Perform the long division for . Identify the repeating block of digits. Does it show cyclic properties if you evaluate ? Now compute , etc. What do you notice?
Long Division of 1/13: 1/13 = 0.076923076923... = 0.0̄7̄6̄9̄2̄3̄
The repeating block is 076923 (6 digits).
Long Division of 2/13: 2/13 = 0.153846153846... = 0.1̄5̄3̄8̄4̄6̄
The repeating block is 153846. This is a different cyclic block from 076923, so 2/13 does not show the same cyclic digits as 1/13, but it forms its own cyclic family.
Computing other fractions:
- 3/13 = 0.230769... (cyclic shift of 076923)
- 4/13 = 0.307692... (cyclic shift of 076923)
- 5/13 = 0.384615... (cyclic shift of 153846)
- 6/13 = 0.461538... (cyclic shift of 153846)
- 7/13 = 0.538461... (cyclic shift of 153846)
- 8/13 = 0.615384... (cyclic shift of 153846)
- 9/13 = 0.692307... (cyclic shift of 076923)
- 10/13 = 0.769230... (cyclic shift of 076923)
- 11/13 = 0.846153... (cyclic shift of 153846)
- 12/13 = 0.923076... (cyclic shift of 076923)
Observation: The fractions form two distinct cyclic groups. Group 1 (1/13, 3/13, 4/13, 9/13, 10/13, 12/13) shares the repeating block 076923, where digits shift cyclically. Group 2 (2/13, 5/13, 6/13, 7/13, 8/13, 11/13) shares the repeating block 153846, where digits also shift cyclically. This is similar to the cyclic number property of 1/7 (142857), but 1/13 produces two separate cycles of 6 digits each instead of one cycle of 6 digits.
Explanation
The textbook introduces cyclic numbers through 1/7 = 0.̄1̄4̄2̄8̄5̄7̄, where multiplying 142857 by 1-6 gives cyclic permutations. For 1/13, since 13-1=12 and the period is 6, we get two separate cyclic families. The remainders 1, 3, 4, 9, 10, 12 form one cycle (076923), and remainders 2, 5, 6, 7, 8, 11 form another (153846). Students should observe that within each group, the same digits appear in cyclic order, demonstrating the beautiful internal structure of rational number decimal expansions.
Solution Steps
Step 1: Perform long division of 1/13. Dividing 1 by 13: 10→0 rem 10, 100→7 rem 9, 90→6 rem 12, 120→9 rem 3, 30→2 rem 4, 40→3 rem 1. The remainder 1 repeats, so 1/13 = 0.076923... Repeating block = 076923.
Step 2: Perform long division of 2/13. Dividing 2 by 13: 20→1 rem 7, 70→5 rem 5, 50→3 rem 11, 110→8 rem 6, 60→4 rem 8, 80→6 rem 2. The remainder 2 repeats, so 2/13 = 0.153846... Repeating block = 153846 (a different cyclic block).
Step 3: Compute 3/13 = 0.230769..., 4/13 = 0.307692..., 5/13 = 0.384615..., 6/13 = 0.461538..., 7/13 = 0.538461..., 8/13 = 0.615384..., 9/13 = 0.692307..., 10/13 = 0.769230..., 11/13 = 0.846153..., 12/13 = 0.923076...
Step 4: Observe that fractions with numerator 1, 3, 4, 9, 10, 12 all have repeating block 076923 in cyclic shifts, and fractions with numerator 2, 5, 6, 7, 8, 11 all have repeating block 153846 in cyclic shifts. Two distinct cyclic families exist.