6.6 Approximation of irrational numbers[29Q]

In the next exercises we will use the following definitions.

Definition 207

[0BS] For \(x∈ ℝ\) we define \(⌊ x ⌋\) to be the floor function defined as the greatest integer less than or equal to \(x\), as in

\[ ⌊ x ⌋{\stackrel{.}{=}}\max \{ n∈ℤ:n≤ x\} \quad . \]

Definition 208

[0BT] \(x-⌊ x ⌋\) is the fractional part of \(x\).

(We define \(𝜑(x)=x-⌊ x ⌋\), note that \(𝜑(3,1415)=0,1415\) but \(𝜑(-4,222)=0,778\) because \(⌊ -4,222 ⌋=-5\)).

E208

[0BV]Note that \(k=⌊ x ⌋\) is the only integer for which you have \(k≤ x {\lt} k+1\) or equivalently \(0≤ (x-k){\lt}1\) or equivalently \(x-1{\lt}k≤ x\).

E208

[0BW] Prerequisites:207.Given \(x ∈ \mathbb {R}\) and \(N ∈ \mathbb {N}, N≥ 2\), prove that at least one element of the set \(\{ x, 2x, \ldots , (N-1)x\} \) is at most distance \(1/N\) from an integer, that is, there exist \(n,m∈ℤ\) with \(1≤ n≤ N-1\) such that \(|nx-m|≤ 1/N\).

Hidden solution: [UNACCESSIBLE UUID ’0BX’]

E208

[0BY] Prerequisites:207,2.Given \(x,b ∈ ℝ\) with \(x≠ 0\) irrational, and \(\varepsilon {\gt}0\), prove that there is a natural \(M\) such that \(M x-b\) is at most \(\varepsilon \) from an integer.

Let \(𝜑(x)=x-⌊ x ⌋\) be the fractional part of \(x\), we have \(𝜑(x)∈[0,1)\). The above result implies that the sequence \(𝜑(nx)\) is dense in the interval \([0,1]\).

Note that instead if \(x≠ 0\) is rational i.e. \(x=n/d\) with \(n,d\) coprime integers and \(d{\gt}0\), then the sequence \(𝜑(nx)\) assumes all and only the values \(\{ 0,1/d,2/d,\ldots (d-1)/d\} \).

(This is demonstrated by the Bézout’s lemma [ 35 ] ).

Hidden solution: [UNACCESSIBLE UUID ’0BZ’]

[UNACCESSIBLE UUID ’0C0’]

[0C1]Prerequisites:2. (Dirichlet’s approximation theorem) Given an irrational number \(x\), show that there are infinitely many rationals \(𝛼\) such that we can represent \(𝛼=m/n\) in order to satisfy the relation

\[ \left| x - \frac m n \right| {\lt} \frac 1{n^ 2}\quad . \]

Some comments.

  • Note for every fixed \(n≥ 2\) there is at most an \(m\) for which the previous relation holds; but there may not be one.

  • Note that if the relation holds for a rational \(𝛼\), there are only finite choices of representations for which it holds,

  • and certainly it holds for the "canonical" representation with \(n,m\) coprimes.

Hidden solution: [UNACCESSIBLE UUID ’0C2’]

[2B0]Note that Hurwitz’s theorem [ 40 ] states that for every irrational number \(𝜉\) there are infinitely many coprime integers \(m, n∈ℤ\) such that

\[ \left|𝜉 -{\frac{m}{n}}\right|{\lt}{\frac{1}{{\sqrt{5}}\, n^{2}}}. \]

[0C3]Fixed \(k{\gt}0\), \(\varepsilon {\gt}0\) and a rational number \(x\), prove that there exist only finitely many rationals \(𝛼\) that can be represented as \(𝛼=m/n\) in order to satisfy the relation

\[ \left| x - \frac m n \right| ≤ \frac k{n^{1+\varepsilon }}\quad . \]

Hidden solution: [UNACCESSIBLE UUID ’0C4’] [0C5]Prove that for every rational \(m/n\) you have

\[ \left| \sqrt{2} - \frac m n \right| {\gt} \frac 1{4n^ 2}. \]

We obtain that the set \(A = ⋃_{m∈ℤ,n ∈ \mathbb {N}^*} \left( \frac m n - \frac 1{4n^ 2} , \frac m n + \frac 1{4n^ 2}\right)\) is an open set that contains every rational number, but \(A≠ℝ\).

Hidden solution: [UNACCESSIBLE UUID ’0C6’]