properties of integers: mathematical inductionlyuu/dm/2012/20120315.pdf · properties of integers:...

Properties of Integers: Mathematical

Induction

c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University 60

Mathematics is a deductive science.

— Bertrand Russell (1872–1970),

Introduction to Mathematical Philosophy (1919)

c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 161

Common Setsa

N = {0, 1, . . .},

Z = {. . . ,−2,−1, 0, 1, 2, . . .},

Z+ = {1, 2, . . .},

R = set of real numbers,

Q = set of rational numbers.

a0 is a natural number!


The Well-Ordering Principlea

Theorem 36 (The well-ordering principle) Every

nonempty subset of Z+ contains a smallest element. (Z+ is

said to be well-ordered.)

• Real numbers are not well-ordered.

– {x ∈ R : x > 1} does not contain a smallest element.

• Rational numbers are not well-ordered.

– {x ∈ Q : x > 1} does not contain a smallest element.

aDefined by Cantor in 1883 and proved by Ernst Zermelo (1871–1953)

in 1904.


Mathematical Inductiona

Theorem 37 Let S(n) denote an (open) mathematical

statement containing references to a positive integer n such

that

• S(1) is true and

• S(k + 1) is true whenever S(k) is true for arbitrarily

chosen k ∈ Z+.

Then S(n) is true for all n ∈ Z+.

aDedekind and Peano.


Richard Dedekind (1831–1916)


Giuseppe Peano (1858–1932)


The Proof

• Intuitively,

– If S(1) is true, then S(2) is true.

– But if S(2) is true, then S(3) is true.

– · · ·– So S(n) must be true for all n ∈ Z+?

• We need a proof based on more foundational principles.


The Proof (continued)

• Let F = { t ∈ Z+ : S(t) is false }.

• Assume that F ̸= ∅.

• F has a least element ℓ by the well-ordering principle.

– So S(ℓ) is false.

• Clearly ℓ > 1 and, hence, ℓ− 1 ∈ Z+.

• Because ℓ− 1 ̸∈ F , S(ℓ− 1) is true.

• It follows that S(ℓ) is true, a contradiction.

• So F = ∅.


Mathematical Induction and the Well-OrderingPrinciple

• The proof of induction says that the well-ordering

principle (p. 163) implies mathematical induction.

• Now we prove the converse.

• Let T ⊆ Z+ and T ̸= ∅.

• It suffices to show that T contains a smallest element.

• Let S(n) be the (open) statement:

“no element of T is smaller than n.”



• S(1) is true as no positive integers are smaller than 1.

• Suppose S(k + 1) holds whenever S(k) does.

• By mathematical induction, S(n) is true for all n ∈ Z+.

• This means for any n ∈ Z+, no element of T is smaller

than n.

• But this is impossible as any integer in T must be

smaller than some integer.

• Hence there is a k ∈ Z+ such that S(k) is true but

S(k + 1) is not.


The Proof (concluded)

• As S(k) holds, no element of T is smaller than k.

• As S(k + 1) does not hold, some elements of T are

smaller than k + 1.

• But as S(k) holds, these elements must equal k.

• Hence the smallest element of T exists and is k.


Philosophical Issues

• Mathematical induction has nothing to do with

induction in the physical and empirical sciences.

– Sun rises on Monday, on Tuesday, etc., so it must rise

every day from now?

• Mathematical induction is merely a property of integers.

• It is deductive.


Compositions of Positive Integers Revisited

• Recall that a composition for m is a sum of positive

integers whose order is relevant and which sum to m

(p. 82).

• Next we use mathematical induction to reprove the

number of compositions for m is 2m−1.

• The statement clearly holds when m = 1.

• So assume it holds for general m and now consider the

compositions of m+ 1.



• Suppose the last summand is n > 1.

– Replace the last summand by n− 1.

– The result is a composition of m.

– This correspondence is one-to-one (why?).

– So there are 2m−1 compositions in this case.



• Now suppose the last summand is 1.

– Remove the last summand.

– The result is a composition of m.

– This correspondence is also one-to-one (why?).

– So there are 2m−1 compositions in this case.

• The total number of compositions of m+ 1 is hence

2m−1 + 2m−1 = 2m.

• This is consistent with Theorem 21 (p. 83).


Do You Really Need Induction?

• In the proof that the total number of compositions of m

is 2m−1, induction was used.

• Can we do away with it by, say, an indirect proof?

• So we set out to obtain a contradiction by assuming the

desired number is not 2m−1.

• What next?

c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University 6

Do You Really Need Induction? (concluded)

• It is typical to work on the smallest m such that the

desired number does not equal 2m−1.

• This m cannot be 1 by inspection.

• But if m > 1, we obtain another contradiction because ...

• This proof relies, if surreptitiously, on the well-ordering

principle (p. 163)!

• But that principle is equivalent to mathematical

induction (p. 169).


Identities for Summations

Theorem 38 For n ∈ Z+,

n2(n+ 1)2

4=

n∑i=1

i3 =

(n∑

i=1

i

)2

.

• The identities are clearly true when n = 1.

• Assume the identities are true for n = k.

• By induction,

(k + 1)2(k + 2)2

4

=k2(k + 1)2

4+ (k + 1)3 =

k∑i=1

i3 + (k + 1)3 =k+1∑i=1

i3.



• So the first identity holds.

• As for the second identity,(k+1∑i=1

i

)2

=

(k∑

i=1

i

)2

+ 2(k + 1)k∑

i=1

i+ (k + 1)2

=

(k∑

i=1

i

)2

+ 2(k + 1)k(k + 1)

2+ (k + 1)2

=

(k∑

i=1

i

)2

+ (k + 1)3

=k∑

i=1

i3 + (k + 1)3 =k+1∑i=1

i3.


Fibonaccia Numbers (1202)

• Let F0 = 0 and F1 = 1.

• Let Fn = Fn−1 + Fn−2 for n ≥ 2.b

– F2 = 0 + 1 = 1.

– F3 = 1 + 1 = 2.

– F4 = 1 + 2 = 3.

• Innumerable applications in surprisingly diverse fields.

aLeonardo Fibonacci (1175–1250).bThis is called a recursive definition; useful when it is simpler or when

an explicit formula is unavailable.


An Identity for Fibonacci Numbers

Theorem 39∑n

i=0 F2i = FnFn+1 for n ∈ Z+.

• For n = 1, F 20 + F 2

1 = 1 = F1F2.

• Inductively,

k+1∑i=0

F 2i

=

(k∑

i=0

F 2i

)+ F 2

k+1 = FkFk+1 + F 2k+1

= Fk+1(Fk + Fk+1) = Fk+1Fk+2.


Another Identity for Fibonacci Numbers

Theorem 40∑n

i=0 Fi = Fn+2 − 1 for n ∈ N.

• For n = 0, F0 = 0 = 1− 1 = F2 − 1.

• Inductively,

k+1∑i=0

Fi

=

(k∑

i=0

Fi

)+ Fk+1

= Fk+2 − 1 + Fk+1

= Fk+3 − 1.


Fundamental Integer Arithmetics

• b|a means that b divides a, where a, b ∈ Z and b ̸= 0.

– b is a divisor or factor of a; a is a multiple of b.

• If a, b ∈ Z with b > 0, then there exist unique q, r ∈ Zsuch that a = qb+ r, where 0 ≤ r < b.

• gcd(a, b) > 0 denotes the greatest common divisor of

a, b ∈ Z, where a ̸= 0 or b ̸= 0 (hence gcd(a, 0) = | a |).

• A prime is a positive integer larger than 1 whose only

divisors are itself and 1.

• There are infinitely many primes (p. 123).


Application: d(n), Number of Positive Divisors

Theorem 41 Let n = pe11 pe22 · · · pett be the prime

factorization of n. Then the number of positive divisors of n

equals d(n) = (e1 + 1)(e2 + 1) · · · (et + 1).

• A positive divisor of n is of form n = ps11 ps22 · · · pstt ,

where 0 ≤ si ≤ ei.

• There are e1 + 1 choices for s1, e2 + 1 choices for s2, etc.


Infinite Sets

• A set is countable if it is finite or if it can be put in

one-to-one correspondence with N (in which case it is

called countably infinite).

– Set of integers Z.∗ 0 ↔ 0, 1 ↔ 1, 2 ↔ 3, 3 ↔ 5, . . . ,−1 ↔ 2,−2 ↔

4,−3 ↔ 6, . . ..

– Set of positive integers Z+: i− 1 ↔ i.

– Set of positive odd integers: (i− 1)/2 ↔ i.

– Set of (positive) rational numbers: See next page.

– Set of squared integers: i ↔√i .a

aGalileo’s (1564–1642) paradox (1638).


Rational Numbers Are Countable

5/2 5/1

1/5 1/2 1/1 1/3 1/4

2/1 2/2 2/3 2/4

3/1 3/2 3/3 3/4

4/1 4/2 4/3

1/6

2/5

6/1


Cardinality

• For any set A, define |A | as A’s cardinality (size).

• Two sets are said to have the same cardinality (written

as |A | = |B | or A ∼ B) if there exists a one-to-one

correspondence between their elements.

• 2A denotes set A’s power set, that is {B : B ⊆ A}.– If |A | = k, then | 2A | = 2k.

– So |A | < | 2A | when A is finite (see p. 189 for proof).


Cardinality (concluded)

• |A | ≤ |B | if there is a one-to-one correspondence

between A and one of B’s subsets.

• |A | < |B | if |A | ≤ |B | but |A | ̸= |B |.

• If A ⊆ B, then |A | ≤ |B |.

• But if A ( B, then |A | < |B |?


A Combinatorial Proof for k < 2k

• Let

A = {1, 2, . . . , k}

be a set with k elements.

• |2A| = 2k.

• But

2A = {{1}, {2}, . . . , {k}, {1, 2}, . . . , {1, 2, . . . , k}}.

• Hence k < |2A| = 2k.


Cardinality and Infinite Setsa

• Suppose A and B are infinite sets.

• Then it is possible that A ( B and yet |A | = |B |.– The set of integers properly contains the set of odd

integers.

– But the set of integers has the same cardinality as

the set of odd integers.b

• This is contrary to the axiom of Euclid that the whole is

greater than any of its proper parts.

• A lot of “paradoxes.”aRussell (1914), “it has taken two thousand years to answer.”bLeibniz uses it to “prove” that there are no infinite numbers (Russell,

1914).


The point of philosophy is

to start with something so simple

as not to seem worth stating,

and to end with something

so paradoxical that no one will believe it.

— Bertrand Russell (1872–1970)


Cantor’s Theorem

Theorem 42 The set of all subsets of N (2N) is infinite and

not countable.

• Suppose (2N) is countable with f : N → 2N being a

one-to-one correspondence.a

• Consider the set B = {k ∈ N : k ̸∈ f(k)} ⊆ N.

• Suppose B = f(n) for some n ∈ N.aNote that f(k) is a subset of N.



• If n ∈ f(n) = B, then n ∈ B, but then n ̸∈ B by B’s

definition.

• If n ̸∈ f(n) = B, then n ̸∈ B, but then n ∈ B by B’s

definition.

• Hence B ̸= f(n) for any n.

• f is not a one-to-one correspondence, a contradiction.


Georg Cantor (1845–1918)

Kac and Ulam (1968), “[If] one

had to name a single person

whose work has had the most

decisive influence on the present

spirit of mathematics, it would

almost surely be Georg Cantor.”


Cantor’s Diagonalization Argument Illustrated

f(1)

f(2)

f(3)

f(4)

f(5)

f(6)

B

1 2 3 4 5 6


Other Uncountable Setsa

• Real numbers are not countable.

– So there are real numbers that cannot be defined in a

finite number of words.

• Transcendental numbers are not countable.b

• So in a sense, most real numbers are transcendental.

aCantor.bA number is algebraic if it is a root of a polynomial equation with

integer coefficients. A real number that is not algebraic is transcen-

dental (Euler).


Application: Existence of Uncomputable Problems

• Every program is a finite sequence of 0s and 1s, thus a

nonnegative integer.

• Hence every program corresponds to some integer.

• It follows that the set of programs is as large as N.

• Now consider functions from N to {0, 1}.

• As each function uniquely defines a subset of N, theircardinality is |2N|.– {i : f(i) = 1} ⊆ N.

• As |N| < |2N| (p. 192), there are functions for which no

programs exist.


Relations and Functions


It is unworthy of excellent men

to lose hours like slaves in the labor of

computation.

— Gottfried Wilhelm von Leibniz (1646–1716)


Cartesian Products

• Let A and B be two sets.

• The Cartesian product of A and B is

A×B = {(a, b) : a ∈ A, b ∈ B}.

• If |A | = m and |B | = n, then

|A×B| = m× n.

• In general,

A1 ×A2 × · · · ×Ak = {(a1, a2, . . . , ak) : a1 ∈ A1, . . .}.

• Ak =

k︷︸︸︷A×A× · · · ×A.


Relationsa

• A subset of A×B is called a relation from A to B.

– A relation can be ∅.

• A subset of A×A is called a binary relation on A.

– {(a, b) : a < b,where a, b,∈ Z} ⊆ Z× Z.

– {(a, b) : b = a2,where a, b,∈ Z} ⊆ Z× Z.aPeirce (1870).


Relations (concluded)

• If |A | = m and |B | = n, then there are

2mn

relations from A to B.

– Each one of the mn 2-tuples (a, b) ∈ A×B can be

either in the relation or not.

– Alternative proof: |A×B | = mn, so there are 2mn

subsets of A×B.


a

b

?

A B


Functionsa

• Let A,B be nonempty sets.

• A function (mapping) f from A to B is denoted by

f : A → B.

– A is called the domain and B is called the

codomain of f .

• A function is a relation from A to B in which f(a) is

associated with a unique b ∈ B, written as f(a) = b.

– b is called the image of a under f , whereas a is a

preimage of b.

– The image of set C under f is f(C) = {f(a) : a ∈ C}.– f(A) is the range of f .

aLeibniz (1692).


Number of Functions

If |A | = m and |B | = n, then there are

|B ||A | = nm (18)

functions from A to B.

• For each a in the domain A, there are n choices of b in

the codomain B to make f(a) = b.

• There are m choices of a ∈ A.

• Hence the desired number is

m︷︸︸︷n× n× · · · × n = nm.


a

?

A B


Number of Booleana Functions

• A function from {0, 1}m to {0, 1} is called a boolean

function.

– For example, f(x1, x2) = x1 ∧ x2 is a boolean

function from {0, 1}2 to {0, 1}.

• There are

2(2m) (19)

boolean functions.

– Eq. (18) on p. 205 with |A | = 2m and |B | = 2.

– Alternative proof: The truth table has 2m rows, and

each row can give you 0 or 1.

aGeorge Boole (1815–1864).


Number of Boolean Functions (concluded)

• In general, a boolean function maps {0, 1}m to {0, 1}n.

• There are

2(n2m)

such boolean functions.

– Eq. (18) on p. 205 with |A | = 2m and |B | = 2n.

– Alternative proof: The truth table has 2m rows, and

each row has n bits of output, for a total of n2m bits

to set.


Partial Functions

• A partial function is like a function except that there

are elements in its domain for which it is undefined.

– f(x) = 1/x is not defined when x = 0.

• Formally, a partial function f : A → B is a function

from some A′ to B:

– A′ ̸= ∅ and A′ ( A.

– f(x) is defined for x ∈ A′.

– f(x) is not defined for x ∈ A−A′.


Number of Partial Functions

If |A | = m and |B | = n, then there are

m−1∑i=1

(m

i

)ni

partial functions from A to B.

• There are(mi

)ways to choose an A′ ( A of size i ≥ 1.

• For each such A′, there are ni functions from A′ to B by

Eq. (18) on p. 205.


Monotone Increasing Functions

A function f : {1, 2, . . . ,m} → {1, 2, . . . , n} is monotone

increasing if f(i) ≤ f(j) whenever i < j.

Theorem 43 There are(m+n−1

m

)monotone increasing

functions from {1, 2, . . . ,m} to {1, 2, . . . , n}.

• Extend monotone increasing function f with f(0) = 1

and f(m+ 1) = n as “guards.”

– f is a monotone increasing function if and only if the

extended f is.



• There are m+ 1 increments:

f(i+ 1)− f(i), 0 ≤ i ≤ m.

• They must all be nonnegative.

• We also know the total increment is fixed, at

f(m+ 1)− f(0) = n− 1.

• So the goal is the number of nonnegative integer

solutions of x1 + x2 + · · ·+ xm+1 = n− 1.



0 0 2

1

m 0

f ( i ) n

f ( m ) - f ( m - 1)

n - f ( m )



• Each such function corresponds to a solution of

xi = f(i)− f(i− 1)

and vice versa (why?).

– For m = 3 and n = 6, the equation to solve is

x1 + x2 + x3 + x4 = 5.

– The solution x1 = 2, x2 = 0, x3 = 1, x4 = 2

corresponds to

f(0) = 1, f(1) = f(0) + 2 = 3,

f(2) = f(1) + 0 = 3, f(3) = f(2) + 1 = 4,

f(4) = f(3) + 2 = 6.



• By the equivalency result on p. 72, the desired count is((m+ 1) + (n− 1)− 1

n− 1

)=

(m+ n− 1

n− 1

)=

(m+ n− 1

m

).

• The number is(2n−1

n

)when m = n.


An Example: Monotone Increasing Functions

• Take m = 4 and n = 3.

• The formula says there are(4+3−1

4

)= 15 monotone

increasing functions from {1, 2, 3, 4} to {1, 2, 3}.

Valid (f(1), f(2), f(3), f(4)) values

1,1,1,1 1,1,1,2 1,1,1,3 1,1,2,2

1,1,2,3 1,1,3,3 1,2,2,2 1,2,2,3

1,2,3,3 1,3,3,3 2,2,2,2 2,2,2,3

2,2,3,3 2,3,3,3 3,3,3,3


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