Exercise 7.6.3 (Not all irrational numbers are approximable by rationals to order $c>2$)

Proof.

Assume for the sake of contradiction that, for all irrational numbers $\xi$, there exist infinitely many rational numbers $h∕k$ such that

In particular, this is true for $\xi =\frac{1+\sqrt{5}}{2}$.

Since $c>2$, $\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\frac{1}{k^{c-2}}=0$, so there is some integer $N$ such that for all $k\ge N$, $\frac{1}{k^{c-2}}<\frac{1}{5}$, thus for all integers $k$

If $\left|\xi -\frac{h}{k}\right|<\frac{1}{k^c}$ for some $k<N$, then $-|\xi |+\left|\frac{h}{k}\right|\le \left|\xi -\frac{h}{k}\right|<1$, therefore $\left|\frac{h}{k}\right|<1+|\xi |$, so $|h|<(1+|\xi |)k<(1+|\xi |)N$. Hence there is only a finitely many fractions $h∕k$ with $k<N$ such that $\left|\xi -\frac{h}{k}\right|<\frac{1}{k^c}$. Consequently, there are infinitely many fractions $h∕k$ with $k\ge N$ such that $\left|\xi -\frac{h}{k}\right|<\frac{1}{k^c}$. By (1), this shows that there are infinitely many fractions $h∕k$ such that

But $5>\sqrt{5}$, and the proof of Theorem 7.18 shows that this is false for $\xi =\frac{1+\sqrt{5}}{2}$.

In conclusion, the following is false for any constant $c>2$: Given any irrational number $\xi$, there exist infinitely many rational numbers $h∕k$ such that

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