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Solutions to homework # 1.
1. Let
r
be rational and
x
be irrational. If
r
+
x
were rational, then
x
= (
r
+
x
)

r
would
be rational too, as a diﬀerence of two rational numbers. Hence
r
+
x
is irrational.
Likewise, if
rx
were rational, then, since 1
/r
exists and is rational, the number
x
=
(
rx
)(1
/r
) would be rational too as a product of two rational numbers. Thus
rx
must be
irrational.
2. Suppose there exists a rational number whose square is 12. Write such a number in
lowest terms as
m/n
, where
m
and
n
are integers having no factors in common. Then
m
2
= 12
n
2
= 3
·
2
2
·
n
2
. Since 3 appears in the righthand side, 3 divides
m
2
. But 3 is prime,
so it must divide
m
; hence
m
2
must be divisible by 9. But then 3 divides
n
2
and therefore
n
as well, contrary to the assumption that
m
and
n
have no common factors. Contradiction!
3. Since
E
is nonempty, it has at least one element, say,
x
. As
α
is a lower bound of
E
, we
get
α
≤
x
. Since
β
is an upper bound of
E
,
x
≤
β
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This document was uploaded on 11/17/2010.
 Spring '09

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