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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?

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  • $\begingroup$ It would be good form to give the source of this theorem. Thanks! $\endgroup$ – Matthew Conroy Nov 27 '13 at 20:50
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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 $\displaystyle x = a_0 + \frac{1}{[a_1;a_2,a_3\cdots]} = a_0 + \frac{1}{a_1 + \frac{1}{[a_2;a_3,a_4\cdots]}} < a_0 + \frac{1}{a_1} = C_1$.

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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.

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