MATH2001
Basic Analysis 2015 Tutorial solutions
Basic Analysis 2015 Solutions of Tutorials
Section 1.1
Tutorial 1.1.1 (Theorem 1.3) [Basic field properties: multiplication]
(a) The number 1 is unique.
(b) For all with 0, the number 1 is unique.
(c) For all , with 0, the equation = has a unique solution. This solution is = 1 .
(d) {0}, (1 )1 = .
(e) , {0}, ()1 = 1 1 .
(f) {0}, ()1 = 1 .
(g) 11 = 1.
Proof. (a) Let 1, 1 such that 1 = and 1 = for all . We must show that 1 = 1 :
1 = 1 1
= 1 1
by (M2)
=1
(b) Let {0} and , such that = 1 and = 1. We must show that = :
= 1
by (M3)
= ( )
= ( )
by (M1)
= ( )
by (M2)
= 1
= 1
by (M2)
by (M3)
(c) First we show that = 1 is a solution. So let = 1 .
Then
= (1 )
= (1 )
by (M1)
=1
by (M4)
=1
by (M2)
by (M3)
To show that the solution is unique let such that = .
Then
= 1 = (1 )
by (M3), (M4)
by (M1)
= ()
by (M2)
= 1
= ()
by (M2)
This shows that the solution is unique.
(d) Note that
1 (1 )1 = 1 by (M4).
On the other hand
1 = 1 = 1 by (M2), (M4).
Basic Analysis 2015 Tutorial solutions
By part (b), it follows that
(1 )1 = .
(e)
()(1 1 ) = (()(1 )1
1
by (M1), (M2)
= (( ))
by (M1)
= ( 1)
=
by (M4)
by (M3)
=1
by (M4)
By part (b), ()1 = 1 1 .
(f)
()1 = ()1 1
1
by (M3)
1
= () ( )
by (M4)
= [() (())]
by (M1), Theorem 1.1 (d)
by (M2), Theorem 1.2 (d)
= [() ()]
= (1)
by (M2), (M4)
= 1
by Theorem 1.2 (e)
By part (b), ()1 = 1 1 .
(g) 1 1 = 1 by (M3), so that 11 = 1 by (M4) and (b).
Tutorial 1.1.2
1. Prove Theorem 1.4, (c)(g):
Let , , , . Then
(c) < + < + .
(d) < and < + < + .
(e) < and > 0 < .
(f) 0 < and 0 < < .
(g) < and < 0 > .
Proof. (c)
<>0
by definition of <
( + ) ( + ) > 0
by axioms of addition
+ <+
by definition of <
(d)
< and < > 0 and > 0
by definition of <
( ) + ( ) > 0
by (O2)
( + ) ( + ) > 0
by axioms and properties of addition
+ <+
by definition of <
(e)
< and > 0 > 0 and > 0
by definition of <
( ) > 0
by (O3)
> 0
by axioms and properties of addition and (D)
<
by definition of <
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Basic Analysis 2015 Tutorial solutions
(f)
0 < and 0 < > 0, > 0, > 0, > 0
<
by definition of <
by (e)
if = 0: = 0 = <
<
if > 0: < <
<
(g)
< and < 0 > 0 and > 0
by definition of < and by (a)
()( ) > 0
by (O3)
> 0
by (D) and Theorem 1.2, (d), (f)
>
by definition of >
2. The absolute value function. Define the following function on :
{
if 0,
|| =
if < 0.
Prove the following statements for , :
(a) || 0,
(b) || = || ||,
(c) || < < < ,
(d) | + | || + ||.
Proof. (a) By (O1), we have to consider 3 cases:
Case I: > 0 || = > 0 || 0
Case II: = 0 || = = 0 || 0
Case I: > 0 || = > 0 || 0
(b) Case I: = 0 or = 0: then = 0 by Theorem 1.2 (b). Hence
|| = |0| = 0 = || ||.
Case II: > 0, > 0: Then > 0 by (O3). Hence
|| = = || ||.
Case III: < 0, > 0: Then < 0 by Theorem 1.4 (g). Hence, by Theorem 1.2 (d),
|| = = () = || ||.
Case IV: > 0, < 0: Interchange and and apply Case III.
Case V: < 0, < 0: Then > 0 and > 0 by Theorem 1.4 (a). Applying Theorem 1.2 (f) and Case II give
|| = |()()| = ()() = || ||.
(c) Let || < . Then 0 || < , so that < 0. If 0, then
< 0 = || < ,
so that < < . If < 0, then | | = |(1)| = || by (b), and the above gives < < , so that
Theorem 1.4 (g) gives < () < (), that is, < < .
Conversely, let < < . If 0, then || = < .
If < 0, then < gives < , so that || = < .
(d) If + 0, then
| + | = + || + || (using ||).
If + < 0, then
| + | = ( + ) || + ||
(using ||).
Basic Analysis 2015 Tutorial solutions
MATH2001
3. Let , , . Which of the following statements are true and which are false?
(a) ,
false
(b) 0 < 1 1 ,
true
(c) < < 0
< 1 ,
true
(d)
< 1 < 1,
true
(e) 2 < 1 1 < < 1,
true
(f) 2 > 1 > 1
false.
4. In each of the following questions fill in the
,
<
(a) 3 2
7
7
(b) 1
3
+1
3
,
<
(c) > 1
10
,
1
<
(d) > 1
1
2
1
,
(e) 2
1
2 1
(f) > 3
with < or >.
<
1
,
2
.
1
<
<
5. Let 0 and 0. Show that < 2 < 2 .
Proof. Observe that
2 < 2 2 2 > 0 ( )( + ) > 0
and
< > 0.
In either case, , so that 0 and 0 gives that > 0 or > 0. It follows that + > 0. Hence
> 0 ( )( + ) > 0
and
( )( + ) > 0 = ( )( + )( + )1 > 0.
