Maths Olympiad Prep

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Number theory Difficulty 6.3 National olympiad Prove it

10. Let the fractions of the nn-th order Farey sequence be 0=a1/b1<a2/b2<<ak/bk=10=a_{1} / b_{1}<a_{2} / b_{2}<\cdots<a_{k} / b_{k}=1. Prove:
(i) k=1+m=1nφ(m)k=1+\sum_{m=1}^{n} \varphi(m);
(ii) j=1kaj/bj=k/2\sum_{j=1}^{k} a_{j} / b_{j}=k / 2;
(iii) j=1k11/(bjbj+1)=1\sum_{j=1}^{k-1} 1 /\left(b_{j} b_{j+1}\right)=1;
(iv) max1j<k(aj+1/bj+1aj/bj)=1/n\max _{1 \leqslant j<k}\left(a_{j+1} / b_{j+1}-a_{j} / b_{j}\right)=1 / n,
min1j<k(aj+1/bj+1aj/bj)=1/n(n1)\min _{1 \leqslant j<k}\left(a_{j+1} / b_{j+1}-a_{j} / b_{j}\right)=1 / n(n-1).

Solution

10. (i) follows from Problem 5 (v); (ii) follows from Problem 5 (v); (iii) follows from Problem 5 (i); (iv) using Problem 5 (i), 0,1/n,1/(n1)0,1 / n, 1 /(n-1) are three consecutive fractions in the nn-th Farey sequence, when n2n \geqslant 2, bjbj+1b_{j} \neq b_{j+1}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.