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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.
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