Daily Notes -- Math 290

April 6, 1999

The students were able to determine the length of the repetend of 1 ÷ 251 as 50. More importantly, by examining smaller primes some discovered through patterns a significant principle: the order of an element in a group is a divisor of the order, or size, of the group. This is known as the Lagrange Theorem, named after the late 18th Century mathematician Joseph Louis Lagrange who, among many other things, conducted pioneering research in the branch of mathematics known later as group theory.

Noting that:

then an efficient search using modulo arithmetic quickly produces the order of 10.

The second challenge from the previous class day amounted to why the Lagrange Theorem works for the cyclic additive group . In other words, what is the connection between the order of an element in and n, and how is it calculated? This was a tough challenge for the class and some time was devoted to exploring two example groups in search of the right pattern, initially arriving at this chart:

  

Element

Order
  

Element

Order

0

1
  

0

1

1

10
  

1

12

2

5
  

2

6

3

10
  

3

4

4

5
  

4

3

5

2
  

5

12

6

5
  

6

2

7

10
  

7

12

8

5
  

8

3

9

10
  

9

4
     

10

6
     

11

12

The class sensed a pattern. Indeed, the symmetry of the orders is compelling. Finally, another column was added to each group, the greatest common factor (G.C.F.) of the element and n (10 or 12).

  

Element

Order

G. C. F.
  

Element

Order

G. C. F.

0

1

10
  

0

1

12

1

10

1
  

1

12

1

2

5

2
  

2

6

2

3

10

1
  

3

4

3

4

5

2
  

4

3

4

5

2

5
  

5

12

1

6

5

2
  

6

2

6

7

10

1
  

7

12

1

8

5

2
  

8

3

4

9

10

1
  

9

4

3
       

10

6

2
       

11

12

1

Now the pattern is clear! In an additive cyclic group an element's order may be easily determined by first finding the greatest common factor of n and element x. Dividing n by this G.C.F. produces the order of x. For example, the G.C.F. of 8 and 12 is 4. Since 12 ÷ 4 = 3, then 3 is the order of 8 in .

Formal proof of the above general observation was not pursued and might be beyond the scope of this course.

At last the class arrived at the big picture for repetends of reciprocals of prime numbers. If one could determine the isomorphism between and then it is a simple matter to determine the order of 10 in , and thus the length of the repetend, by looking at the counterpart to 10 in . Has this been done before? We don't know.

At this point we changed gears and asked the following specific question: What prime number(s) p, if any, will yield a repetend of length 5 for 1 ÷ p? Answering this question amounts to finding all prime number solutions to .

Congruence, or modulo, arithmetic obeys some of the same properties as ordinary real-number algebra. It is permissible to multiply both sides of an equation by the same number, and also okay to add or subtract the same number from both sides. Thus,

This means that if is divided by p then the remainder is 0. In this case we say that "p divides ," or symbolically, p \ . Thus

p \ 99,999 .

Now they factored 99,999 into its prime components:

Thus the only solutions to are 3, 41 and 271. But 10 has order 1 modulo 3 (explaining why 1 ÷ 3 has a repetend of length 1). Since 10 obviously does not have order 1 modulo 41 or modulo 271, then the only primes such that 1 ÷ p has a repetend of length 5 are 41 and 271. Indeed,

The class was left with this challenge:

The following calculator program was sent to the class via e-mail and works on the author's TI-81. It may work on all TI graphing calculators.

:Disp "ENTER P"

:Input P

:Lbl A

:Disp "ENTER N"

:Input N

:P*(N/P-IPart (N/P))->X

:Disp X

:Goto A

The program, once P is entered, will calculate P (mod N) , looping in such a way that one can continually enter new values for N. The program may be ended by pushing the "ON" key.

Day 4