Answers to inclass exercises, Aug. 26
1. Prove that
√
2 is irrational.
Proof.
Suppose, on the contrary, that
√
2 were rational, that is,
√
2
=
p
/
q
for some integers
p
and
q
. We may assume without loss of gener
ality that
p
and
q
are in lowest terms (i.e., are relatively prime). Now
2
=
p
2
q
2
,
so
p
2
=
2
q
2
.
Therefore,
p
2
is even. Since the product of two odd integers is odd
(prove this), it must be the case that
p
is even, that is,
p
=
2
m
for some
integer
m
. Now
2
q
2
=
p
2
= (
2
m
)
2
=
4
m
2
,
which implies
q
2
=
2
m
2
. Hence
q
is even, so
p
and
q
are not in lowest
terms, which contradicts the hypothesis.
We conclude that
√
2 cannot be rational.
2. Use induction to prove that
n
∑
i
=
1
i
3
=
n
2
(
n
+
1
)
2
4
.
(1)
Proof.
Basis step: the result is clearly true for
n
=
1.
Inductive step: Suppose that Eq. (1) holds for
n
. (This is not a vacuous
assumption, since it’s true for
n
=
1.) We must show that Eq. (1) holds
with
n
replaced by
n
+
1, that is,
n
+
1
∑
i
=
1
i
3
=
(
n
+
1
)
2
(
n
+
2
)
2
4
.
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 Spring '07
 thieme
 Real Numbers, Addition, Integers, Rational number, 2m, lemmas

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