This preview shows page 1. Sign up to view the full content.
11
Base Step
.If
n
=
1, then
4
n
+
1
+
5
2
n
−
1
=
16
+
5
=
21
,
which is obviously divisible by 21. Since our inductive step will involve
two predecessors, we are obliged to check the case
n
=
2. But 4
3
+
5
3
=
64
+
125
=
189
=
21
×
9.
Inductive Step
.
4
n
+
2
+
5
2
n
+
1
=
4
·
4
n
+
1
+
5
2
·
5
2
n
−
1
=
4
·
4
n
+
1
+
(
4
·
5
2
n
−
1
−
4
·
5
2
n
−
1
)
+
5
2
·
5
2
n
−
1
=
4
(
4
n
+
1
+
5
2
n
−
1
)
+
5
2
n
−
1
(
5
2
−
4
).
Now the last term is divisible by 21; the
f
rst term, by the inductive hypoth
esis, and the second because 5
2
−
4
=
21.
1.21
For any integer
n
≥
2, prove that there are
n
consecutive composite num
bers. Conclude that the gap between consecutive primes can be arbitrarily
large.
Solution.
The proof has nothing to do with induction. If 2
≤
a
≤
n
+
1,
then
a
is a divisor of
(
n
+
1
)
!
; say,
(
n
+
1
)
!=
da
for some integer
d
.I
t
follows that
(
n
+
1
)
!+
a
=
(
d
+
1
)
a
, and so
(
n
+
1
)
!+
a
is composite for
all
a
between 2 and
n
+
1.
1.22
Prove that the
f
rst and second forms of mathematical induction are equiv
alent; that is, prove that Theorem 1.4 is true if and only if Theorem 1.12 is
true.
Solution.
Absent.
1.23
(
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

Click to edit the document details