- #1
Hummingbird25
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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.
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.
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