9. Prove the following.
a. If
a b
, then
a
+
c b
+
c
Proof: Suppose
a b
. Then
a
−
b
0. So that
a
+
c
–
b
+
c
=
a
−
b
0. Hence,
a
+
c b
+
c.
b. If
a b
and
b c
, then
a c.
Proof: Suppose
a b
and
b c
. Then
a
−
b
0 and
b
−
c
0 . So that, by A16 we will find
a
−
b
–
b
−
c
=
a
−
c
0. Hence,
a c.
c. If
a b
and
b c
, then
a c.
Proof: Suppose
a b
and
b c
. Then
a
−
b
0 and
b
−
c
0 . So that, by A16 we will find
a
−
b
–
b
−
c
=
a
−
c
0. Hence,
a c.
d. If
a
0 and
b
0, then
a
+
b
0. Proof :
If
a
0 and
b
0 – a 0 and 0 - b 0 0 - a + 0 - b 0
0 – a + b 0 0 a + b or a + b 0
e. If
a
0 and
b
0, then
a b
0. Proof :
If
a
0 and
b
a 0 and 0 - b 0 a 0 – b 0
a0 – ab 0 0 – ab 0
– ab 0
ab 0 f.
If
a
0 and
b
0, then
a b
0. Proof :
If
a
0 and
b
0 - a 0 and 0 - b 0 -a – b 0
ab 0 g.
If
a b
and
c
0, then
a c bc
. Proof :
If
a b
and
c
b - a 0 and c 0 b - a c 0
bc – ac 0 bc ac or ac bc
h. If
a b
and
c
0, then
a c bc
. Proof:
If
a b
and
c
b - a 0 and 0 - c 0 b - a 0 – c 0
b – a -c 0 -bc + ac 0
ac bc i.
If 0
a b
, then
a
2
b
2 .
Proof:
If 0
a b
0 a and a b and also b 0 a 0 and a b
a
2
ab .......... 1 b 0 and a b
ab b
2
.......…2 From equation 1 and equation 2 we have
a
2
ab b
2
, so a
2
b
2
j. If
a b
0, then
a
2
b
2 .
Proof:
If
a b
0 a , 0 b and a b -a 0 and a b
-a b-a 0 a
2
ab…… 1 -b 0 and a b
- b b –a 0 ab b
2
..... 2 From equation 1 and equation 2 we have
a
2
ab b
2
, so a
2
b
2
k. 1
0.
Proof:
Suppose 1 ≤ 0 There are two possibilities 1 = 0 or 1 0
1 = 0 is false because 1 is identity of multiplication and unique 1 0 if
. 1
.
a or
a a
a
is false because a 0 Therefore 1 0
l. For
a
∈ R, write
a
2 =
a
·
a
. Show that for every
a
∈ R,
a
2 ≥ 0.
Proof:
a R a
2
≥ 0 if a = 0
a. a = 0.0 = 0 if a 0
-a-a = a
2
if a 0 a. a 0
Therefore for every a R, a
2
≥ 0 m.
Explain why the equation
x
2 = −1 has no solution
x
∈ R. Suppose x
2
= 1 has a solution, suppose a is the solution, so we get If a 0, a
2
= a.a= -1 contradiction A18
If a0, a
2
=a.a=-1 contradict 9f If a=0, a.a=-1 contradict thoe 2.3.2
Thus, x
2
=-1 has no solution.
10.
a.
−1
does not exist in ℝ
ans. b.
if 1 then
−1
− → ≡ ∧ Pick
0, we suppose that
−1
0 meaning that
−1
0 or
−1
= 0, we will prove that
−1
−1
. = 1 Property of A13
0. = 1
0 = 1 This contradicts the Property of A15
−1
. = 1 We know that 1 0 meaning that 1 is positive
if 0 and
−1
0 then .
−1 −1
. = 1 , it contradicts the proof of 9.e
c. if 0 , then
−1
We can prove this by using its contraposition if
−1
0, then 0 we don’t use
−1
= 0 since we’ve already proven 10.a
Pick
−1
0, we suppose that 0 meaning that
0 or = 0, we will prove
that
−1
. = 1 Property of A13
−1
. 0 = 1 0 = 1 This contradicts the Property of A15
−1
. = 1 We know that 1 0 meaning that 1 is positive
if
−1
0 and 0 then .
−1 −1
. = 1 , this contradicts the proof of 9.e
Since the contraposition is correct, then
d. 1 if and only if 0
−1
1
firstly 1
→
−1 −1
1
1 →
−1
0 has been already proven by 10.b
1 →
−1
1 1
.
−1
1.
−1
1
−1
Secondly,
−1
1 → 1
−1 −1
1 → 1
−1
0 then we directly know has been proven that 1
−1
1 .
−1
. 1 Property of A8 1
e. if 0 and 1 , then
1 .
. 1 A18 . .
−1
. 1.
−1
A18 . 1 .
−1
f. if , ∈ ℤ and
= 1 , then = = ±1
there are five cases : =
−1, = 0, = 1, 1, and −1 case 1
1 .
= 1 0 .
= 1 = .
−1
0 , this
11. If ,
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