By R. Bruggerman

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**Sample text**

Since bn = an qn + rn = an qn , then rn−1 = an bn = an−1 = rn−2 , and similarly rn−2 rn−3 . Continuing in this fashion,we see that rn−j rn−j−1 for each natural number j < n, so rn−1 r1 r0 = a1 . 7), with j = 0, rn−1 b. Therefore, rn−1 is a common divisor of a and b. Moreover, if d is a common divisor of a and b, then d r0 , but a = a0 = b1 = q1 a1 + r1 = q1 r0 + r1 , so d r1 . Continuing in this fashion we see that d rj for all j < n. 5. In other words, gcd(a, b) = rn−1 . 2. In fact, we have shown something somewhat stronger, namely that gcd(a, b) = gcd(rj , rj+1 ) for any integer j with 0 ≤ j < n.

Then α ∈ Q if and only if α can be written as a finite simple continued fraction. Proof. If α = q 0 ; q 1 , . . , q = 1, then with qi ∈ Z, then we use induction on . If 1 q q +1 = 0 1 ∈ Q. q1 q1 Assume that all simple continued fractions of length less than α = q0 + q 0; q 1, . . , q = q0 + are in Q. Since 1 , q 1; . . , q then by the induction hypothesis q 0 ; q 1 , . . , q ∈ Q. Conversely, assume that b/a ∈ Q with a ∈ N and b ∈ Z. Then we may set a = r0 , b = r−1 and invoke the EA to get the recursive relation rj−1 = rj q j + rj+1 where 0 < rj+1 < rj , for j = 0, 1, .

7 One of the greatest mathematicians who ever lived was Carl Friedrich Gauss (1777–1855). At the age of eight, he astonished his teacher, B¨ uttner, by rapidly adding the integers from 1 to 100 via the observation that the fifty pairs (j +1, 100−j) for j = 0, 1, . . , 49 each sum to 101 for a total of 5050. When still a teenager, he cracked the age-old problem of dividing a circle into 17 equal parts using only straightedge and compass. The ancient Greeks had known about construction of such regular n-gons for the cases where 2 ≤ n ≤ 6, but the case n = 7 eluded solution, since as Gauss showed, the only ones that could be constructed in this fashion are those derivable from Fermat primes — see page 37.