1.1. NUMBERS, SEQUENCES, AND SUMS 3
1.1.14. Let x = a+r and y = b+s, where a and b are integers and r and s are real numbers such that 0 ≤ r, s <
1. By Exercise 14, [2x] + [2y] = [x] + [x +
1
2
] + [y] + [y +
1
2
]. We now need to show that [x +
1
2
] + [y +
1
2
] ≥
[x + y]. Suppose 0 ≤ r, s <
1
2
. Then [x +
1
2
] + [y +
1
2
] = a + b + [r +
1
2
] + [s +
1
2
] = a + b, and [x + y] =
a+b+[r+s] = a+b, as desired. Suppose that
1
2
≤ r, s < 1. Then [x+
1
2
]+[y+
1
2
] = a+b+[r+
1
2
]+[s+
1
2
] =
a + b + 2, and [x + y] = a + b + [r + s] = a + b + 1, as desired. Suppose that 0 ≤ r <
1
2
≤ s < 1. Then
[x +
1
2
] + [y +
1
2
] = a + b + 1, and [x + y] ≤ a + b + 1.
1.1.15. Let x = a + r and y = b + s, where a and b are integers and r and s are real numbers such that 0 ≤
r, s < 1. Then [xy] = [ab + as + br + sr] = ab + [as + br + sr], whereas [x][y] = ab. Thus we have [xy] ≥
[x][y]. If x and y are both negative, then [xy] ≤ [x][y]. If one of x and y is positive and the other negative,
then the inequality could go either direction. For examples take x = −1.5, y = 5 and x = −1, y = 5.5. In
the first case we have [−1.5 · 5] = [−7.5] = −8 > [−1.5][5] = −2 · 5 = −10. In the second case we have
[−1 · 5.5] = [−5.5] = −6 < [−1][5.5] = −1 · 5 = −5.
1.1.16. If x is an integer then −[−x] = −(−x) = x, which certainly is the least integer greater than or equal
to x. Let x = a + r, where a is an integer and 0 < r < 1. Then −[−x] = −[−a − r] = −(−a + [−r]) =
a − [−r] = a + 1, as desired.
1.1.17. Let x = [x] + r. Since 0 ≤ r < 1, x +
1
2
= [x] + r +
1
2
. If r <
1
2
, then [x] is the integer nearest to x and
[x+
1
2
] = [x] since [x] ≤ x+
1
2
= [x]+r +
1
2
< [x]+1. If r ≥
1
2
, then [x]+1 is the integer nearest to x (choos-
ing this integer if x is midway between [x] and [x+1]) and [x+
1
2
] = [x]+1 since [x]+1 ≤ x+r+
1
2
< [x]+2.
1.1.18. Let y = x + n. Then [y] = [x] + n, since n is an integer. Therefore the problem is equivalent to proving
that [y/m] = [[y]/m] which was done in Example 1.34.
1.1.19. Let x = k + ² where k is an integer and 0 ≤ ² < 1. Further, let k = a
2
+ b, where a is the largest integer
such that a
2
≤ k. Then a
2
≤ k = a
2
+ b ≤ x = a
2
+ b + ² < (a + 1)
2
. Then [
√
x] = a and [
p
[x]] = [
√
k] =
a also, proving the theorem.
1.1.20. Let x = k + ² where k is an integer and 0 ≤ ² < 1. Choose w from 0, 1, 2, . . . , m − 1 such that w/m ≤
² < (w + 1)/m. Then w ≤ m² < w + 1. Then [mx] = [mk + m²] = mk + [m²] = mk + w. On the
other hand, the same inequality gives us (w + j)/m ≤ ² + j/m < (w + 1 + j)/m, for any integer j =
0, 1, 2, . . . , m −1. Note that this implies [² + j/m] = [(w + j)/m] which is either 0 or 1 for j in this range.
Indeed, it equals 1 precisely when w+j ≥ m, which happens for exactly w values of j in this range. Now
we compute
P
m−1
j=0
[x + j/m] =
P
m−1
j=0
[k + ² + j/m] =
P
m−1
j=0
k + [² + j/m] = mk +
P
m−1
j=0
[(w + j)/m] =
mk +
P
m−1
j=m−w
1 = mk + w which is the same as the value above.
1.1.21. a. Since the difference between any two consecutive terms of this sequence is 8, we may compute the
nth term by adding 8 to the first term n − 1 times. That is, a
n
= 3 + (n − 1)8 = 8n − 5.
b. For each n, we have a
n
−a
n−1
= 2
n−1
, so we may compute the nth term of this sequence by adding
all the powers of 2, up to the (n − 1)th, to the first term. Hence a
n
= 5 + 2 + 2
2
+ 2
3
+ ··· + 2
n−1
=
5 + 2
n
− 2 = 2
n
+ 3.
c. The nth term of this sequence appears to be zero, unless n is a perfect square, in which case the term
is 1. If n is not a perfect square, then [
√
n] <
√
n, where [x] represents the greatest integer function.
If n is a perfect square, then [
√
n] =
√
n. Therefore, [[
√
n]/
√
n] equals 1 if n is a perfect square and 0
otherwise, as desired.
d. This is a Fibonacci-like sequence, with a
n
= a
n−1
+ a
n−2
, for n ≥ 3, and a
1
= 1, and a
2
= 3.
1.1.22. a. Each term given is 3 times the preceding term, so we conjecture that the nth term is the first term
multiplied by 3, n − 1 times. So a
n
= 2 · 3
n−1
.
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