Let $ a_0,a_1,\cdots a_n$ and $b_0,b_1,\cdots,b_{n+1}$ be positive integers. State the conditions for $$[a_0;a_1,\cdots, a_n] < [b_0; b_1,\cdots, b_{n+1}]$$
The above problem is from the book "Ivan Niven, Herbert S. Zuckerman, Hugh L. Montgomery - An introduction to the theory of numbers-Wiley", chapter 7, exercice 7.2.
For $n=0$, the above holds true if $a_0 \le b_0 $. For $n=1, $ $a_0+\dfrac{1}{a_1} < b_0+\dfrac{1}{b_1+\dfrac{1}{b_2}}$ holds if $a_0 < b_0$ or $ a_0 = b_0 \text{ and } a_1 > b_1$. Similarly for $n=2, $ $a_0+\dfrac{1}{a_1+\frac{1}{a_2}} < b_0+\dfrac{1}{b_1+\dfrac{1}{b_2+\frac{1}{b_3}}}$ holds if $a_0 < b_0$ or $ a_0 = b_0 \text{ and } a_1 > b_1$ or $ a_0 = b_0, a_1=b_1 \text{ and } a_2 \le b_2$.
Using strong induction, we can say that $[a_0;a_1,\cdots, a_n] < [b_0; b_1,\cdots, b_{n+1}]$ holds when $a_0 < b_0$ or $ a_0 = b_0 \text{ and } a_1 > b_1$ or $ a_0 = b_0, a_1=b_1 \text{ and } a_2 < b_2$, $\cdots\cdots$ or $ a_0 = b_0, a_1=b_1 , a_2=b_2, \cdots ,a_{n-1}=b_{n-1} \text{ and } a_n > b_n$ (for odd $n$. For even $n$, $a_{n} \le b_{n}$).
Is this proof correct? or am I missing something?