# Theory of Numbers (V63.0248) Professor M. Hausner Answer sheet ```Theory of Numbers (V63.0248)
Professor M. Hausner
*1. Evaluate h1, 2, 3, 4, 5i.
Ans. We use the algorithm as follows
0
1
1
0
1
1
1
2
3
2
3
10
7
4
43
30
5
225
157
So the continued fraction is 225/157.
*2. Evaluate h1, 3, 2i.
Ans. Let x = h3, 2i. Then x = h3, 2, xi. We evaluate the latter continued fraction as follows:
0
1
1
0
3
3
1
2
7
2
x
7x + 3
2x + 1
So x = (7x +
we get 2x2 + x = 7x + 3 or 2x2 − 6x − 3 = 0. The positive root
√ 3)/(2x + 1). Solving,
√
is x = (6 + 60)/4 = (3 + 15)/2. Finally,
√
√
√
h1, 3, 2i = h1, xi = 1 + 1/x = 1 + 2/( 15 + 3) = 1 + 2( 15 − 3)/6 = 15/3.
√
*3. Write 3 + 15 as an infinite repeating continued fraction.
Ans.
√ The solution is√as follows:
A) 15√+ 3 = 6 + (√15 − 3)
√
B) 1/(√15 − 3) = (√15 + 3)/6 = 1 + ( 15 − 3)/6
√
C) 6/( 15 − 3) = ( 15 + 3). This brings us back to A). So 3 + 15 = h6, 1i
*4. Let ha0 , a1 , . . . , an , . . .i be a continued fractions, where each an is a positive integer
√ for n ≥ 0. Let
1
+
5
the n-th convergent be pn /qn in lowest terms. Prove that pn ≥ φn where φ =
(= 1.618 . . .)
2
is the positive solution of the equation x2 − x − 1 = 0. Hint: Use induction.
Ans. For n = 0, p0 = a0 ≥ 1 = φ0 . For n = 1, p1 = a1 p0 + a−1 ≥ 1 + 1 φ1 . Let n &gt; 1. Assume the
result for n − 1 and n − 2. Then
pn = an pn−1 + pn−2 ≥ pn−1 + pn−2 ≥ φn−1 + φn−2 = φn−2 (φ + 1) = φn−2 φ2 = φn .
This is the result for n, proving the result by induction.
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