22
(x)
If
a
and
b
are natural numbers, there there are natural numbers
s
and
t
with gcd
(
a
,
b
)
=
sa
+
tb
.
Solution.
False.
1.47
Given integers
a
and
b
(possibly negative) with
a
=
0, prove that there
exist unique integers
q
and
r
with
b
=
qa
+
r
and 0
≤
r
<

a

.
Solution.
We have already proved this when
a
>
0 and
b
≥
0. Assume
now that
a
>
0 and
b
<
0. Now
−
b
>
0, and so there are integers
q
and
r
with
−
b
=
qa
+
r
and 0
≤
r
<
a
; it follows that
b
= −
qb
−
r
. If
r
=
0,
we are done; if
r
>
0, then
b
=
(
−
q
−
1
)
a
+
(
a
−
r
)
and 0
<
a
−
r
<
a
(by Proposition A.2(ii),
−
r
<
0 implies
a
−
r
<
a
.
Now assume that
a
<
0, so that
−
a
>
0 (and so

a
 = −
a
. By what we
have proved so far, there are integers
q
and
r
with
b
=
q
(
−
a
)
+
r
, where
0
≤
r
<
−
a
; that is,
b
=
(
−
q
)
a
+
r
, where 0
≤
r
<

a

.
1.48
Prove that
√
2 is irrational using Proposition 1.14 instead of Euclid
’
s lemma.
Solution.
Assume, on the contrary, that
√
2
=
a
/
b
, where
a
and
b
are inte
gers. By Proposition 1.14, we have
a
=
2
k
m
and
b
=
2
n
, where
k
,
≥
0
and
m
,
n
are odd. If
k
≥
, then we may cancel to obtain
√
2
=
2
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 Fall '11
 KeithCornell
 Integers, Natural Numbers, Prime number, nonzero integers

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