Test 1 solutions
Answer FOUR questions. Be sure to give a reason for each step; if that
reason is one of the axioms, identify which one by its symbol.
1. Show that if
a
∈
R
then 0
a
= 0.
2. Let
a, b
∈
R
.
(i) Show that (

a
)
b
=

(
ab
) =
a
(

b
).
(ii) Show that (

a
)(

b
) =
ab
.
3. Let
a, b
∈
R
.
(i) Show that
a
2
+
b
2
>
0.
(ii) Show that if
a
2
+
b
2
= 0 then
a
=
b
= 0.
4. Show that if
a
∈
R
and if
b
∈
R
is nonzero then
±
±
±
a
b
±
±
±
=

a


b

.
Note
: It may
be assumed that
x > y
>
0
⇒
x
2
> y
2
.
5. Deﬁne
N
as a subset of
R
and prove that no element
n
of
N
satisﬁes
0
< n <
1.
Solutions
:
1. Note ﬁrst that
0
a
AID
= (0 + 0)
a
D
= 0
a
+ 0
a
whence addition of

0
a
yields
0
AIV
= (

0
a
)+0
a
= (

0
a
)+(0
a
+0
a
)
AA
= ((

0
a
)+0
a
)+0
a
AIV
= 0+0
a
AID
= 0
a
and concludes the proof.
2. (i) Note that
ab
+ (

a
)
b
D
= (
a
+ (

a
))
b
AID
= 0
b
1
= 0
so that (

a
)
b
is indeed the additive inverse

(
ab
) of
ab
; similarly,
a
(

b
)
equals

(
ab
).
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Staff
 Logic, Multiplicative inverse, Inverse element, additive inverse, a2 + b2

Click to edit the document details