The WellOrdering Property

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The Well-Ordering
Property Every nonempty set of nonnegative
integers has a least element.
Theorem. If a is an integer and d is a
positive integer, then there are unique
integers q and r such that 0 ? r lt d and a dq
r.
Proof Let S a-dq a-dq ? 0 and q is an
integer .
If q -a, then a-dq ? 0 and hence S is
nonempty.
Let r a-dq0 be the least element of S.
Suppose that r ? d. Then r-d a-dq0 -d a-d(q0
1) ? 0
This means that r-d is in S and r gt r-d which
contradicts the assumption that r is the least
element of S.
Thus 0 ? r lt d.
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Mathematical Induction
Mathematical Induction is a method for proving
statements of the form P(n) where n ranges over
the positive integers.
The Principle of Mathematical Induction can be
derived from the Well-Ordering Property. Its
most formal statement is
P(1) ? ?k ( P(k) ? P(k1)) ? ?n P(n)
A proof by mathematical induction that P(n) is
true for every positive integer consists of two
steps

• Basis Step
• The proposition P(1) is shown to be true.

• Inductive Step
• The implication P(k) ? P(k1) is shown
• to be true for every positive integer k.

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In the inductive step, where you must prove
that P(k) ?
P(k1) is a true implication, the statement
P(k) is called the inductive hypothesis ( or IH).
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Example Proof Using Mathematical Induction
Proposition For every n ? 1, the sum of the
first n odd integers is n2
Proof.
Basis Step The sum of the first 1 odd integers
is 1. 12 1. Thus P(1) is true.
Inductive Step
First state P(k) 1 3 . . . (2k-1) k2
This is our inductive hypothesis, i.e., we assume
P(k) is true.
Now state what we must prove, i.e., P(k1)
1 3 . . . (2(k1)-1) (k1)2
Proof of inductive step
1 3 . . . (2(k1)-1) 1 3 . . .
(2k1) 1 3 . . . (2k-1) (2k1) k2
(2k1) (by
the IH) (k 1)2
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Example Proof Using Mathematical Induction
Proposition For every n ? 1, 2n gt n
Proof.
Basis Step 21 2 gt 1 Thus P(1) is true.
Inductive Step
First state the inductive hypothesis P(k) 2k gt
k
Now state what we must prove, i.e., P(k1)
2k1 gt k1
Proof of inductive step
2 k1 22k
gt 2k by the IH
? k1 since k ? 1
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Example Proof Using Mathematical Induction
Proposition For every n ? 1, n3 -n is
divisible by 3
Proof.
Basis Step 13 -1 0, which is divisible by 3
Thus P(1) is true.
Inductive Step
Inductive hypothesis P(k) k3 k 3t for some
integer t
Must prove P(k1) (k1)3 - (k1) is
divisible by 3
Proof of inductive step
(k1)3 - (k1) (k3 3k2 3k 1) (k1)
(k3 k) 3(k2 k)
3t 3(k2 k) by the IH
3(t k2 k)
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Harmonic Numbers
For each positive integer k , the harmonic
number Hk is defined by Hk 1 1/2 1/3
??? 1/k
For instance, H4 1 1/2 1/3 ¼ 25/12
Proposition If k 2n and n ? 0, then Hk ? 1
n/2
Proof.
Basis Step If n 0, then 2n 20 1 and H1
1 ? 1 0/2.
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Harmonic Numbers, Inductive Proof
Inductive Step
Inductive hypothesis P(n) If k 2n, then Hk ?
1 n/2
Must prove P(n1) If r 2n1, then Hr ? 1
(n1)/2
Proof of inductive step
Hr (1 1/2 1/3 ??? 1/2n) (1/(2n1)
??? 1/2n1)
Hk (1/(2n1) ??? 1/2n1), k 2n
? 1 n/2 (1/(2n1) ??? 1/(2n2n), by the
IH
? 1 n/2 2n ?1/2n1
1 n/2 1/2
1 (n1)/2
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Example Proof Using Mathematical Induction
Theorem For n ? 1, every 2n?2n chessboard
with one square removed can be covered using
non-overlapping L-shaped pieces, where each such
piece is made up of three squares of the same
size as the squares of the chessboard.
Tiling of the chessboard less one square
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Example Proof of Tiling Result
Basis Step, n 1 A 2 by 2 chessboard with one
square removed is
an L, so the result holds.
Inductive Step
Inductive Hypothesis P(n) Every 2n?2n
chessboard with one
square removed can be L-tiled
Must prove P(k1) Every 2n1?2n1 chessboard
with one
square removed can be L-tiled
Proof split the 2n1?2n1 chessboard into 4
quadrants each of these quadrants will be a
2n?2n chessboard, one of which will have one
square removed. Remove one square from each of
the other 3 quadrants so that the 3 missing
squares form an L. Each of the quadrants is now a
2n?2n chessboard with one square removed and the
result follows by the inductive hypothesis
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Second Principle of Mathematical Induction
A proposition P(n) is true for all positive
integers n if
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Example proof using the second form of
mathematical induction
Theorem Every integer ? 2 is either prime or
can be written as a product
of primes
Proof. Basis Step 2 is prime.
Inductive Step
Inductive Hypothesis every j with 2 ? j ? k is
either prime or a
product of primes.
To Prove k1 is either
prime or a product of primes
Proof if k1 is prime, the conclusion follows
trivially. If not, then k1 ab for some a,b ?
2. Moreover, each of a,b is less than k1. By
the IH, each of a and b are either prime or a
product of primes, so k1 is a product of
primes.
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1. Study Section 3.2 of the text, especially the
example proofs
2. Do the following Exercises from that
section 4, 6, 12, 14, 20