Let $\theta$ be an irrational number with continued fraction expansion $[a_0; a_1, a_2, \cdots]$.

Suppose $P_n/Q_n = [a_0; a_1, \cdots , a_n]$ is the $n^{th}$ convergent. Then how do I show that $P_0=a_0$.

I have that $P_0/Q_0= [a_0]$ but where do I go from here?

  • $\begingroup$ Since $[a_0] = a_0$ is an integer, it follows immediately that $Q_0 = 1$, doesn't it? $\endgroup$ – A.P. May 13 '15 at 20:11
  • $\begingroup$ is [a0] supposed to be the floor of a0? $\endgroup$ – cf12418 May 13 '15 at 20:45
  • $\begingroup$ Definitely not. $[a_0; a_1,a_2,\dotsc]$ is just a shorthand notation for the continued fraction of a number $\alpha$, with digits $a_0,a_1,a_2,\dotsc$. It is worth noting that the ';' separates the integer part of $\alpha$, $a_0$, from the digits of the fractional part of $\alpha$. $\endgroup$ – A.P. May 13 '15 at 20:49
  • $\begingroup$ It is worth noting that there are multiple different notations for continued fractions. $\endgroup$ – A.P. May 13 '15 at 20:50

As you observed, the $n$-th convergent of the continued fraction $[a_0; a_1,a_2,\dotsc]$ is the rational number $$ c_n = [a_0; a_1,\dotsc,a_n] = a_0 + \cfrac{1}{a_1 + \cfrac{1}{\ddots + \cfrac{1}{a_n}}} $$ and the integers $p_n,q_n$ are simply numerator and denominator of $c_n$, respectively.

In particular, $c_0 = a_0 \in \Bbb{Z}$ has denominator $1$, so $p_0 = c_0$.

  • 1
    $\begingroup$ Could you please upvote and/or accept this answer if it satisfied you? That way this question won't pop up in the "unanswered" tab anymore. $\endgroup$ – A.P. May 14 '15 at 15:57

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