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Math 5615H: Introduction to Analysis I.
Fall 2011
Homework #1. Problems and short Solutions.
#1.
Prove that
√
6 and
√
2 +
√
3 are NOT rational.
Proof.
If
p
:=
√
6
∈
Q
, then
p
2
= 6, and we get a contradiction in the same way as in
Example 1.1 in the textbook.
If
q
:=
√
2 +
√
3
∈
Q
, then
q
2
= 5 + 2
√
6, and
√
6 = (
q
2

5)
/
2
∈
Q
, in contradiction with
the previous part.
#2.
Let
A
and
B
be bounded sets in
R
. Consider the
algebraic sum
of
A
and
B
,
A
+
B
:=
{
x
∈
R
:
x
=
a
+
b
for some
a
∈
A
and
b
∈
B
}
.
Show that sup(
A
+
B
) = sup
A
+ sup
B
.
Proof.
Denote
α
:= sup
A, β
:= sup
B, γ
:= sup(
A
+
B
).
∀
x
∈
A
+
B,
∃
a
∈
A
and
b
∈
B
such that
x
=
a
+
b.
Then
x
=
a
+
b
≤
α
+
β
, and
γ
≤
α
+
β
.
On the other hand,
∀
a
∈
A,
∀
b
∈
B
, we have
a
+
b
∈
A
+
B
=
⇒
a
+
b
≤
γ
=
⇒
α
+
b
≤
γ
=
⇒
α
+
β
≤
γ.
Therefore,
γ
=
α
+
β.
#3.
Find
sup
A,
where
A
:=
{
x
∈
R
:
x
2
<
3
x

2
}
.
Solution.
We can write
A
= (1
,
2). By verifying the properties (i) and (ii) in Deﬁnition 1.8,
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 Fall '09
 Math

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