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Induction II:Inductive Pictures
Great Theoretical Ideas In Computer Science
Steven Rudich
CS 15-251 Spring 2004
Lecture 14 Feb 26, 2004 Carnegie Mellon University
Inductive Proof: “Standard” Induction
“Least Counter-example”“All Previous” Induction
Inductive Definition:Recurrences
Recursive Programming
Theorem? (k¸0)1+2+4+8+…+2k = 2k+1 -1
Try it out on small examples:20 = 21 -120 + 21 = 22 -120 + 21 + 22 = 23 -1
Sk´ “1+2+4+8+…+2k = 2k+1 -1”
Use induction to prove k¸0, Sk
Establish “Base Case”: S0. We have already check it.
Establish “Domino Property”: k¸0, Sk ) Sk+1
“Inductive Hypothesis” Sk:
1+2+4+8+…+2k = 2k+1 -1
Add 2k+1 to both sides:
1+2+4+8+…+2k + 2k+1= 2k+1 +2k+1 -11+2+4+8+…+2k + 2k+1= 2k+2 -1
FUNDAMENTAL LEMMA OF THE POWERS OF TWO:
The sum of the first n powers of 2, is one less than the next power of 2.
Yet another way of packaging inductive
reasoning is to define an “invariant”.
Invariant (adj.) 1. Not varying; constant.
2. (mathematics) Unaffected
by a designated operation, as a transformation of coordinates.
Yet another way of packaging inductive
reasoning is to define an “invariant”.
Invariant (adj.)
3. (programming) A rule, such as the ordering an ordered list or heap, that applies throughout the life of a data structure or procedure.
Each change to the data structuremust maintain the correctness of the invariant.
Invariant InductionSuppose we have a time varying
world state: W0, W1, W2, …Each state change is assumed to come from a list of permissible
operations. We seek to prove that statement S is true of all future
worlds. Argue that S is true of the initial world.
Show that if S is true of some world – then S remains true after one permissible operation is performed.
Odd/Even Handshaking Theorem: Odd/Even Handshaking Theorem: At any party, at any point in time, define a At any party, at any point in time, define a person’s parity as ODD/EVEN according to person’s parity as ODD/EVEN according to the number of hands they have shaken.the number of hands they have shaken. Statement: The number of people of odd Statement: The number of people of odd
parity must be even.parity must be even.
Initial case: Zero hands have been shaken at the start of a party, so zero people have odd parity.
If 2 people of different parities shake, then they both swap parities and the odd parity count is unchanged. If 2 people of the same parity shake, they both change. But then the odd parity count changes by 2, and remains even.
Inductive Definition of n!
[said n “factorial”]
0! = 1; n! = n*(n-1)!
0! = 1; n! = n*(n-1)!
F:=1;For x = 1 to n do
F:=F*x;Return F
Program for n! ?
0! = 1; n! = n*(n-1)!
F:=1;For x = 1 to n do
F:=F*x;Return F
Program for n! ?
n=0 returns 1n=1 returns 1n=2 returns 2
0! = 1; n! = n*(n-1)!
F:=1;For x = 1 to n do
F:=F*x;Return F
Loop Invariant: F=x!True for x=0. If true after k times through – true after k+1 times through.
Inductive Definition of T(n)
T(1) = 1T(n) = 4 T(n/2) + n
Notice that T(n) is inductively defined for positive powers of 2, and undefined on other values.
Inductive Definition of T(n)
T(1) = 1T(n) = 4T(n/2) + n
Notice that T(n) is inductively defined for positive powers of 2, and undefined on other values.
T(1)=1 T(2)=6 T(4)=28 T(8)=120
Guess a closed form formula for T(n).
Guess G(n)
G(n) = 2n2 - nLet the domain of G be the powers of two.
Two equivalent functions?
G(n) = 2n2 - nLet the domain of G be the powers of two.
T(1) = 1T(n) = 4 T(n/2) + nDomain of T are the powers of two.
Inductive Proof of EquivalenceInductive Proof of Equivalence
Base: G(1) = 1 and T(1) = 1
Induction Hypothesis:T(x) = G(x) for x < n
Hence: T(n/2) = G(n/2) = 2(n/2)2 – n/2
T(n) = 4 T(n/2) + n= 4 G(n/2) + n
= 4 [2(n/2)2 – n/2] + n= 2n2 – 2n + n= 2n2 – n= G(n)
G(n) = 2n2 - n
T(1) = 1T(n) = 4 T(n/2) + n
We inductively proved the assertion that
G(n) =T(n).
Giving a formula for T with no sums or
recurrences is called solving the recurrence
T.
Solving RecurrencesSolving RecurrencesGuess and VerifyGuess and Verify
Guess: G(n) = 2n2 – n
Verify: G(1) = 1 and G(n) = 4 G(n/2) + n
Similarly:T(1) = 1 and T(n) = 4 T(n/2) + n
So T(n) = G(n)
Technique 2Technique 2Guess Form and Calculate CoefficientsGuess Form and Calculate Coefficients
Guess: T(n) = an2 + bn + c for some a,b,c
Calculate: T(1) = 1 a + b + c = 1
T(n) = 4 T(n/2) + n an2 + bn + c = 4 [a(n/2)2 + b(n/2) + c] + n = an2 + 2bn + 4c + n (b+1)n + 3c = 0
Therefore: b=-1 c=0 a=2
A computer scientist not only deals with
numbers, but also with
•Finite Strings of symbols
•Very visual objects called graphs
•And especially, especially the special graphs called trees
GRAPHSGRAPHS
b
roota
Definition: GraphsDefinition: Graphs
A graph G = (V,E) consists of a finite set V of vertices (nodes) and a finite set E of edges. Each edge is a set {a, b} of two different vertices.
A graph may not have self loops or multiple edges.
Definition: Directed GraphsDefinition: Directed Graphs
A graph G = (V,E) consists of a finite set V of vertices (nodes) and a finite set E of edges. Each edge is an ordered pair <a,b> of two different vertices.
Unless we say otherwise, our directed graphs will not have multi-edges, or self loops.
Definition: TreeDefinition: TreeA tree is a directed graph with one special node called the root and the property that each node must a unique path from the root to itself.
Child: If <u,v>2E, we sav is a child of uParent: If <u,v>2E, we say u is the parent of uLeaf: If u has no children, we say u is leaf.Siblings: If u and v have the same parent, they are siblings.Descendants of u: The set of nodes reachable from u (including u). Sub-tree rooted at u: Descendants of u and all the edges between them where u has been designated as a root.
Classic Visualization: TreeClassic Visualization: Tree
Inductive rule:If G is a single node
Viz(G) =
If G consists of root r with sub-trees T1, T2, …, Tk
Viz(G) =Vuz(T1)Viz(T2) Viz(Tk)
…..
I own 3 beanies and 2 ties. How many beanie/tie
combos might I wear to the ball tonight?
A choice tree is a tree with an object called a “choice”
associated with each edge and a label on each leaf.
Choice TreeChoice Tree
Definition: Labeled TreeDefinition: Labeled Tree
A tree node labeled by S is a tree T = <V,E> with an associated functionLabel1: V to S
A tree edge labeled by S is a tree T = <V,E> with an associated functionLabel2: E to S
was very illuminating.
Let’s do something similar to illuminate the nature of
T(1)=1; T(n)= 4T(n/2) + n
T(1)=1; T(n)= 4T(n/2) + n
For each n (power of 2),we will define a tree W(n) node labeled by Natural
numbers. W(n) will give us an incisive picture of T(n).
Inductive Definition Of Labeled Tree W(n)Inductive Definition Of Labeled Tree W(n)
T(n) = n + 4 T(n/2)
n
W(n/2) W(n/2) W(n/2) W(n/2)
W(n) =
W(1)
T(1) = 1
1=
Inductive Definition Of Labeled Tree W(n)Inductive Definition Of Labeled Tree W(n)
T(2) = 6
2W(2) =
W(1)
T(1) = 1
1=
1 1 1 1
Inductive Definition Of Labeled Tree W(n)Inductive Definition Of Labeled Tree W(n)
T(4) = 28
4W(4) =
2
1 1 1 1
2
1 1 1 1
2
1 1 1 1
2
1 1 1 1
Inductive Definition Of Labeled Tree W(n)Inductive Definition Of Labeled Tree W(n)
T(4) = 28
4W(4) =
2
1 1 1 1
2
1 1 1 1
2
1 1 1 1
2
1 1 1 1
NOTE: When we sum all the node labels on W(n), we get
T(n)
Invariant: LabelSum W(n) = T(n)Invariant: LabelSum W(n) = T(n)
T(n) = n + 4 T(n/2)
n
W(n/2) W(n/2) W(n/2) W(n/2)
W(n) =
W(1)
T(1) = 1
1=
n
W(n/2) W(n/2) W(n/2) W(n/2)
W(n) =
n
W(n/2) W(n/2) W(n/2)
W(n) =
n/2
W(n/4)W(n/4)W(n/4)W(n/4)
nW(n) =
n/2
W(n/4)W(n/4)W(n/4)W(n/4)
n/2
W(n/4)W(n/4)W(n/4)W(n/4)
n/2
W(n/4)W(n/4)W(n/4)W(n/4)
n/2
W(n/4)W(n/4)W(n/4)W(n/4)
nW(n) =
n/2 n/2 n/2n/2
11111111111111111111111111111111 . . . . . . 111111111111111111111111111111111
n/4 n/4 n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4 n/4
n
n/2 + n/2 + n/2 + n/2
Level i is the sum of 4i copies of n/2i
. . . . . . . . . . . . . . . . . . . . . . . . . .
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4
0
1
2
i
log n2
= 1n
= 2n
= 4n
= 2in
= (n)n
n(1+2+4+8+ . . . +n) = n(2n-1) = 2n2-n
Level i is the sum of 4 i copies of n/ 2 i
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
. . . . . . . . . . . . . . . . . . . . . . . . . .
n/ 2 + n/ 2 + n/ 2 + n/ 2
n
n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4
0
1
2
i
lo g n2
Instead of T(1)=1; T(n)= 4T(n/2) + n
We could illuminateT(1)=1; T(n) = 2T(n/2) + 2
T(n) = n + 2 T(n/2)
n
W(n/2) W(n/2)
W(n) =
W(1)
T(1) = 1
1=
n
n/2 + n/2
Level i is the sum of 2i copies of n/2i
. . . . . . . . . . . . . . . . . . . . . . . . . .
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
0
1
2
i
log n2
n
n
n
. . . . . . . . . . . . . . . . . . . . . . . . . .
n
0
1
2
i
log n2
T(1)=1; T(n) = 2T(n/2) + n
Has closed form: nlog2(n) where n is a power of 2
Representing a recurrence relation as a labeled tree is one of the basics tools you will have to put recurrence
relations in closed form.
The Lindenmayer GameThe Lindenmayer Game
= {a,b}Start word: a
SUB(a) = ab SUB(b) = aFor each w = w1 w2 … wn
NEXT(w) = SUB(w1)SUB(w2)..SUB(wn)
The Lindenmayer GameThe Lindenmayer Game
SUB(a) = ab SUB(b) = aFor each w = w1 w2 … wn
NEXT(w) = SUB(w1)SUB(w2)..SUB(wn)
Time 1: aTime 2: abTime 3: abaTime 4: abaabTime 5: abaababa
The Lindenmayer GameThe Lindenmayer Game
SUB(a) = ab SUB(b) = aFor each w = w1 w2 … wn
NEXT(w) = SUB(w1)SUB(w2)..SUB(wn)
Time 1: aTime 2: abTime 3: abaTime 4: abaabTime 5: abaababa
How long are the strings as a function of
time?
Aristid Lindenmayer (1925-1989)Aristid Lindenmayer (1925-1989)
1968 Invents L-systems in Theoretical Botany
Time 1: aTime 2: abTime 3: abaTime 4: abaabTime 5: abaababa
FIBONACCI(n) cells at time n
The Koch GameThe Koch Game
= {F,+,-}Start word: F
SUB(F) = F+F--F+F SUB(+)=+ SUB(-)=-For each w = w1 w2 … wn
NEXT(w) = SUB(w1)SUB(w2)..SUB(wn)
The Koch GameThe Koch Game
Gen 0:FGen 1: F+F--F+FGen 2: F+F--F+F+F+F--F+F--F+F--F+F+F+F--F+F
The Koch GameThe Koch Game
Picture representation:
F draw forward one unit+ turn 60 degree left - turn 60 degrees right.
Gen 0: FGen 1: F+F--F+FGen 2: F+F--F+F+F+F--F+F--F+F--F+F+F+F--F+F
The Koch GameThe Koch Game
F+F--F+F F+F--F+F
The Koch GameThe Koch Game
F+F--F+F+F+F--F+F--F+F--F+F+F+F--F+F F+F--F+F+F+F--F+F--F+F--F+F+F+F--F+F
Koch CurveKoch Curve
Dragon Game
SUB(X) =X+YF+SUB(Y) = -FX-Y Dragon
Curve:
Hilbert GameSUB(L)= +RF-LFL-FR+SUB(R)= -LF+RFR+FL-
Hilbert Curve:
Note: Make 90 degree turns instead of 60 degrees.
Hilbert’s Space Filling CurveHilbert’s Space Filling Curve
Peano-Gossamer CurvePeano-Gossamer Curve
Sierpinski TriangleSierpinski Triangle
Lindenmayer 1968Lindenmayer 1968
SUB(F) = F[-F]F[+F][F]
Interpret the stuff inside brackets as a branch.
Lindenmayer 1968Lindenmayer 1968
Inductive LeafInductive Leaf
Here is another kind of inductive definition called a formal grammar.
Formal Grammar:Formal Grammar:G = (T, V, S, P)G = (T, V, S, P)
A finite alphabet T called the terminal symbols.
A disjoint finite alphabet N of non-terminal symbols (or variable symbols). One symbol S in V is called the start symbol.
A finite set P of production rules of the form , where , 2 (N [ T)* and contains at least one non-terminal symbol.
Example.Example.
Terminal Symbols: {a,b,c}Non-terminal symbols: {S, X}Start symbol: S
Production rules:S SSS aSS b
Formal Grammar:Formal Grammar:G = (T, V, S, P)G = (T, V, S, P)
A finite set P of production rules of the form , where , 2 (N [ T)* and contains at least one non-terminal symbol.
Let x and y be strings over N[T.We say that y is produced in one step from x if there is a rule such that one occurrence of the substring is replaced with to obtain the string y. We denote this x 1 y
S S 11 SSS SSS 11 bS bS 1 1 baba
Terminal Symbols: {a,b,c}Non-terminal symbols: {S, X}Start symbol: S
Production rules:S SSS aSS b
Definition of x Definition of x kk y y
We know what x 1 y means.Define x 0 y only when x = y
Define x ->k to mean 9 z s.t. x->1z and z->k-1y
Define x*y to mean that there exists a k such that xk y .
Language Produced By Language Produced By Formal Grammar GFormal Grammar G
LG = {x 2 T* | S*x }
= {x 2 T* | 9 k s.t. Skx }
The language of strings of terminal symbols that are eventually produced by the start symbol.
Language = {a,b}Language = {a,b}++
Terminal Symbols: {a,b,c}Non-terminal symbols: {S, X}Start symbol: S
Production rules:S SSS aSS b
Evaluating A Formal DerivationEvaluating A Formal Derivation
Let G be a formal grammar.A sequence D = S1, S2, …., Sk, f strings is called a formal derivation of if:
For 1· q · k-1 Sq 1 Sq+1
Once you agree on a specific grammar G, it is trivial to verify proofs of membership in the language.
G = BalancedG = Balanced
S ()S SSS (S)
Now let’s show Symbol Dude the following derivation:S, (S), (SS), (SS), (S()), ((())())
Yo, dude! ((())()) is like totally Yo, dude! ((())()) is like totally balancedbalanced
XXh42TEhha
Expressions in Propositional LogicExpressions in Propositional Logic
E True | False
x | y | z
E
(E Ç E)
(E Æ E)
Yo, dude! Yo, dude! ::(x(xÆÆy) is like totally an y) is like totally an expression in propositional logic.expression in propositional logic.
XXh42TEhha
Meaning:Meaning:The Part Symbol Dude IgnoresThe Part Symbol Dude Ignores
E True | False (constants)
x | y | z (variables)
E1 (not E1)
(E1 Ç E2) (E1 or E2)
(E1 Æ E2) (E1 and E2)
Inductively Associate a Inductively Associate a MEANING M(E) of any expression EMEANING M(E) of any expression E
M(T) = T M(F)=F
M(: E) = not M(E)
M( (E1 Æ E2) = M(E1) and M(E2)
M( (E1 Ç E2) = M(E1) or M(E2)
not T = F not F =T
AND.
F and F = F F and T = F T and F – F T and T= T
Expressions in Expressions in First OrderFirst Order Logic Logic
T x | y | z f(T1, T2, …, Tk) (functions) T = Terms
E R(T1, T2, …, Tk) (relation on terms)
E1 | (E1 Ç E2) | (E1 Æ E2)
8 x. E1 (universal quantifier)
9 x. E1 (existential quantifier)
E.g., Number TheoryE.g., Number Theory
T x | y | z 0 (the zero function)
S(T1) (successor function)
(T1+T2) | (T1*T2)| (T1^T2) E (T1 = T2) | (T1 < T2 ) (relations between terms)
8 x. E1 | 9 x. E1
8 x. (S(S(x)) < S(x^S(S(0))))
= 8 x. ((x+2) < (x2 + 1))
What about something even
more complex, like Java?
You can write down the entire grammar!
Statements in JavaStatements in Java
S ;
Expression ;
{ S } | { }
if ( Expression ) then S1 else S2
do S while ( Expression ) ;
while ( Expression ) S ;
return ; | return Expression ;
…
Expressions in JavaExpressions in Java
Expression Numeric Expression
Logical Expression
String Expression
…
Numerical Expressions in JavaNumerical Expressions in Java
NE - NE | --NE | ++NE
NE -- | NE ++
NE += NE | NE -= NE
NE + NE | NE - NE
( NE )
Identifier…
Programs in JavaPrograms in Java
Type Decleration
Class Decleration | Interface Decleration
Class Decleration
class Identifier extends ClassName
{ FieldDeclerationList }
Compilation Unit
Import Statement Type Decleration
Now we will expand our notion of a legal move a bit to include things that are very intuitive.
Symbol TemplatesSymbol Templates
Fix a finite alphabet .Fix a finite alphabet .Let {#} be a symbol not in [
Define a template for to be any finite string over [ .
Template Example.Template Example.
= {a, b}= {X, Y}
Templates:
aaXXa XYXYabaXbaaa
Template ExamplesTemplate Examples
= {a, b}= {X, Y, Z}
XYX matches aaabababaaabaaaabababaaaba has the form XYX
Algebra Game TemplatesAlgebra Game Templates
X+Y = Y+X(X+y)+z = x+(y+z)(x*y)*z=x*(y*z)X*Y =Y*XX*(Y+Z) = (X*Y) + (X*Z)X+0 = XX*0= 0
Algebra Game AllowsAlgebra Game Allowssubstitution too!substitution too!
Example UBLExample UBL = { S, =} = { S, =}
One Axiom A: { S=S }
One Inference Rule:r() = SS
Proof: S=S, SS=SS, SSS=SSS
Example UBL Example UBL = { S, =} = { S, =}
One Axiom A: { S=S }One Inference Rule: r() = SS
Proof that SSS=SSS: S=S, SS=SS, SSS=SSS
Example UBL Example UBL = { S, =} = { S, =}
One Axiom A: { S=S }One Inference Rule: r() = SS
Proof that premise SSS=S implies SSSSS=SSS: SSS=S, SSSS=SS, SSSSS=SSS
Example UBL Example UBL = { S, =} = { S, =}
One Axiom A: { S=S }One Inference Rule: r() = SSUBL is completely defined in syntactic terms. UBL is just a game of symbol pushing. Does it have a natural semantics?
UBLUBL
UBL can be interpreted as
UNARY BALANCE LOGIC
Propositional LogicPropositional Logic
Axioms = all strings of the form
Inference rules [Hilbert]:r1( ) =
r2() = [this is a relation, actually]
r3(( ) ) = ( )
Br( , : ) =
Example: p, Example: p, :: p p q implies q q implies q
p, p q, q, :: p p q, q q, q q , q q , q
Inference rules [Hilbert] :r1( ) =
r2() = [this is a relation, actually]
r3(( ) ) = ( )
Br( , : ) =
Propositional LogicPropositional Logic
Propostional logic and proof can be viewed as a symbol game, or view semantically.
: means NOT, means OR.An n-variable formula is a tautology if it evaluates to TRUE under all 2n truth assignments.
Propositional LogicPropositional Logic
SOUND: Any formula that can be derived by the manipulation rules is indeed a tautology.
COMPLETE: All tautologies can be derived by the logic.
First-order PredicatesFirst-order Predicates
Enhance our syntax to be give us the possibility of a richer semantics.
Inductive Grammar Notation Inductive Grammar Notation
<integer> ::= <natural> | -<natural>
<natural> ::= <decimal digit> | <natural> <decimal digit>
<decimal digit> ::= 0|1|2|3|4|5|6|7|8|9
Grammar: Grammar:
<operator> ::= + | - | * | /<expression> ::= <natural> | <natural> <operator> <natural>
<expression>::= (<expression) | - (<expression>) |(<expression>)<operator>(<expression>)
Computer Language Syntax Computer Language Syntax
The syntax of computer languages can be defined by an inductive grammar. In fact, most languages can be defined by a restricted form of grammar, called BNF form.
The Natural NumbersThe Natural Numbers
= { 0, S0, SS0, SSS0, . . .}
Inductive definition:
0 is a natural number called zero.Set notation: 0 2
If X is a natural number, then SX is a natural number called successor of X.Set notation: X 2 ) SX 2
Inductive Definition of +Inductive Definition of +
= { 0, S0, SS0, SSS0, . . .}
Inductive definition of addition (+):
X, Y 2 )X “+” 0 = XX “+” SY = S(X”+”Y)
Defining One to TenDefining One to Ten
“1” = S0; “2”= SS0; “3” = SSS0. “4” = SSSS0. “5” = SSSSS0; “6” = SSSSSS0; “7” = SSSSSSS0. “8” = SSSSSSSS0; “9=SSSSSSSSS0. “10” = SSSSSSSSSS0
Or…. “1” = S0. “2” = S1. “3” = S2. “4” = S5 …
1 + 1 = 21 + 1 = 2
Proof:
1 + 1 = S0 + S0 = S(S0 + 0) = S(S0) = SS0 = 2
X, Y 2 )X “+” 0 = XX “+” SY = S(X”+”Y)
Inductive Definition of *Inductive Definition of *
= { 0, S0, SS0, SSS0, . . .}
Inductive definition of times (*):
X, Y 2 )X “*” 0 = 0X “*” SY = (X”*”Y) + X
Inductive Definition of ^Inductive Definition of ^
= { 0, S0, SS0, SSS0, . . .}
Inductive definition of raised to the (^):
X, Y 2 )X “^” 0 = 1 [or X0 = 1 ]X “^” SY = (X”^”Y) * X [or XSY = XY * X]
Inductive Definition of Base 10 Inductive Definition of Base 10 Notation.Notation.
Base10( 10*X + 0) = Base10(X) “0”Base10( 10*X + 1) = Base10(X) “1” …Base10( 10*X + 1) = Base10(X) “9”
Base10(SSSSSSSSSSSS0) = Base10(SSSSSSSSSS0) 2 =1 2
= { 0, 1, 2, 3, . . .}= { 0, 1, 2, 3, . . .}
Defining < for :
8 x,y 2 “x > y” is TRUE “y < x” is TRUE“x > y” is TRUE “y > x” is FALSE
“x+1 > 0” is TRUE“x+1 > y+1” is TRUE ) “x > y” is TRUE
= { 0, 1, 2, 3, . . .}= { 0, 1, 2, 3, . . .}
Defining partial minus for :
8 x,y 2 x-0 = x x>y )
(x+1) – (y+1) = x-y
a =a = [ [a DIV ba DIV b]*b + [a mod b]]*b + [a mod b]
Defining DIV and MOD for :
8 a,b 2 a<b )
a DIV b = 0 a¸b>0 )
a DIV b = 1 + (a-b) DIV b
a MOD b = a – [b*(a DIV b)]
The maximum number of times b goes into a without going over.
The remainder when a is divided by b.
45 =45 = [ [45 DIV 1045 DIV 10]*10 + [ 45 MOD 10]]*10 + [ 45 MOD 10]= 4*10 + 5= 4*10 + 5
Defining DIV and MOD for :
8 a,b 2 a<b )
a DIV b = 0 a¸b>0 )
a DIV b = 1 + (a-b) DIV b
a MOD b = a – [b*(a DIV b)]
Giuseppe Peano [1889]Giuseppe Peano [1889]Axiom’s For Axiom’s For
A1. Sx 0
A2. [Sx = Sy] ) [x=y]
A3. For any proposition P(x) where x2 . Mathematical Induction Applies To P:
[P(0) and 8 x2 , P(x)) P(Sx)]) 8 x P(x)
Giuseppe Peano [1889]Giuseppe Peano [1889]Axiom’s For Axiom’s For
A1. Sx 0
A2. [Sx = Sy] ) [x=y]
A3. For any proposition P(x) where x2 .
Mathematical Induction Applies To P:[P(0) and 8 x2 , P(x)) P(Sx)]
) 8 x P(x) Inductive Definition of +:
x + 0 = xx + Sy = S(x + y)
Let’s prove the
Commutativity of addition: x +y = y+ x
Lemma: 0 + x = xLemma: 0 + x = x
Let P(x) be the proposition that “0 + x = x”
P(0) is “0 + 0 = 0”Assume P(x): “0 + x = x” Show P(Sx):0+Sx = S(0+x) = S(x) = Sx
Lemma: Sx + y = S(x+y)Lemma: Sx + y = S(x+y)
Let P(y) be the proposition that “8 x, Sx + y = S(x+y)”
P(0) is “8 x, Sx + 0 = S(x+0) = Sx”Assume P(y): “8 x, Sx+y = S(x+y)” Show P(Sy):Sx+Sy = S(Sx+y) = S(S(x+y) ) = S(x+Sy)
Theorem: Commutative Theorem: Commutative Property Of Addition: x + y = Property Of Addition: x + y =
y + xy + x
Let P(y) be the proposition that “8 x, x + y = y + x”
P(0) is “8 x, x + 0 = 0 + x”Assume P(y): “8 x, x + y = y + x” Show P(Sy):x+Sy = S(x+y) = S(y+x) = Sy + x
““God Made Induction On The God Made Induction On The Naturals. Everything Else Is The Naturals. Everything Else Is The
Work Of Man.”Work Of Man.”
= { 0, 1, 2, 3, . . .}
Peano Axioms: Gives us inductive reasoning.
Inductively Define: all of arithmetic, high school math, and beyond…
Peano
ReferencesReferences
The Heritage of Thales, by W. S. Anglin and F. Lambek
The Book Of Numbers, by J. Conway and R. Guy
Programming Pearls, by J. Bentley
History of Mathematics, Histories of Problems, by The Inter-IREM Commission