Number theory continued fractions?

Question

Find the continued fractions for the following numbers:
(a) 3 + 2root2;
(b) root11 − 10;
(c) (1 + root5)/4.

Answer 1

a ) I assume you want the workings for these, which get a little difficult to follow sometimes, and I don't know whether this text box will preserve a decent setting-out of the working.



For 3 + 2√2, the 2√2 is the awkward bit, so concentrate on this to begin with.



The whole-number part of √2 is 2, so write √2 as 2 + (√2 - 2)



2 + (√2 - 2) = 2 + (2√2 - 2) (2√2 + 2) / (2√2 + 2)



= 2 + (8 - 4) / (2√2 + 2) = 2 + 1 / (2√2 + 2) /4



This is the first part of the c.f., 2 + 1 / X, where X = (2√2 + 2) /4



Now look at (2√2 + 2) /4, which = {4 + (2√2 - 2)} /4 = 1 + (2√2 - 2) /4



The 1 is our next denominator, so looking at the (2√2 - 2) /4 part,



(2√2 - 2) /4 = (2√2 - 2) (2√2 + 2) / 4 (2√2 + 2) = (8 - 4) / 4 (2√2 + 2) = 1 / (2√2 + 2)



So our X = 1 + 1 /(2√2 + 2) , and so far we have 2 + (1/1+) (1/(2√2 + 2) +)



Now, extracting the whole-number part of √2 again,



(2√2 + 2) = 4 + (2√2 - 2) = 4 + 4 / (2√2 + 2) = 4 + 1 / (2√2 + 2) /4



4 is the next denominator, and we are back with the expression we called X above. So these two steps, (1/1+) (1/4+) repeat again and again.



So the c.f. for 2√2 = 2 + (1/1+) (1/4+) (1/1+) (1/4+) .......



(or, in the alternative notation, <2; 1, 4, 1, 4,...>



But the expression you want is ( 3 + 2√2 ), so you have to add 3 to the initial 2, giving



( 3 + 2√2 ) = 5 + (1/1+) (1/4+) (1/1+) (1/4+) .......

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b ) √11 + 10 . Start with the √11 bit again.



The whole-number part of √11 is 3, so write √11 as 3 + (√11 - 3)



And (√11 - 3) = (11 - 9) / (√11 + 3) = 2 / (√11 + 3) = 1 / { (√11 + 3) / 2}



So the first part of the c.f. is 3 + 1 / X, where X = (√11 + 3) / 2



(√11 + 3) / 2 = { 3 + (√11 - 3) + 3 } / 2 = { 6 + (√11 - 3 } / 2



= 3 + { (√11 - 3) (√11 + 3) } / 2 (√11 + 3) = 3 + 2 / 2 (√11 + 3) = 3 + 1 / (√11 + 3)



So 3 is the first denominator, we have so far 3 + (1/3+) (1/ (√11 + 3))



(√11 + 3) = 6 + (√11 - 3) ........ (I am shortening the steps a bit now)



= 6 + (√11 - 3) (√11 + 3) / (√11 + 3) = 6 + 2 / (√11 + 3) = 6 + 1 / { (√11 + 3) /2 }



The second denominator is 6, and we are back at X again, so these two steps repeat continuously.



Therefore, √11 = 3 + (1/3+) (1/6+) (1/3+) (1/6+)......



(or, if you prefer, <3; 3. 6, 3, 6, .....>



But you want √11 - 10, so just subtract 10, to give



<-7 ; 3, 6, 3, 6, ....>



(There is more than one way of expressing a continued fraction, and it is likely that there is another expression for this which has a uniformity of signs, but this seems to me the simplest way of doing this one.)

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c ) (1 + √5) / 4 = { 1 + 2 + (√5 - 2) } /4 = ¾ + (√5 - 2) / 4



(√5 - 2) / 4 = (√5 - 2) (√5 + 2) / 4 (√5 + 2) = 1 / 4(√5 + 2)



= 1 / 4 { 4 + (√5 - 2) } = 1 / { 16 + 4 (√5 - 2) }



16 is the first denominator, so we have so far



(1 + √5) / 4 = ¾ + (1 / 16 + X), where X = 4 (√5 - 2)



4 (√5 - 2) = 4 (√5 - 2) (√5 + 2) / (√5 + 2) = 4 / (√5 + 2) = 1 / { (√5 + 2) / 4 }



(√5 + 2) / 4 = [ { 2 + (√5 - 2) } + 2 ] / 4 = { 4 + (√5 - 2) } / 4 = 1 + (√5 - 2) / 4



1 is the second denominator, so X = 1 / (1 + (√5 - 2) / 4 )





And we already know that (√5 - 2) / 4 = (1/16+) (1/1+)..... .....(see above)



So altogether we have (1 + √5 ) / 4 = ¾ + (1/16+) (1/1+) (1/16+) (1/1+) .....



(or, < ¾; 16, 1, 16, 1, .....>

Answer 2

By grinding out the answer with a calculator, starting with an 10-digit representation of √2 you can get an answer for a) that has the continued fraction part repeating 1 and 4 alternately and apparently indefinitely. A similar thing happens in b) using √11 and in c) root five degenerates to a string of 4's



√5 = < 2; 4, 4, 4,....>



That leads me to think there must be an analytic solution. But I have no knowledge of the deeper theory of continued fractions and perhaps you are meant to grind the answers out like this :-(

Answer 3

(a) Let x=3+2√2

x²=9+12√2+8

x²=17+12√2

x²=6(3+2√2)-1

x²=6x-1

Dividing both sides by x:

x=6-1/x

So the continued fraction is:

x=6-1/(6-1/6-1/...

Answer 4

the respond isn't 7/29 yet 29/7. The partial quotients are 2,28,4,28,4,28,... the 1st 4 convergents are 2/a million fifty seven/28 230/113 and 6497/3192. i ask your self whether it is the only answer.