Graphics Reference
In-Depth Information
15
proposition
justification
a
0
given
0
a
or 0
a
trichotomy, qn 5
0
a
or 0
(
a
)
positiveclosurefor
multiplication
a
(
a
)
algebra
0
a
16 0
1
1. So
1 is negative, and no square is negative from question
15.
17 Given 0
from question
14. Both these conclusions conflict with what is given so
a
b
,by
trichotomy.
a
and 0
b
;
a
b
a
b
;
b
a
b
a
19
(i) 0
a
b
a
a
b
ab
b
.
(ii) Given 0
from
part (i). Both these conclusions conflict with what is given, so
a
b
, by trichotomy.
a
and 0
b
;
a
b
a
b
;
b
a
b
a
20
(i) The result has already been established for
n
1 (trivial),
n
2
(question 14) and
n
3 (question 19(i)).
a
b
a
ab
.
a
b
ab
b
.So
a
b
by transitivity. Thersult
follows by induction.
(ii) Given 0
a
and 0
b
;
a
b
a
b
;
b
a
b
a
by part
(i). Both these conclusions conflict with what is given, so
a
b
,
by trichotomy.
21 The equivalence holds for odd integers
n
.
22 0
a
and 0 · (1/
a
)
0
1
a
· (1/
a
), from question 16, implies 0
1/
a
from question 12.
23 0
1/
a
and 0
1/
b
from question 22. So
a
· (1/
a
)· (1/
b
)
b
· (1/
a
) · (1/
b
),
from question 11. Now 0
1/
b
1/
a
, claiming transitivity.
24
a
0 and
a
· (
1/
a
)
1
0
0 · (
1/
a
) (from question 16) implies
0
1/
a
(question 12). So 1/
a
0 from question 3.
25 1/
a
0 and 1/
b
0 from question 24. So 0
(
1/
a
)(
1/
b
)
1/
ab
from question 4 and positive closure. Thus
a
b
implies
a
· (1/
ab
)
b
· (1/
ab
)or1/
b
1/
a
and by transitivity 1/
b
1/
a
0.
27 Apply trichotomy to 1
x
.1
x
0 is inadmissible.
1
x
0
1/(1
x
)
0. 1
x
0 and
1/(1
x
)
1
1
1
x
0
x
,so
1
x
0.
Now
1
x
0
0
1
x
1
1
1/(1
x
).
28
(i) True.
(ii) False, try
a
3 and
b
2
.
(iii) True.
(iv) False, try
a
b
2.
