Exercise 7.1.4 (Inequalities between continued fractions)

Question

Given positive integers $b,c,d$ with $c>d$, prove that $\langle a,c \rangle < \langle a,d \rangle$, but $\langle a,b, c \rangle > \langle a,b,d \rangle$ for any integer $a$.

Richard Ganaye · Accepted Answer

\begin{proof} Since $c>d>0$, then $1/c < 1/d$, so $$\langle a, c \rangle = a + \frac{1}{c} < a + \frac{1}{d} = \langle a, d \rangle.$$ Replacing $a$ with $b$, we obtain $0 < \langle b, c \rangle < \langle b,d \rangle,$ therefore, for any integer $a$, $$\langle a,b, c \rangle = a + \frac{1}{\langle b, c \rangle} > a + \frac{1}{\langle b, d \rangle} = \langle a,b, d \rangle.$$ \end{proof}

Exercise 7.1.4 (Inequalities between continued fractions)

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Exercise 7.1.4 (Inequalities between continued fractions)

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