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By Peter J. Cameron

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INFINITE CONTINUED FRACTIONS Note that is y is approximable to order n, then it is approximable to any smaller order, since if m < n then c/qn ≤ c/qm for positive integer q. We will see that algebraic numbers (roots of polynomials over the integers) are not approximable to arbitrarily high order. Then, by writing down a number which is approximable to order n for every n, we will have exhibited a transcendental number (one which is not a root of a polynomial over Z). 7 (a) Positive rational numbers are approximable to order 1 and no higher.

We say that a rational number p/q is a best approximation to y if |y − p/q| < |y − a/b| for any rational number a/b with b < q. We see that the convergents from c2 on are best approximations to an irrational number. The proof involves quite a bit of work, which we isolate in a preliminary lemma. 6 Let [a0 ; a1 , a2 , . ] be the continued fraction for the irrational number y, and let [a0 ; a1 , . . , an ] = cn = pn /qn be the nth convergent. If gcd(p, q) = 1 and q ≤ qn , then |qy − p| ≥ |qn−1 y − pn−1 |, with equality if and only if p/q = pn−1 /qn−1 .

For which the limit of the sequence of convergents of [a0 ; a1 , . ] is y. Proof We take a0 = y , so that 0 < y − a0 < 1. Then we put y1 = 1/(y − a0 ), so that y1 is an irrational nummber greater than 1, and continue the process: ai = yi , yi+1 = 1 . yi − ai 32 CHAPTER 4. INFINITE CONTINUED FRACTIONS Then a0 , a1 , a2 , . . are positive integers and y0 = y, y1 , y2 are irrational numbers greater than 1, so the process continues infinitely and produces an infinite continued fraction [a0 ; a1 , a2 ].

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