Continued fraction explanation 
This is about simple infinite continued fraction. I don't understand the line '...then $C_0 < x < C_1$'. $C_k$ here refers to $C_k=[a_0;a_1,a_2,...,a_k]$ where $1 \leq k \leq n$. $C_o=a_0$.
Can anyone explain it to me why is the inequality true?
 A: Certainly $C_0 < x$, since $\displaystyle x = a_0 + \frac{1}{[a_1;a_2,a_3\cdots]}$ and the fraction is positive.
On the other hand, $\displaystyle \frac{1}{a_1 + \frac{1}{[a_2;a_3,a_4\cdots]}} < \frac{1}{a_1}$ since the fraction in the denominator is positive.  Thus $\displaystyle x = a_0 + \frac{1}{[a_1;a_2,a_3\cdots]} = a_0 + \frac{1}{a_1 + \frac{1}{[a_2;a_3,a_4\cdots]}} < a_0 + \frac{1}{a_1} = C_1$.
A: The convergents of a continued fraction alternate above and below the final value.  There are more formal ways of showing this, but for now simply consider that a continued fraction can be formed by truncating the integer part and taking the reciprocal of the remainder at each step.  Since truncating always takes a lower value, it is apparent that the first convergent, $C_0=a_0$, is less than $x$; however truncating a reciprocal will take a higher value, thus $C_1 \left( =a_0+\dfrac1{a_1} \right) > x$.  Since the reciprocal operation reverses the direction that truncating takes at each step, the convergents will continue to alternate around the ultimate value, unless the CF is finite.
