Math 52B
Fall 2012
Due Monday, December 3, 2012
at the beginning of class.
1. Prove: if x  C , the Cantor set, then 1  x  C . Hint: think ternary.
2. To determine the rational numbers in the Canfor set that have terminating decimal representations, let x be a
rational number in C. Then x has a terminating or repeating ternary representation. To have a terminating
ternary representation, x must be a finite sum of powers of 1/3, and none of these rationals have a
terminating decimal expansion. If x has a repeating ternary expansion, then
x  (0. s1s2 s3
sm t1t2
where each si and t j is 0 or 2. The part s1s2 s3
t1t2
t1t2
tk t1t2
tk t1t2
tk
)3
sm is the non-repeating part before the repeating part
tk starts. For example, if x  0.20220 2002 2002 2002 , then s1s2 s3 sm  20220 and
tk  2002 . For a rational number p/q in reduced form to have a terminating decimal expansion, q
must be the product of a power of 2 and a power of 5. Verify that 3  3 , 9  3 and 81  3 are the only
powers of 3 (up to the 10th power) for which (3k  1) has only 2’s and 5’s as factors. (This is in fact true
for k up to at least 100.)
1
3. (Continuing problem 2.) Let y  3m x  ( s1s2 s3
sm . t1t2
tk t1t2
t k t1t 2
2
4
tk
has no non-repeating part in its ternary representation. Show that (3k  1) y  ( s1s2 s3
integer p . Since y 
)3 . Note that y
sm t1t2
tk )3 , an
p
p
, x m k
, so x has a terminating decimal representation if and only y
3 1
3 (3  1)
k
does. Hence, without loss of generality, we may assume x has no non-repeating part after the decimal point
in its ternary representation.
4. (Continuing problem 2.) From problem 2, for a repeating base 3 rational number that also has a
terminating decimal representation, the repeating part must be of length at most 4. Let
x  (0 . t1t2t3t4 t1t2t3t4 t1t2t3t4
)3 assume x is in the Cantor set C. (a) Show that 81x  x  (t1t2t3t4 )3
. (b) By letting (t1t2t3t4 )3 cycle through the 16 possible ternary strings using just 0 and 2, create the list
of rational numbers in the Cantor set which have a terminating decimal expansion.
xn  12
3x ,

5. Define  xn n 0 by: x0 is given and xn 1   n
, n  0 . (a) Prove that xn  C implies
1
 3(1  xn ), xn  2
xn 1  C for all n  0 . (b) Let P be the prisoner set associated with the iteration process defining
xn  .
Show that C  P by showing ( x0  C )  ( xn  C, for all n)  ( x0  P) . (c) Conversely,
prove x0  C  x0  P (so x0  P  x0  C ). Hint: x0  C means that
x0  (0 . t1t2t3t4
tk 1***
)3 -- that is, a 1 appears for the first time somewhere. The process used to
)3 . What happens when xk 1 is computed. These prove that
prove part (b) shows that xk  (0 .1***
P C.