Exercise 3. For any integer n∈N∗, we denote v3(n) as the 3-adic valuation of n, which is the greatest integer k such that n is divisible by 3k. We set u1=2 and un=4v3(n)+2−un−12 for all n⩾2 (provided that un−1 is defined and non-zero).
Show that, for any positive rational number q, there exists a unique integer n⩾1 such that un=q.
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Official solution
## Solution to Exercise 3
First, note that u1=2,u2=1,u3=3,u4=23,u5=32 and u6=3. We prove by induction that, for all integers n≥2, we have 00, we set qp=p+q. Then:
- φ(qp)=q−pp=p<p+q=qp if 0<p<q; - φ(1+qp)=2q−pp≤2q<(p+q)+q=1+qp if 0<p<q; - φ(2+qp)=qp=p+q<(p+2q)+q=2+qp if 0<p and 0<q.
In all cases, if x is a strictly positive rational number such that x∈/{1,2}, we have ∥φ(x)∥<∥x∥. Now, for all strictly positive rational numbers x and for all integers n≥1:
- if 0<x<1, then un=φ(x)⇔u3n+2=x; - if 1<x<2, then un=φ(x)⇔u3n+1=x; - if 2<x, then un=φ(x)⇔u3n=x.
An induction on ∥x∥ immediately shows that, for all strictly positive rational numbers x, there exists a unique integer n≥1 such that un=x, which concludes the exercise.
Source: NuminaMath-1.5,
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