Graphics Reference
In-Depth Information
Answers and comments
1
a
2,
b
3,
c
0,
d
0,
e
1,
f
2.
2
a
0,
b
2,
c
2,
d
1,
e
1,
f
1.
2775/999 999
5
· (
)
· (
) · (
) · (
).
3 Every positive rational number, different from 1, may be expressed in a
unique way as a product of prime numbers and of reciprocals of prime
numbers.
4 Thenumbrs of
m
/2
m
Z
are evenly spaced at intervals of 1/2
.
5 2
65 · 73
2
, so 1/2
1/(65· 73). [2
·
]
7182, so
7183/2
.
6 [
a
· 2
]
a
· 2
[
a
· 2
]
1
[
a
· 2
]/2
a
[
a
· 2
]/2
1/2
b
.
7 Sincethreis an
m
/2
number in
x
a
x
b
there must be two, one
in
x
a
x
(
a
b
)/2
and onein
x
(
a
b
)/2
x
b
. The
argument may be repeated.
8 By qn 7,
would be smaller. Contradiction. Likewise if
a
were smallest,
a
/2 would
besmallr. Contradiction.
T
is dense on the number line. If
m
/2
were smallest,
m
/2
9
m
/2
(
m
· 5
)/10
. Since
T
is dense and
D
contains
T
,
D
is dense.
.
m
/2
5
m
2
2 is theonly primefactor of 5
m
.
Contradiction.
(i) (ii) (iii) Yes.
(iv) No, (
m
/2
)/(
k
/2
)
m
/
k
, both
m
and
k
may beodd.
Sameanswrs for
D
.
10 All terms of the sequence in
T
by qn 9 parts (i), (ii) and (iii).
11
m
/10
3
m
10
, so 2 and 5 aretheonly primefactors of 3
m
.
Contradiction.
([10
/3]/10
)
0.3, 0.33, 0.333, 0.3333, . . .
12
a
decimal point followed by
n
9s. So 10
a
10
1, and
a
1
1/10
.
Thus
a
1
1/10
, and (
a
1) is a null sequence.
13
13
100
·
1
10
1
10
1
a
1
...
10
1
10
1
13
100
·
13
99
·
1
10
1
.
1
10
1
