### 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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