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Section 20
5.
Show that
F
n
>
1
:
6
n
once
n
is big enough
.
The
basis step
for two consecutive values of
n
. That±s because the induction can±t go
from
n
1
to
n
, we generate
F
n
=
F
n
2
+
F
n
1
from the
two
preceding
values. Computing
F
n
and
1
:
6
n
for lots of values of
n
F
29
= 832040
>
830767
:
5 = 1
:
6
29
F
30
= 1346269
>
1329228 = 1
:
6
30
The
induction step
is to show that if
F
k
2
>
1
:
6
k
2
and
F
k
1
>
1
:
6
k
1
, then
F
k
>
1
:
6
k
. That is, if the inequality is true when
n
is any
number less than
k
(and, in fact, we need only assume it for
n
=
k
2
and
n
=
k
1
) then it is true for
n
=
k
. In fact
F
k
=
F
k
2
+
F
k
1
>
1
:
6
k
2
+1
:
6
k
1
= 1
:
6
k
2
(2
:
6)
>
1
:
6
k
2
(1
:
6)
2
= 1
:
6
k
So
F
n
>
1
:
6
n
for all
n
±
29
.
Section 21
5.
in the line is a man, then somewhere in the line a woman is directly in
front of a man
.
The
basis step
is to examine the shortest line, which has length two
and consists of one woman in front of one man. Clearly the statement
is true in this case.
The
induction step
is to pass from the truth of the statement for lines
of length
k
to the truth of the statement for lines of length
k
+1
. Look
at a line of length
k
+ 1
(the front of the line is to the right)
M X
_ _
:::
_
W
If
X
, the next to the last person in line, is a woman, then there is a
woman directly in front of the last person in line, who is a man. If
X
is a man, then look at the shorter line obtained by removing
M
,
the last man. This is a line of length
k
that starts with a woman and
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View Full Document ends with the man
X
. By the induction hypothesis, our statement is
true for lines of the length
k
. So in that shorter line, there is a woman
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This note was uploaded on 07/13/2011 for the course MAD 2104 taught by Professor Staff during the Spring '08 term at FAU.
 Spring '08
 STAFF

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