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Language, Proof and
Logic
Mathematical Induction
Chapter 16
16.0
Induction – a powerful method of proof
P(1)
x (P(x)P(x+1))
------------------------------------x P(x)
...
1
2
3
4
5
6
7
8
9
10
11
12
...
If
• Person #1 knows the secret
• For all n, if person # n knows the secret, then so does person # n+1
-----------------------------------------------------------------------------------Then
• For all n, person # n knows the secret
16.1.a
Inductive definitions and inductive proofs
Example of an inductive definition: wff
1. Every atomic wff is a wff
2. If P is a wff, so is P
3. If P1,…,Pn are wffs, so are (P1…Pn) and (P1…Pn)
4. If P and Q are wffs, so are (PQ) and (PQ)
5. If P is a wff and x is a variable, xP and xP are wffs;
6. Nothing is a wff unless it is generated by repeated applications of 1-5.
REMEMBER
An inductive definition of a set consists of:
• a base clause, which specifies the basic elements of the defined set,
• one or more inductive clauses, which tell us how to generate additional
elements, and
• a final clause, which tells us that all the elements are either basic or
generated by the inductive clauses.
16.1.b
Inductive definitions and inductive proofs
Example of an inductive definition: pal
1. Each letter in the alphabet (a,b,...,z) is a pal.
2. If a string  is a pal, so is the result of putting any letter of the
alphabet both in front and in back of  (e.g. aa, bb, etc.)
3. Nothing is a pal unless it is generated by repeated applications of 1-2.
REMEMBER
An inductive definition of a set consists of:
• a base clause, which specifies the basic elements of the defined set,
• one or more inductive clauses, which tell us how to generate additional
elements, and
• a final clause, which tells us that all the elements are either basic or
generated by the inductive clauses.
16.1.c
Inductive definitions and inductive proofs
Example of an inductive definition: pal
1. Each letter in the alphabet (a,b,...,z) is a pal.
2. If a string  is a pal, so is the result of putting any letter of the
alphabet both in front and in back of  (e.g. aa, bb, etc.)
3. Nothing is a pal unless it is generated by repeated applications of 1-2.
REMEMBER
Given an inductive definition of a set S, an inductive proof of the fact
that a certain property holds of all elements of S requires:
• a basis step, which shows that the property holds of the basic
elements, and
• an inductive step, which shows that if the property holds of some
elements, then it holds of any elements generated from them by
the inductive clauses.
The assumption that begins the inductive step is called the inductive
hypothesis.
16.1.d
Inductive definitions and inductive proofs
Example of an inductive definition: pal
1. Each letter in the alphabet (a,b,...,z) is a pal.
2. If a string  is a pal, so is the result of putting any letter of the
alphabet both in front and in back of  (e.g. aa, bb, etc.)
3. Nothing is a pal unless it is generated by repeated applications of 1-2.
By induction, prove that:
a) Every pal has an odd length
b) Every pal is a palindrome
16.2
Inductive definitions in set theory
Making the final clause of an inductive definition more precise:
The set P of pals is the smallest set --- i.e. the intersection of all sets --such that:
1. Each letter in the alphabet (a,b,...,z) is in P.
2. If a string  is in P, so is the result of putting any letter of the
alphabet both in front and in back of  (e.g. aa, bb, etc.)
16.3.a
Induction on the natural numbers
The inductive definition of natural numbers:
1. 0 is a natural number.
2. If n is a natural number, then n+1 is a natural number.
3. Nothing is a natural number except in virtue of repeated
applications of (1) and (2).
OR:
The set N of natural numbers is the smallest set satisfying:
1. 0N
2. If nN, then n+1N.
To prove by induction that P(x) is true of all natural numbers:
1. Prove P(0) (basis step)
2. Prove x[P(x)P(x+1)] (inductive step)
16.3.b
Induction on the natural numbers
Proposition 4. For every natural number n, the sum of the first n
natural numbers is n(n-1)/2.
Proof.
Basis: The sum of the first 0 natural numbers is indeed 0.
Inductive step: Assume the sum of the first k natural numbers is
k(k-1)/2 (inductive hypothesis). We want to show that then the same
is true for k+1 instead of k, that is, the sum of the first k+1 natural
numbers is (k+1)((k+1)-1)/2, i.e. it is k(k+1)/2.
But indeed, the sum of the first k+1 natural numbers is X+k, where
X is the sum of the fist k natural numbers. By the inductive hypothesis,
X= k(k-1)/2 . Thus, the sought X+k is
k(k-1)/2+k = (k(k-1)+2k)/2 = k(k-1+2)/2 = k(k+1)/2, as desired.
Axiomatizing the natural numbers
Peano Arithmetic PA
Language: =, 0, s, +,  (s(a) means a+1)
16.4.a
Axioms:
1. x(s(x)0)
2. xy (s(x)=s(y)  x=y)
Gödel’s Incompleteness Theorem:
These axioms are not sufficient to
prove every true arithmetical sentence.
Nor would any bigger set of axioms be
sufficient.
3. x (x+0 = x)
5. x (x0 = 0)
Axiom 7 is a scheme, for every
Q. If Q contains additional
variables z1,...,zn, then the whole
thing should be prefixed with
z1... zn
6. xy [xs(y) = (xy)+x]
This axiom is called the
induction scheme
4. xy [x+s(y) = s(x+y)]
7. [Q(0)  x (Q(x)  Q(s(x)))]  xQ(x)
Informal proof of x(s(x)=s(0)+x)
16.4.b
What we need:
Axiom 3: x (x+0 = x)
Axiom 4: xy [x+s(y) = s(x+y)]
Axiom 7: [Q(0)  x (Q(x)  Q(s(x)))]  xQ(x)
with s(x)=s(0)+x in the role of Q(x)
Basis: Q(0), i.e. s(0)=s(0)+0. Follows from Axiom 3.
Inductive step:
Assume (induction hypothesis) Q(n), i.e. s(n)=s(0)+n
We want to show that then
Q(s(n)), i.e. s(s(n))=s(0)+s(n)
But indeed:
s(s(n)) = s(s(0)+n) by induction hypothesis.
= s(0)+s(n) by Axiom 4.
Induction in Fitch
16.5
Peano Induction:
P(0)
n P(n)
…
P(s(n))
xP(x)
Where n does not occur outside
the subproof where it is introduced
16.6
Ordering the Natural Numbers
A binary relation R is said to be a total strict ordering iff it is:
1. Irreflexive:
2. Transitive:
3. Trichotomous: xy (xRy  x=y  yRx)
The ordinary relation < on natural numbers is a total strict ordering.
In PA, “x<y” can be treated as an abbreviation of “z(x+s(z)=y)”.
An alternative approach, taken in Fitch, is to treat < as a legitimate
symbol of the language, defined by the (additional Peano) axiom
xy[x<y  z(x+s(z)=y)]
16.7
Strong Induction
Strong induction (other names: complete induction; course of values
induction) in Fitch takes the following form:
n x[x<n P(x)]
…
P(n)
xP(x)
Where n does not occur outside
the subproof where it is introduced
This principle does not allow us to prove anything new because it is
equivalent to ordinary induction (either one follows from the other --see the textbook). However, it can often offer greater convenience.
Example: Using strong induction, prove the so called Fundamental
Theorem of Arithmetic, according to which every natural number
greater than 1 is either prime or the product of some primes.