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Why is the reciprocal of the fractional part (vs the integer part) taken at each step? Perhaps this is obvious, if so I don't see it. Appreciate your guidance.

Example from Wiki showing reciprocal taken at each step:

cf

Nick
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  • Why would taking the reciprocal of the integer part be useful? –  Nov 19 '20 at 14:49
  • They need to get a 1 in the numerator, so a remainder of 49/200 is the same as 1/(200/49). The 49 is not useful in the numerator but the 1 is. – ruferd Nov 19 '20 at 14:49
  • If you did that, the next step would have 0 integer part and then the algorithm would crash because you can't take the reciprocal of 0. –  Nov 19 '20 at 14:51
  • @SenZen yes, but how is $1/f = r$? In step 1 I thought $r$ is $49/200$, but it is really $f$. – Nick Nov 23 '20 at 13:36
  • @Nick In step 1, $r = 649/200$ and $f = 49/200$??? –  Nov 23 '20 at 18:03
  • @ruferd there is not a value of $1/(200/49)$ like you mention above. Using the example graphic, first $f = 1/(49/200) = 200/49$. I read the Wiki page on continued fractions and still cannot find the reasoning behind the reciprocal. Any insight appreciated – Nick Nov 26 '20 at 16:06

1 Answers1

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It's just the Euclidean algorithm.

$$\cfrac{649}{200}=3+\cfrac{49}{200}=3+\cfrac{1}{\cfrac{200}{49}}=3+\cfrac{1}{4+\cfrac{4}{49}}=3+\cfrac{1}{4+\cfrac{1}{\cfrac{49}{4}}}=3+\cfrac{1}{4+\cfrac{1}{12+\cfrac{1}{4}}}$$

metamorphy
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ruferd
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