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Project a – Farey tree for the rational fractions in (0, 1) We can enumerate all rational fractions in the unit interval (0, 1) by the Farey tree scheme. To do this, however, requires a strange operation called the mediant. Let us begin with two fractions a/b and c/d
and assume a/b is smaller than c/d, i.e. a/b
< c/d. Then, we define the mediant of a/b and c/d by
To convince yourself that the above inequality relationship holds true, it is simplest to cross-multiply and then reduce (A) into ordinary inequalities. For instance, we get the reference inequality ad < bc from a/b < c/d to prove (A). We now begin with the parent fractions 0/1 and 1/1, located at the end points of (0, 1) in figure A. From 0/1
and 1/1, the first generation mediant is
Figure A. Farey tree In figure A, we identify the first sequence { 0/1, 1/3, 1/2, 2/3, 1/1} of denominator "3," the second { 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1} of denominator "4," and the third { 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1} of denominator "5." Obviously, the first sequence is contained in the second, which in turn is contained in the third sequence. Hence, as the Farey tree grows, it will include more and more sequences of the denominator of high order. In this way, we can eventually account for all the rational numbers in (0, 1) and every rational fraction should have a leaf somewhere in the Farey tree. Return to main text |
. It is interesting to note that mediant 
, which lies in between 0 and 1, i.e., 0/1<
1/2 < 1/1. Next, there are two mediants, 
and 
, of the second generation and four mediants, 0/1
< 1/3 <1/2 and
1/2 < 2/3 < 1/1 and 1/4,
2/5, 3/5, and 3/4 of the third generation. Can you fill in the fourth generation mediants in figure A?