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Section 0.2
7
23. By Exercise 22, there exists an integer
k
such that
n
=
4k
+
1 or
n
=
4k
+
3.
Case 1:
n
=
4k
+
1.
If
k
is even, there exists an integer m such that
k
=
2m,
so
n
=
4(2m)
+
1
= 8m
+
1, and
the desired conclusion is true.
If
k
is odd, there exists an integer m such that
k
= 2m
+
1, so
n
=
4(2m
+
1)
+
1
=
8m
+
5, and the desired conclusion is true.
Case 2:
n
=
4k
+
3.
If
k
is even, there exists an integer m such that
k
=
2m,
so
n
=
4(2m)
+
3
= 8m
+
3, and
the desired conclusion is true.
If
k
is odd, there exists an integer m such that
k
= 2m
+
1,
so
n
=
4(2m
+
1)
+
3
=
8m
+
7, and the desired conclusion is true. In all cases, the desired conclusion
is true.
24. [BB]
If
the statement is false, then there does exist a smallest positive real number r. Since
~r
is
positive and smaller than r, we have reached an absurdity. So the statement must be true.
25. We give a proof by contradiction.
If
the result is false, then both
a
>
Vn
and
b
>
Vn.
(Note that the
negation of an "or" statement is an "and" statement.) But then
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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