# Ternary Expansions and Cantor Set

• Nov 2nd 2009, 12:48 PM
eg37se
Ternary Expansions and Cantor Set
The problem is to find out for what values of p (for integers between 0 and 13), is p/13 in the cantor set. I know that they are only in the cantor set if they can be expressed in the ternary expansion to base 3 using only 2's and 0's. I have a lot of problems with these ternary expansions so I was wondering if you guys could check my work quick, and let me know where I have gone wrong.

THANKS

1/13=\$\displaystyle (.002002...)_3\$
2/13=2x1/13=(.004004...)=\$\displaystyle (.011011...)_3\$
3/13=3x1=\$\displaystyle 3(.002002...)_3=(.02002...)_3\$
4/13=2x2/13=\$\displaystyle 2(.011011...)_3=(.022022...)_3\$
5/13=4/13+1/13=\$\displaystyle (.002002...)_3+(.022022...)_3=(.02110211)_3\$
6/13=3x2/13=\$\displaystyle 3(.011011...)_3=(.11011...)_3\$
7/13=6/13+1/13=\$\displaystyle (.11011...)_3+(.002002...)_3=(.112112...)_3\$
8/13=7/13+1/13=\$\displaystyle (.002002...)_3+(.112112...)_3=(.111111...)_3\$
9/13=3x3/13=\$\displaystyle 3(.02002...)_3=(.2002002...)_3\$
10/13=9/13+1/13=\$\displaystyle (.2002002...)_3+(.002002...)_3=(.2022022...)_3\$
11/13=2/13+9/13=\$\displaystyle (.011011...)_3+(.2002002...)_3=(.211211)_3\$
12/13=3x4/13=\$\displaystyle 3(.022022...)_3=(.22022)_3\$
• Nov 3rd 2009, 12:50 AM
Opalg
Quote:

Originally Posted by eg37se
The problem is to find out for what values of p (for integers between 0 and 13), is p/13 in the cantor set. I know that they are only in the cantor set if they can be expressed in the ternary expansion to base 3 using only 2's and 0's. I have a lot of problems with these ternary expansions so I was wondering if you guys could check my work quick, and let me know where I have gone wrong.

THANKS

1/13=\$\displaystyle (.002002...)_3\$
2/13=2x1/13=(.004004...)=\$\displaystyle (.011011...)_3\$
3/13=3x1=\$\displaystyle 3(.002002...)_3=(.02002...)_3\$
4/13=2x2/13=\$\displaystyle 2(.011011...)_3=(.022022...)_3\$
5/13=4/13+1/13=\$\displaystyle (.002002...)_3+(.022022...)_3=(.02110211)_3\$ Should be \$\displaystyle \color{red}(.101101...)_3\$.
6/13=3x2/13=\$\displaystyle 3(.011011...)_3=(.11011...)_3\$
7/13=6/13+1/13=\$\displaystyle (.11011...)_3+(.002002...)_3=(.112112...)_3\$
8/13=7/13+1/13=\$\displaystyle (.002002...)_3+(.112112...)_3=(.111111...)_3\$ Should be \$\displaystyle \color{red}(.121121...)_3\$.
9/13=3x3/13=\$\displaystyle 3(.02002...)_3=(.2002002...)_3\$
10/13=9/13+1/13=\$\displaystyle (.2002002...)_3+(.002002...)_3=(.2022022...)_3\$
11/13=2/13+9/13=\$\displaystyle (.011011...)_3+(.2002002...)_3=(.211211{\color{red }...})_3\$
12/13=3x4/13=\$\displaystyle 3(.022022...)_3=(.22022{\color{red}0...})_3\$

The rest are correct.