good work by an outstanding student
leaving only the “typing it up worse
for the net” to me… the -set
homework i’m about to loosely base my
plain-text solutions on is quite a lovely
thing to look upon… as my versions will
not be. now. make it so.
6.1.3
Let a/b, a’/b’, and a”/b” be consecutive fractions
in the Farey sequence of order n.
(i.e., in
F_n = {0/1, 1/n, …, 1/2, … , (n-1)/n, 1/1}
[for the record… it’s convenient to
have a “symbol” for this sequence;
F-subscript-n is standard but the
text doesn’t include it.]
)
We will show that these fractions
satisfy the “Mediant Property”:
a’/b’ = (a+a”)/(b+b”).
Consecutive “Farey fractions”
satisfy Theorem 6.1 and so
a’b – ab’ = 1 and
a”b’ – a’b” = 1.
Equating the left-hand sides,
a’b – ab’ = a”b’ – a’b”.
Regroup; factor;
“divide through by b'(b+b”)”:
a'(b+b”) = b'(a”+a)
a’/b’ = (a+a”)/(b+b”).
(We are done.)
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