Download Complete Induction Lecture 12 Robb T. Koether Mon, Mar 21, 2016

Complete Induction
Lecture 12
Robb T. Koether
Hampden-Sydney College
Mon, Mar 21, 2016
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
1 / 16
Outline
1
The Principle of Complete Mathematical Induction
Second-Order Recursion
2
Examples
3
Other Examples
4
Assignment
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
2 / 16
Outline
1
The Principle of Complete Mathematical Induction
Second-Order Recursion
2
Examples
3
Other Examples
4
Assignment
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
3 / 16
The Principle of Complete Mathematical Induction
The Principle of Complete Mathematical Induction
Let P(n) be a statement about an arbitrary natural number n.
To prove that P(n) is true for all n ∈ N, it suffices to show that
P(1) is true.
For all natural numbers k ≥ 1, if P(j) is true for all j ∈ N with j ≤ k ,
then so is P(k + 1).
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
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Second-Order Recursion
A recursive sequence is second order if each term (except the first
two) is defined in terms of two previous terms.
The Fibonacci is the best-known second-order recursive
sequence.
f1 = 1,
f2 = 1,
fn = fn−1 + fn−2 for n ≥ 3.
The sequence is 1, 1, 2, 3, 5, 8, 13, 21, . . ..
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
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Second-Order Recursion
Theorem
Let {fn }n∈N be the Fibonacci sequence. Then for all n ∈ N,
n
X
fi = fn+2 − 1.
i=1
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
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Second-Order Recursion
Theorem
Let {fn }n∈N be the Fibonacci sequence and let α =
β=
√
1− 5
2 .
√
1+ 5
2
and
Then for all n ∈ N,
fn =
Robb T. Koether (Hampden-Sydney College)
αn − β n
.
α−β
Complete Induction
Mon, Mar 21, 2016
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Outline
1
The Principle of Complete Mathematical Induction
Second-Order Recursion
2
Examples
3
Other Examples
4
Assignment
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
8 / 16
Example
Example (2" and 5" Blocks)
Suppose we have an unlimited number of blocks 2 inches long
and blocks 5 inches long.
Then we can assemble them end-to-end to form a length n inches
long for any n ≥ 4.
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
9 / 16
Outline
1
The Principle of Complete Mathematical Induction
Second-Order Recursion
2
Examples
3
Other Examples
4
Assignment
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
10 / 16
Example
Theorem
For any natural number n, consider n blue dots and n red dots
arranged in the plane in such a way that no 3 dots are collinear. Then
a line L can be drawn such that each side of L has an equal number
blue dots and red dots.
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
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Example
Theorem
For any natural number n, given n lines in the plane, no 3 of which are
coincident, these lines divide the plane into n(n+1)
+ 1 regions.
2
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
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Example
Theorem
For any natural number n, given n planes in space, no 4 of which are
coincident, these planes divide the plane into ??? regions.
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
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Outline
1
The Principle of Complete Mathematical Induction
Second-Order Recursion
2
Examples
3
Other Examples
4
Assignment
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
14 / 16
Assignment
Assignment
Presentations this week:
3.22, 3.23, 3.24, 3.25, 3.26.
Write up the following two problems and turn them in Wednesday,
March 30.
P3.27 Prove that every natural number n can be factored
into a product of primes.
Robb T. Koether (Hampden-Sydney College)
Complete Induction
Mon, Mar 21, 2016
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Assignment
Assignment
P3.28 Prove that, for any n ∈ N, a 2n × 2n square with the lower-left 1 × 1
square removed can be tiled with L-shaped pieces consisting of
three 1 × 1 squares.
2n
2n
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Complete Induction
Mon, Mar 21, 2016
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