Exercise: 30
b
a. 218
1 2 0
Answers
Explanation:
140 Answers and Solutions
.....-_______ ---,D
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BL..---...L..-----~
F
FIGURE 1.3.
for each positive integer m, so that la - r2m-ll and la - r2m I are both less than
1 1
Ir2m - r2m-ll .:::: 42(m-l) Ir2 - rll = 24m - 3 ·
It follows that limn--->oo rn = limn--->oo rn-l = a. Taking limits in the recursion in
(d) leads to
1
a=I+--
l+a
which reduces to a 2 = 2. Since rn > 0 for each n, a > O.
1.4(c). 347/19 = 18 + 1/3 + 1/1 + 1/4.
1.4(d). Since v'2 = 1 + (v'2 - 1) = 1 + 1/1 + v'2, it follows that the process
does not terminate, since we can always replace the final 1 + v'2 by 2 + 1/1 + v'2
to get the continued fraction in Exercise 1.3.
1.7(b). This can be established from Exercise 1.5(a) by induction.
1.7(c). For each positive integer k, let P2k = Xk and q2k = 2Yk. Then, from
Exercise 1.5(b), xf - 8Yf = P~k - 2q~k = 1.
1.7(d). If X2 - 8y2 = -1, then x would have to be odd. But then x2 == 1 (mod
8), yielding a contradiction.
2.4(a). Since qn + qn-l = (qn - qn-l) + 2qn-l = Pn-l + (Pn - Pn-l) = Pn,
Pn(qn - qn-l) = PnPn-l = Pn-l (qn + qn-d.
2.4(b). From Exercise 2.3(c),
Pn+lqn+l = 4pnqn + 2(Pnqn-l + Pn-lqn) + Pn-lqn-l
= 4pnqn + 2(Pnqn - Pn-lqn-d + Pn-lqn-l = 6pnqn - Pn-lqn-l·
3.1(b). Yes. It suffices to consider the case for which the greatest common divisor
of a, b, cis 1. Since each odd square leaves a remainer 1 and no square leaves a1.1(a). The point E can be obtained by folding the square along the bisector of
angle BAC, so that AB is folded onto the diagonal with B landing on E. If F is
the point where the bisector intersects B C, then B F folds to E F. Thus E F -1 AC
and BF = EF. Since L.EFC = 45° = L.ECF, FE = EC.
l.1(b). WC! = IBC! - IBFI = IAEI - ICEI = 21AEI - lAC! and IEC! =
lAC! - IAEI = lAC! - IBC!.
l.3(a).
n Pn qn rn
1 1 1 1
2 3 2 3/2 = 1.5
3 7 5 7/5 = 1.4
4 17 12 17/12 = 1.416666
5 41 29 41/29 = 1.413793
1.3(c). Clearly, rn > 1 for each n, so that from (b), rn+1 - rn and rn - rn-I have
opposite signs and
1
Irn+l - rnl < 41rn - rn-II·
Since r2 > 1 = rl, it follows that rl < r3 < r2. Suppose as an induction
hypothesis that
rl < r3 < ... < r2m-1 < r2m < ... < r2.
Then 0 < r2m - r2m+1 < r2m - r2m-1 and 0 < r2m+2 - r2m+1 < r2m - r2m+J.
so that r2m-1 < r2m+1 < r2m+2 < r2m. Let k, 1 be any positive integers. Then, for
m > k, I, r2k+1 < r2m-1 < r2m < r21.
l.3(d). Leta be the least upper bound (i.e., the smallest number at least as great as
all) of the numbers in the set {rl, r3, r5, ... , r2k+l, ... }. Then r2m-1 ::::: a ::::: r2m
Answer:
PLEASE BRAIN LIST MY ANSWER
Explanation:
A) 218 IS THE ANSWER