Anna Medvedovsky
[email protected]
Math 23b / Spring 2009
HW #1 solutions
1.
A
=
{
j
2

j
:
j
∈
Z
}
and
B
=
{
k
2
+
k
:
k
∈
Z
,k
≥
0
}
. Prove that
A
=
B
.
First, I claim that
A
is a subset of
B
: If
a
∈
A
, then there exists
j
∈
Z
with
j
2

j
=
a
.
If
j
≥
1, then
j

1
∈
Z
≥
0
; reexpressing
a
as
a
= (
j

1)
2
+ (
j

1), we see that
a
∈
B
. If
j
≤
0, then

j
∈
Z
≥
0
; reexpressing
a
as
a
= (

j
)
2
+ (

j
) we see that
a
is again in
B
. So
a
∈
B
regardless of
j
, and therefore
A
is a subset of
B
.
Now I claim that
B
is a subset of
A
: If
b
∈
B
, then there exists
k
∈
Z
≥
0
with
k
2
+
k
=
b
.
But then
b
= (
k
+ 1)
2

(
k
+ 1); since
k
+ 1
∈
Z
, we see that
b
∈
B
.
Since each set is a subset of the other, the two sets are equal, as desired.
2.
A
=
{
x
2

y
2
:
x,y
∈
Z
}
and
B
=
{
n
∈
Z
:
n
even or
n
divisible by
4
}
.
To show:
A
=
B
.
First, I claim that
A
is a subset of
B
. Indeed, if
a
∈
A
, then
a
has the form
a
=
x
2

y
2
= (
x
+
y
)(
x

y
)
for some
x
and
y
in
Z
. If
x
and
y
have the same parity (both odd or both even) then
x
+
y
and
x

y
are both even numbers, so their product
a
= (
x
+
y
)(
x

y
) is divisible by 4 and
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 Spring '09
 BongLian
 Math, Xu, Negative and nonnegative numbers, Parity, xu + yv

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