Question #5bc5b

1 Answer
May 18, 2017

Answer:

#F_5={0/1, 1/5, 1/4, 1/3, 1/2, 2/3, 3/4, 4/5, 1/1}#
#F_6={0/1, 1/6, 1/5, 1/4, 1/3, 1/2, 2/3, 3/4, 4/5, 5/6, 1/1}#
#F_7={0/1, 1/7, 1/6, 1/5, 1/4, 1/3, 1/2, 2/3, 3/4, 4/5, 5/6, 6/7, 1/1}#

Explanation:

The trick to recognizing these Farey sequences is to first look at the denominators.

For #F_3#, the denominators (in order) are:

#{1,3,2,3,1}#

The number 3 is the biggest denominator and there are 5 terms.

For #F_4#, the denominators are:

#{1,4,3,2,3,4,1}#

The number 4 is the biggest denominator and there are 7 terms.

The pattern for the denominators of all subsequent Farey sequences should be obvious.

The number of terms is always increased by 2. So, for #F_5#, the number of terms will be 9. And for #F_6#, the number of terms will be 11.

The first term is always #0//1#. The last term is always #1//1#. The middle term is always #1//2#. After 0, the numerators up to #1//2# will always be #1#. The numerators after #1//2# will simply count from 2 to the max number.

It may be easier to see if you let the last term #1/1# be equal to the maximum denominator, divided by itself. So,

Last term of #F_3# is #3//3#
Last term of #F_4# is #4//4#
Last term of #F_5# is #5//5#, etc.

There is always two more terms shoved in each subsequent Farey sequence. Let #m# be the max denominator. Always, just after the zero term (the first term in the sequence) will be #1//m#. And finally, right before the final 1 term (the next to last in the sequence) will be #(m-1)//m#.

I'll let you explore the "questions to consider" on your own!