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For each part of this problem, convert the fraction to a "decimal" in the specified base.

(a) Convert the fraction 1/2 to base 3

(b) Convert the fraction 1/7 to base 8

(c) Convert the fraction 1/(b-1) to base b

 

Thank you in advance!  :)

 Jun 19, 2020
 #1
avatar+8352 
+2

Because \(\dfrac13 + \dfrac1{3^2} + \dfrac1{3^3} + \cdots = \dfrac{\dfrac13}{1 - \dfrac13} = \dfrac12\),

 

\(\dfrac12 = 0.\overline{1}_3\)

 

By a similar approach,

 

\(\dfrac17 = 0.\overline{1}_8\)

and 

\(\dfrac1{b - 1} = 0.\overline{1}_b\)

 Jun 19, 2020
 #2
avatar+55 
+2

Thank you for replying!  I do not really know how you got the first equation, I might be missing something really simple, but if you have the time, could you explain it to me? Thanks!

ConfuzzledKitten  Jun 19, 2020
 #3
avatar+8352 
+2

It is the sum of geometric series.

 

It looks like this:

For -1 < r < 1, and an arbitrary a,

\(a+ar+ar^2+ar^3+\cdots+ar^n+\cdots=\dfrac{a}{1 - r}\)\(\)

MaxWong  Jun 19, 2020
 #4
avatar+55 
+2

Thank you!  I searched up a proof on Google and I think I understand now! :-)

 Jun 19, 2020

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