alt="f(3)\:=\[2\cdot3 - 5]\cdot3 + 3)\cdot3 - 1" />

    Now "read" the steps we will take: . \begin{array}{cccccc}\text{2 times 3}\\ \text{minus 5} \\ \text{times 3} \\ \text{plus 3} \\ \text{times 3} \\ \text{minus 1}\end{array} . \begin{array}{cccccc}6\\ 1\\3\\6\\18\\\boxed{17}\end{array}

    We could have done this mentally ... no squaring or cubing involved.

    But if all that factoring is required, it's not much of a shortcut, is it?


    Look again at the factored polynomial: . f(x)\:=\:\left([2x - 5]\[\cdot x + 3\right)\cdot x - 1

    "Drop" the grouping symbols: . 2x - 5\cdot x + 3\cdot x - 1 \;= \;2x - 5x + 3x - 1

    This is the original polynomial with the exponents removed . . .


    If we are given: . g(x)\;=\;3x^4 - 7x^3 + 4x^2 - 9x + 8
    . . we can (mentally) drop the exponents: . 3\cdot x - 7\cdot x + 4\cdot x - 9\cdot x + 8
    . . and we have the factored form (minus the grouping symbols).

    For g(2): 3 times 2, minus 7, times 2, plus 4, times 2, minus 9, times 2, plus 8

    Important: We must complete each addition/subtraction before the next multiplication.

    On our calculator, it would be:
    . . 3\;\boxed{\times}\;2\;\boxed{-}\;7\;\boxed{=}\;\;\boxed{\times}\;2\;\boxed{+}\;4  \;\boxed{=}\;\;\boxed{\times}\;2\;\boxed{-}\;9\;\boxed{=}\;\;\boxed{\times}\;2\;\boxed{+}\;8  \;\boxed{=}

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

    Back to base-five . . .

    Example: Convert 3024_5 to base-ten.

    The value is: . 3\cdot5^3 + 0\cdot5^2 + 2\cdot5 + 4 .and we have a "polynomial in 5".


    So we have: . 3\cdot5 + 0\cdot 5 + 2\cdot 5 + 4\quad\Rightarrow\quad\begin{array}{cccccc}\text{3 times 5} \\ \text{plus 0}\\ \text{times 5}\\ \text{plus 2}\\ \text{times 5}\\ \text{plus 4}\end{array} \begin{array}{cccccc}15\\15\\75\\77\\385\\\boxed{3  89}\end{array}

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

    This procedure is even more spectacular in base-two.


    Example: Write 101011_2 in base-ten.

    We have: . 1\times2 + 0\times2 + 1\times 2 +0\times2 + 1\times2 + 1
    . . . . . . . . . . . 2\quad\,2\quad\;4\quad5\;\;\;10\;\;\;10\;\;\;20\; \;\;21\;\;42\;\;\:\boxed{43}<br /> <br />

    You have the answer while others are discovering that it begins with 2^5.

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  7. #7
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    Quote Originally Posted by Soroban
    But has anyone ever seen this method?
    It is Horner's Rule.

    And reading that I see "The Horner scheme is often used to convert between different positional numeral systems in which case x is the base of the number system, and the ai coefficients are the digits of the base-x representation of a given number." And I didn't know that connection. So thanks, Soroban!
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  8. #8
    dan
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    Question

    So, I have a question...
    Is it possible to write a given number x in a given base y where y is not a integer i.e. write 1220 is base e. The problem I'm having is that to use the traditional method for converting bases with remainders, your remainder is a decimal. Also the practicality seems to run down when you realize that you have e (2.71828182846...) numbers in your system.
    Dan
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    Quote Originally Posted by dan
    So, I have a question...
    Is it possible to write a given number x in a given base y where y is not a integer i.e. write 1220 is base e. The problem I'm having is that to use the traditional method for converting bases with remainders, your remainder is a decimal. Also the practicality seems to run down when you realize that you have e (2.71828182846...) numbers in your system.
    Dan
    No, for example you cannot express,
    \pi = a_0+a_1e+a_2e^2+...+a_ne^n
    Because, they are algebraically independent.
    Meaning there is not polynomial with rational coefficients that when evaluavated at e gives \pi.
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by dan
    So, I have a question...
    Is it possible to write a given number x in a given base y where y is not a integer i.e. write 1220 is base e. The problem I'm having is that to use the traditional method for converting bases with remainders, your remainder is a decimal. Also the practicality seems to run down when you realize that you have e (2.71828182846...) numbers in your system.
    Dan
    One system: Golden Ratio Base.

    I think this is dealt with in general in volume 2 of Don Knuth's TAOCP "SemiNumerical Algorithms", but my copy is at work so I can't check it now.

    RonL
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    Eater of Worlds
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    Quote Originally Posted by ThePerfectHacker
    No, for example you cannot express,
    \pi = a_0+a_1e+a_2e^2+...+a_ne^n
    Because, they are algebraically independent.
    Meaning there is not polynomial with rational coefficients that when evaluavated at e gives \pi.
    I believe those are called 'transcendental numbers'. Pi and e are classic examples.
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  12. #12
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    Quote Originally Posted by ThePerfectHacker
    No, for example you cannot express,
    \pi = a_0+a_1e+a_2e^2+...+a_ne^n
    Because, they are algebraically independent.
    Meaning there is not polynomial with rational coefficients that when evaluavated at e gives \pi.
    Have I missed something? Why would you expect the base e representation of \pi to terminate?

    RonL
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  13. #13
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    Quote Originally Posted by CaptainBlack
    Have I missed something? Why would you expect the base e representation of \pi to terminate?

    RonL
    No I said you cannot express it.
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  14. #14
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    Quote Originally Posted by ThePerfectHacker
    No I said you cannot express it.
    To express \pi in base e we would like a representation in the form:

    <br />
\pi = \sum_{r=n_0}^{\infty} a_r\ e^{-r}<br />

    where n_0, may be negative.

    In fact does the greedy algorithm not give a construction with all the a_is are either 0,\ 1 or 2?

    RonL
    Last edited by CaptainBlack; July 30th 2006 at 10:44 AM.
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  15. #15
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack
    To express \pi in base e we would like a representation in the form:

    <br />
\pi = \sum_{r=n_0}^{\infty} a_r\ e^{-r}<br />

    where n_0, may be negative.

    In fact does the greedy algorithm not give a construction with all the a_is are either 0,\ 1 or 2?

    RonL
    It appears that in base e one representation of \pi is:

    <br />
\pi \approx 10.101002020..._{\mbox{base e}}<br />

    Is this unique? I think so, but haven't checked.

    RonL
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