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By Peter J. Cameron

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Ak−1 ] Proof We have y + a0 = [2a0 , a1 , . . , al−1 ] = [2a0 , a1 , . . , al−1 , y + a0 ] (y + a0 )pl−1 + pl−2 = , (y + a0 )ql−1 + ql−2 where pk /qk are convergents to y + a0 . Now 2a0 ql−1 + ql−2 = 2a0 [a1 , . . , al−1 ] + [a1 , . . , al−2 ] = 2a0 [a1 , . . , al−1 ] + [a2 , . . , al−1 ] = pl−1 , 56 CHAPTER 6. LAGRANGE AND PELL using the fact that [a1 , . . , al−2 ] = [al−1 , . . , a2 ] = [a2 , . . , al−1 , ] the first equality because a1 = al−1 , . . 5)(a)) in Notes 2). Hence y + a0 = (y + a0 )pl−1 + pl−2 .

So u = 0. Similarly, if v = 0, then pn−1 u = p, qn−1 u = q, so p/q = pn−1 /qn−1 , which is the extremal case. So we can also assume that v = 0, and we have to prove that strict inequality holds in the conclusion. 5. 6 41 Now if v < 0, then qn−1 u = q − qn v, so u > 0; and if v > 0, then qn−1 u = q − qn v < 0 since q < qn and v ≥ 1, so u < 0. Thus u and v have opposite signs. Now y lies between the two consecutive convergents cn−1 = pn−1 /qn−1 and cn = pn /qn . So qn−1 y − pn−1 and qn y − pn have opposite signs.

The nth convergent is cn = pn /qn = [a0 ; a1 , . . , an ]. As we saw on p. 4, y = [a0 ; a1 , . . , an−1 , yn ], where yn = [an ; an+1 , . ]. Also, y− pn 1 1 = ≤ . 4 For all n ≥ 2, we have (a) |qn y − pn | < |qn−1 y − pn−1 |; (b) |y − cn | < |y − cn−1 |. Proof First we show that (a) implies (b). We have qn |y − cn | = |qn y − pn |. Also, qn > qn−1 . So, if we show that |qn y − pn | < |qn−1 y − pn−1 |, then we will be able to conclude that |y − cn | = |qn y − pn | |qn−1 y − pn−1 | < = |y − cn−1 |.

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A Course on Number Theory [Lecture notes] by Peter J. Cameron


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