Let $x=[a_0;a_1,a_2]$ be shorthand notation for the continued fraction $x=a_0+\frac{1}{a_1+\frac{1}{a_2}}.$ Then every $x\in\mathbb{Q}$ can be represented as a finite continued fraction $[a_0;a_1,a_2,\dots,a_n]$, where $a_0\in\mathbb{Z}$ and $a_1,a_2,\dots,a_n\in\mathbb{N}$. Moreover, let $\begin{matrix} p_0=a_0 & q_0=1\\ p_1=a_0a_1+1 & q_1=a_1\\ p_k=a_kp_{k-1}+p_{k-2} & q_k=a_kq_{k-1}+q_{k-2} \end{matrix}$ and define $C_k=\frac{p_k}{q_k}$to be the $k$th convergent of $x$.
I am trying to show that for any finite continued fraction, all even-numbered convergents are less than the fraction's value.
For example, let $x=43/30$. Then $x=1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}},$ or $x=[1;2,3,4]$. It follows that $C_0=1$, $C_1=3/2$, $C_2=10/7$, and $C_3=43/30$, and both $C_0$ and $C_2$ are less than $x$, as stated.
How can I prove this by induction?