Graphics Reference
In-Depth Information
49 The first and last inequalities were established in qns 45 and 48. The
middle one only needs 1
1
1/
n
. As a consequence any
a
is less
than any
b
.
b
(1
1/5)
(6/5)
46 656/15 625
46 875/15 625
3.
50 (i) Since0
1, this involves repeated use of qn 11(i).
51 From qn 49 (1
1/
n
)
3, so for
n
3, (1
1/
n
)
n
.
((
n
1)/
n
)
n
(
n
1)
n
.
52
0
53 When
a
0,
a
a
a
. When
a
is positive,
a
0
a
a
.
When
a
is negative,
a
0
a
a
.
54
a
a
follows from qn 53. So also
a
a
a
a
, from qn 6.
55 The result follows from the definition when
a
0. When
a
0, use
qn 4.
56
a
(
a
)(
a
)
a
(
a
)(
a
).
57
a
max(
a
,
a
)
max(
a
,
a
)
a
.
a
)(
b
)
58 (
(
ab
) sums up the four non-zero cases.
59 Provided
b
0,
a
(
a
/
b
)·
b
(
a
/
b
)
·
b
by qn 58.
60 When
x
lies between
b
.
If
b
a
b
and
b
a
b
, then max(
a
,
a
)
b
and
a
b
.
If
b
. This last
inequality implies
b
a
, so together we have
b
a
b
.
The work in this question proves that
a
b
b
a
b
.
a
b
, then max(
a
,
a
)
b
,so
a
b
and
a
61
a
a
a
and
b
b
b
together imply
a
b
a
b
a
b
, so from qn 60,
a
b
a
b
.
62 For thefirst inequality put
b
for
b
in thetriangleinequality.
a
b
c
a
b
c
a
b
c
.
63 Thersult
c
a
c
a
follows from qn 60.
64
.
(ii) Keep the negative sign on square rooting the left-hand side;
multiply by 2 and add
a
(i) 0
(
aq
bp
)
b
p
q
, to both sides.