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Urgend Assistance needed: Abstract Algebra

  1. Sep 27, 2006 #1
    Hi I'm fairly new at abstract algebra and have therefore got stuck with this assignment.

    Hope there is somebody here who can help me complete it, because I have been ill these last couple of weeks.

    Its goes something like this

    b is a number written in base 10

    [tex]b\;= \;b_010^0 + b_110^1 + b_210^2 + \hdots + b_n10^n[/tex]

    where [tex]0 \leq b_j \leq 10[/tex]

    (a) prove that 2 divides b if a only if 2 divides b_0.

    My Solution:

    Let [tex]b[/tex] be a number written in base 10 as:

    [tex]b\;= \;b_010^0 + b_110^1 + b_210^2 + \hdots + b_n10^n[/tex] where [tex]0 \leq b_i < 10[/tex]

    Prove that: .[tex]2|b \;\Longleftrightarrow \;2|b_0[/tex]
    [/quote]
    Given: .[tex]2|b[/tex], we have:

    . . [tex]b \;=\;10^nb_n + 10^{n-1}b_{n-1} + 10^{n-2}b_{n-2} + \hdots + 10^2b_2 + 10b_1 + b_o \;=\;2k[/tex] for some integer [tex]k.[/tex]

    . . [tex]b_o \;=\;2k - \left(10^nb_n + 10^{n-1}b_{n-1} + 10^{n-2}b_{n-2} + \hdots + 10^2b_2 + 10b_1\right)[/tex]

    . . [tex]b_o \;= \;2k\:-\:\left(2\!\cdot\!5\1\cdot\!10^{n-1}a_n + 2\!\cdot5\1\cdot\!10^{n-1}b_{n-1} + \hdots + 2\!\cdot\!5\!\cdot\!10b_2 + 2\!\cdot\!5b_1\right)[/tex]

    . . [tex]b_o \;= \;2k\:-\:2\left(5\!\cdot\!10^{n-1}b_n + 5\!\cdot\!10^{n-1}b_{n-1} + \hdots + 5\!\cdot\!10b_2 + 5b_1\right)[/tex]

    . . [tex]b_o \;= \;2\left(k - 5\!\cdot\!10^{n-1}b_n - 5\!\cdot\!10^{n-1}b_{n-1} - \hdots - 5\!\cdot\!10b_2 - 5b_1\right)[/tex]

    The right side is a multiple of 2, hence the left side is a multiple of 2.

    Therefore: .[tex]2|b_o[/tex]



    Given: .[tex]2|b_o[/tex], then [tex]b_o = 2k[/tex] for some integer [tex]k.[/tex]

    Then: .[tex]b \;=\;10^nb_n + 10^{n-1}b_{n-1} + 10^{n-2}b_{n-2} + \hdots + 10^2b_2 + 10b_1 + 2k[/tex]

    . . . . . [tex]b\;=\;2\!\cdot\!5\!\cdot\!10^{n-1}b_n + 2\!\cdot\!5\!\cdot\!10^{n-2}b_{n-1} + \hdots + 2\!\cdot\!5\!\cdot\!10b_2 + 2\!\cdot\!5\!\cdot\! b_1 + 2k [/tex]

    . . . . . [tex]b\;=\;2\left(5\!\cdot\!10^{n-1}b_n + 5\!\cdot\!10^{n-2}b_{n-1} + \hdots + 5\!\cdot\!10b_2 + 5\!\cdot\!b_1 + k\right) [/tex]

    The right side is a multiple of 2, hence the left side is a multiple of 2.

    Therefore: .[tex]2|b[/tex]



    Does this look okay ?

    Sincerely Yours
    Hummingbird25.
     
    Last edited: Sep 27, 2006
  2. jcsd
  3. Sep 27, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is abstract algebra? I hope it is an introductory problem!
    (by the way, it should be [itex]0\le b_j < 10[/itex].)

    Hint: 2 divides 10.
     
  4. Sep 27, 2006 #3
    Would You say it enough to show that 2 divides 10 ???

    Sincerley

    Hummingbird

    p.s. I'm going into treatment for a heart condition tomorrow, and have to have these calculations finished by the end of today.

    So therefore I hope that somebody in there can assist.

    Sincerley

    Brenda(Hummingbird25)
     
    Last edited: Sep 27, 2006
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