Mathematical Reasoning,Induction,and Recursion

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Mathematical Reasoning,Induction,and Recursion

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Title: Mathematical Reasoning,Induction,and Recursion

1
Chapter 3

Mathematical Reasoning,Induction,and Recursion

2
3.1 PROOF STRATEGY

FORWARD AND BACKWARD REASONING
Example 3.1.1 Backward Reasoning
Let a,b?R, with a?b. Then
Proof Conclusion is true if
which is true if (ab)2gt4ab,
which is true if a22abb2gt4ab,
which is true if a2-2abb2gt0,
which is true if (a-b)2gt0,which is true.

Example 3.1.2Foreward Reasoning
Let n ? N, such that n is not divisible by 2 or
3.
Then n2 - 1 is divisible by 24.
Proof Since n is not divisible by 2 or 3,
it follows that n 6k 1 or 6k 5.
Case 1. (6k 1)2 - 1 ? 36k2 12k mod 24
? 12k2 12k ? 0 mod 24, since 2 divides k2 k.
Case 2. (6k 5)2 - 1 ? 36k2 60k 24 mod 24
? 12k2 12k ? 0 mod 24, as above.

4
TWO FAMOUS PROBLEMS

Fermats Last ThmFor n gt 2, the equation
anbncn
has no solutions for non-zero integers a, b, c.
Status Proved by Andrew Wiles.
Goldbach Conjecture Every even number
larger than 2 is the sum of two odd primes.
Status Open.

5
3.2 SEQUENCES AND SUMMATIONS

DEF A sequence in a set A is a function f
from a subset of the integers (usually 0, 1, 2,
. . . or 1, 2, 3, . . .) to A. The values of a
sequence
are also called terms or entries.
NOTATION The value f(n) is usually denoted an.
A sequence is often written a0, a1, a2, . . ..

Example 3.2.1 Two sequences.
Example 3.2.2 Five ubiquitous sequences.

7
STRINGS

DEF A set of characters is called an alphabet.
Example 3.2.3 Some common alphabets
0, 1 the binary alphabet
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 the decimal digits
0, 1, 2, 3, 4, 5, 6, 7, 8, 9,A,B, C,D,E, F
the hexadecimal digits
A,B,C,D, . . . , X,Y,Z English uppercase
ASCII

DEF A string is a sequence in an alphabet.
notation Usually a string is written without
commas, so that consecutive characters are
juxtaposed.
Example 3.2.4 If f(0) M,f(1) A,
f(2) T, and f(3) H, then write MATH.

9
SPECIFYING a RULE

Problem Given some initial terms a0, a1, ..., ak
of a sequence, try to construct a rule that is
consistent with those initial terms.
Approaches There are two standard kinds of
rule for calculating a generic term an.
DEF A recursion for an is a function whose
arguments are earlier terms in the sequence.
DEF A closed form for an is a formula whose
argument is the subscript n.

Example 3.2.5 1, 3, 5, 7, 9, 11, . . .
recursion a0 1 an an-1 2 for n ? 1
closed form an 2n 1
The differences between consecutive terms
often suggest a recursion. Finding a recursion is
usually easier than finding a closed formula.

Example 3.2.6 1, 3, 7, 13, 21, 31, 43, . . .
recursion b0 1 bn bn-1 2n for n ? 1
closed form bn n2 n 1
Sometimes, it is significantly harder to
construct
a closed formula.

Example 3.2.7 1, 1, 2, 3, 5, 8, 13, 21, 34, . .
.
recursion c0 1, c1 1
cn cn-1 cn-2 for n ? 1
closed form
where and

13
INFERRING a RULE

The ESSENCE of science is inferring rules from
partial data.
Example 3.2.8 Sit under apple tree.
Infer gravity.
Example 3.2.9 Watch starlight move 0.15
arc-seconds in total eclipse. Infer relativity.
Example 3.2.10 Observe biological species.
Infer DNA.

Important life skill Given a di?cult general
problem, start with special cases you can solve.
Example 3.2.11 Find a recursion and a
closed form for the arithmetic progression
c, c d, c 2d, c 3d, . . .
recursion a0 c an an-1 d
closed form an c nd.
Q How would you decide that a given sequence
is an arithmetic progression?
A Calculate di?erences betw consec terms.

DEF The di?erence sequence for a sequence
an is the sequence an an - an-1 for n 1.
Example 3.2.5 redux
Analysis Since an is constant, the sequence is
speci?ed by this recursion
a0 1 an an-1 2 for n 1.
Moreover, it has this closed form
an a0 a1 a2 an
a0 22 2 1 2n

If you dont get a constant sequence on the ?rst
di?erence, then try reiterating.
Revisit Example 1.7.6 1, 3, 7, 13, 21, 31, 43, .
. .
bn 1 3 7 13 21 31 43
bn 2 4 6 8 10 12
bn 2 2 2 2 2
Analysis Since bn is constant, we have
bn 2 2n
Therefore,

Consolation Prize Without knowing about
?nite sums, you can still extend the sequence
bn 1 3 7 13 21 31 43 57
bn 2 4 6 8 10 12 14
bn 2 2 2 2 2 2

18
SUMMATIONS

DEF Let an be a sequence. Then the big-sigma
notation
means the sum
am am1 am2 an-1 an
TERMINOLOGY j is the index of summation
TERMINOLOGY m is the lower limit
TERMINOLOGY n is the upper limit
TERMINOLOGY aj is the summand

Theorem 3.2.1. These formulas for summing
falling powers are provable by induction (see
3.3)

Example 3.2.12 True Love and Thm 3.2.1
On the jth day ... True Love gave me

Corollary 3.2.2. High-powered look-ahead to
formulas for summing jk j 0, 1, ..., n.

22
POTLATCH RULES for CARDINALITY

DEF nondominating cardinality Let A and
B be sets. Then A ? B means that ? one-to-
one function f A ? B.
DEF Set A and B have equal cardinality
(write A B) if ? bijection f A ? B,
which
obviously implies that A ? B and B ? A.
DEF strictly dominating cardinality Let
A and B be sets. Then A lt B means that
A ? B and A ? B.

DEF The cardinality of a set A is n if
A 1, 2, . . . , n and 0 if A ?. Such
cardi-
nalitiesare called finite. NOTATION A n.
DEF The cardinality of N is ? (omega), or
alternatively, N0(aleph null).
def A set is countable if it is finite or ?.
Remark N0 is the smallest in.nite cardinality.
The real numbers have cardinality N1 (aleph
one), which is larger than N0, for reasons to be
given.

24
INFINITE CARDINALITIES

Proposition 3.2.3. There are as many even
nonnegative numbers as non-negative numbers.
Proof f(2n) n is a bijection.
Theorem 3.2.4. There are as many positive
integers as rational fractions.
Proof
Example 3.2.13

Theorem 3.2.5. (G. Cantor) There are more
positive real numbers than positive integers.
Semi-proof A putative bijection f Z ? R
would generate a sequence in which each real
number appears somewhere as an infinite decimal
fraction, like this
Let f(n)k be the kth digit of f(n), and let ? be
the permutation 0 ? 9, 1 ? 0, . . . 9 ? 8. Then
the infinite decimal fraction whose kth digit is
? (f(n)k) is not in the sequence. Therefore, the
function f is not onto, and accordingly, not a
bijection.

26
3.3 MATHEMATICAL INDUCTION

From modus ponens
p basis assertion
p ? q conditional assertion
q conclusion
we can easily derive double modus ponens
p0 basis assertion
p0 ? p1 conditional assertion
p1 ? p2 conditional assertion
p2 conclusion
We might also derive triple modus ponens,
quadruple modus ponens, and so on. Thus, we
have no trouble proving assertions about arbi-
trarily large integers. For instance,
The initial domino falls.
If any of the first 999 dominoes falls,
then so does its successor.
Therefore, the first 1000 dominoes all fall
down.

The induction axiom for the integers may be
characterized as
THE GREAT LEAP TO INFINITY
Given a countably infinite row of dominoes,
suppose that
(1) The initial domino falls.
(2) If domino n, then so domino n 1.
Conclusion All the dominoes all fall down.

Example 3.3.1 a proof by induction
Calculate the sum of the first k odd numbers
1 3 5 (2k - 1)
Practical Method for General Problem Solving.
Special Case Deriving a Formula
Step 1. Calculate the result for some small
cases.
Step 2. Guess a formula to match all those cases.
Step 3. Verify your guess in the general case.

Step 1. examine small cases
(empty sum) 0
1 1
1 3 4
1 3 5 9
1 3 5 7 16
Step 2. It sure looks like 13...(2k-1) k2.
Step 3. Try to prove this assertion by induction.
(see next page for proof)

Basis Step. when k 0
Ind Hyp. when k n
Ind Step. Consider the case k n 1.

Why is induction important to CS majors?
It is the method used to prove that a loop or a
recursively defined function correctly calculates
the intended result. (just for a start)

Example 3.3.2 another proof by induction
Calculate the sum of the first k numbers
1 2 3 k
Step 1. examine small cases
(empty sum) 0 0 1/2
1 1 1 2/2
1 2 3 2 3/2
1 2 3 6 3 4/2
1 2 3 4 10 4 5/2
Step 2. Infer pattern
Step 3. Use induction proof to verify pattern.
See next page.

Proposition 3.3.1.
Basis Step. when k 0.
Ind Hyp. when k n.
Ind. Step.

34
NONALGEBRAIC APPLICATIONS of INDUCTION

Consider tiling a 2k-by-2k chessboard.
(k 3 in the figure below)
with L-shaped tiles, so that one corner-square is
left uncovered.
Basis Step. You can do this when k 0.
Ind Hyp. Assume you can do this for k n.
Ind. Step. Prove you can do it for k n1.

35
ALTERNATIVE FORMS of INDUCTION

In a proof by induction, verifying the inductive
premise means you show that the antecedent of
the quantified statement implies the conclusion.
DEF In a proof by mathematical induction, the
inductive hypothesis is the antecedent of the
inductive premise.

Standard 0-based inductive rule of inference
0 ? S basis premise
(?n)n ? 0 ? n ? S ? n 1 ? S ind prem____
(?n)n ? 0 ? n ? S conclusion
Alternative Form 1. Using an integer other than
zero as a basis.
b ? S basis premise
(?n)n ? b ? n ? S ? n 1 ? S ind prem____
(?n)n ? b ? n ? S conclusion

Example 3.3.3 using 5 as the basis
n2 gt 2n 1 for all n ? 5
Basis Step. 52 gt 2 5 1
Ind Hyp. Assume k2 gt 2k 1 for k ? 5.
Ind. Step.
(k 1)2 k2 2k 1 by arithmetic
gt (2k 1) 2k 1 by ind hyp
4k 2 by
arithmetic
2(k 1) 2k by
arithmetic
? 2(k 1) 10 since k ?
5
? 2(k 1) 1 since 10
? 1

Example 3.3.4 2n gt n2 for all n ? 5.
Basis Step. 25 gt 52
Ind Hyp. Assume 2k gt k2 for k ? 5
Ind. Step.
2k1 2 2k arithmetic
2k 2k arithmetic
gt k2 k2 ind. hyp.
gt k2 (2k 1) by Example 3.3.3
(k 1)2 arithmetic

Example 3.3.5 Prove that any postage of
8 cents or more can be created from nothing but
3-cent and 5-cent stamps.
Basis Step. 8 1 3 1 5
Ind Hyp. Assume n possible from 3s and 5s.
Ind. Step. Try to make (n 1) postage.
Suppose that n r 3 s
Case 1 s ? 1. Then n 1 . . .
Case 2 s 0. Then n 1 . . .

Alternative Form 2. Inductive hyposthesis is that
the first n dominoes all fall down.
b ? S basis premise
(?n)n ? b ? (? k ? n)k ? S ? n 1 ? S ind p
(?n)n ? b ? n ? S conclusion

Example 3.3.6 Prove that every integer n gt 0
is the product of finitely many primes.
Basis Step. 1 is the empty product.
Ind Hyp. Assume that 1, . . . , n are each a
prod-
uct of finitely many primes.
Ind Step.
(1) Either n 1 is prime, or ?b, c ? Z such that
n 1 bc. (law of excl middle, def of prime)
(2) But b and c are the products of finitely many
primes. (by Ind Hyp)
(3) Thus, so is bc.

42
Mind-Benders re Induction

1. 2/3 ancestry
2. All solid billiard balls are the same color.
3. Everyone is essentially bald.

43
3.4 RECURSIVE DEFINITIONS

Functions can be defined recursively. The
simplest form of recursive definition of a
function f on the natural numbers speci.es a
basis rule
(B) the value f(0)
and a recursion rule
(R) how to obtain f(n) from f(n - 1), ?n ? 1

Example 3.4.1 n-factorial n!
(B) 0! 1
(R) (n 1)! (n 1) n!
However, recursive de.nitions often take some-
What more general forms.

Example 3.4.2 mergesort (A1 . . . 2n real)
if n 0
return(A)
otherwise
return(merge (msort(1st half), msort(2nd
half)))

Since a sequence is defined to be a special kind
of a function, some sequences can be specified
recursively.
Example 3.4.3 Hanoi sequence
0, 1, 3, 7, 15, 31, . . .
h0 0
hn 2hn-1 1 for n ? 1
Example 3.4.4 Fibonacci seq
1, 1, 2, 3, 5, 8, 13, . . .
f0 1
f1 1
fn fn-1 fn-2 for n ? 2

Example 3.4.5 partial sums of sequences
Example 3.4.6 Catalan sequence
1, 1, 2, 5, 14, 42, . . .
c0 1
cn c0cn-1 c1cn-2 cn-1c0 for n ? 1

48
RECURSIVE DEFINITION of SETS

DEF A recursive definition of a set S com-
prises the following
(B) a basis clause that specifies a set of
primitive elements
(R) a recursive clause that specifies how ele-
ments of the set may be constructed from ele-
ments already known to be in set S there may
be several recursive subclauses
(E) an implicit exclusion clause that anything
not in the set as a result of the basis clause or
the recursive clause is not in set S.

Backus Normal Form (BNF) is an example of
a context-free grammar that is useful for giving
resursive definitions of sets. In W3261, you will
learn that context-free languages are
recognizable
by pushdown automata.

Example 3.4.7 a rec. def. set of integers
(B) 7, 10 ? S
(R) if r ? S then r 7, r 10 ? S
This reminds us of the postage stamp problem.
Claim (?n ? 54)n ? S
Basis 54 2 7 4 10
Ind Hyp Assume n r 7 s 10 with n ? 54.
Ind Step Two cases.
Case 1 r ? 7. Then n1 (r-7)7(s5)10.
Case 2 r lt 7 ? r 7 ? 42 ? s ? 2.
Then n 1 (r 3) 7 (s . 2) 10.
In computer science, we often use recursive
definitions of sets of strings.

51
RECURSIVE DEFINITION of STRINGS

NOTATION The set of all strings in the alphabet
? is generally denoted ?
Example 3.4.8 0, 1 denotes the set of all
binary strings.

DEF string in an alphabet ?
(B) (empty string) ? is a string
(R) If s is a string and b ? ?, then sb is a
string.
Railroad Normal Form for strings
Example 3.4.9 BNF for strings
ltstringgt ? ltstringgtltcharactergt

53
RECURSIVE DEFINITION of IDENTIFIERS

DEF An identifier is (for some programming
languages) either
(B) a letter, or
(R) an identifier followed by a digit or a
letter.

Example 3.4.10 BNF for identifiers
ltlowercase_lettergt a b z
ltuppercase_lettergt A B Z
ltlettergtltlowercase_lettergtltuppercase_lettergt
ltdigitgt 0 1 9
ltidentifiergt ltlettergt ltidentifiergtltlettergt
ltidentifiergtltdigitgt

55
ARITHMETIC EXPRESSIONS

DEF arithmetic expressions
(B) A numeral is an arithmetic expression.
(R) If e1 and e2 are arithmetic expressions, then
all of the following are arithmetic expressions
e1 e2, e1 - e2, e1 e2, e1/e2, e1 e2, (e1)

Example 3.4.11 Backus Normal Form
ltexpressiongt ltnumeralgt
ltexpressiongt ltexpressiongt
ltexpressiongt - ltexpressiongt
ltexpressiongt ltexpressiongt
ltexpressiongt / ltexpressiongt
ltexpressiongt ltexpressiongt
(ltexpressiongt)

57
SUBCLASSES of STRINGS

Example 3.4.12 binary strings of even length
(B) ? ? S
(R) If b ? S, then b00, b01, b10, b11 ? S.
Example 3.4.13 binary strings of even length
that start with 1
(B) 10, 11 ? S
(R) If b ? S, then b00, b01, b10, b11 ? S.

DEF A strict palindrome is a character string
that is identical to its reverse. (In natural
lan-
guage,blanks and other punctuation are ignored,
as is the distinction between upper and lower
case letters.)
Able was I ere I saw Elba.
Madam, Im Adam.
Eve.
Example 3.4.14 set of binary palindromes
(B) ?, 0, 1 ? S
(R) If x ? S then 0x0, 1x1 ? S.

59
LOGICAL PROPOSITIONS

DEF propositional forms
(B) p, q, r, s, t, u, v,w are propositional forms
(R) If x and y are propositional forms, then so
are ?x, x ? y, x ? y, x ? y, x ? y and (x).
Propositional forms under basis clause (B) are
called atomic.
Remark Recursive definition of a set
facilitates proofs by induction about properties
of its elements.

Proposition 3.4.1. Every proposition has an
even number of parentheses.
Proof by induction on the length of the
derivation of a proposition.
Basis Step. All the atomic propositions have
evenly many parentheses.
Ind Step. Assume that propositions x and y have
evenly many parentheses. Then so do proposi-
tions ?x, x ? y, x ? y, x ? y, x ? y and (x).

61
CIRCULAR DEFINITIONS

DEF A would-be recursive definition is
circular if the sequence of iterated
applications it generates fails to terminate in
applications to elements of the basis set.
Example 3.4.15 a circular de.nition from
Index and Glossary of Knuth, Vol 1.
Circular Definition, 260
see Definition, circular
Definition, circular,
see Circular definition

62
3.5 RECURSIVE ALGORITHMS

REVIEW An algorithm is a computational
representation of a function.
Remark Although it is often easier to write a
correct recursive algorithm for a function,
iterative implementations typically run faster,
because they avoid calling the stack.

63
RECURSIVELY DEFINED ARITHMETIC

Example 3.5.1 recursive addition of natural
numbers succ successor, pred predecessor.

Example 3.5.2 iterative addition of natural
numbers

Example 3.5.3 proper subtraction of natural
numbers succ successor, pred predecessor.

Example 3.5.4 natural multiplication

Example 3.5.5 factorial function

NOTATION Hereafter, we mostly use the
infix notations , -, , and ! to mean the
functions sum, diff, prod, and factorial,
respectively.
68
RECURSIVELY DEFINED RELATIONS