MELJUN CORTES Automata Lecture Turing Machines 2

8/21/2019 MELJUN CORTES Automata Lecture Turing Machines 2

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Theory of Computation (With Automata Theory)

* Property of STI

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Types of Proofs

Types of ProofsMathematical Definitions and

Theorems

v Definitions

• A math emat ic al d ef in it io n

describes a concept or an

object in math.

• Definitions must clearly state

what constitutes that object

and what does not.

• After defining a concept or anobject, mathematical

statements can be made about

them usually stating whether

or not they possess a certain

property.


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* Property of STI

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Types of Proofs

v Theorems

• Mathematical statements may

be true or not.

A theorem is a mathematicalstatement that has been

proven to be true.

Lemmas are similar to

theorems except that theyhave minor significance and

are often used in proving

theorems. In other words,

they are stepping stones in the

process of proving a theorem.

Corol lar ies are statements

that can be proven deductively

by applying other theorems.


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* Property of STI

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Types of Proofs

Types of Proof

v Direct Proof

• One of the popular methods of

proving the correctness of

statements is by using

established facts without

making any further

assumptions.

This is usually done by using

proven theorems or lemmas in

establishing the correctness of

the statement.

This type of proof is called the

d irec t p roo f or direct

argument .


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* Property of STI

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Types of Proofs

Examples:

Prove that the statement

“division is transitive”,

meaning, if x is divisible by y ,

and y is divisible by z , then x is

divisible by z .

Proof:

x is divisible by y means

x = (a)y

y is divisible by z means

y = (b)z

where a and b are

integers.

x = (a)y = (a)(b)z

x = (ab)z

Since ab is also an integer,

this means that x is

divisible by z .


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* Property of STI

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Types of Proofs

Prove that If x and y are odd,

then xy is also odd, where x and y are integers.

Proof:

x is an odd integer means:

x = (2a + 1)

y is an odd integer means:

y = (2b + 1)

where a and b are

integers.

xy = (2a + 1 )(2b + 1)

= 4ab + 2a +

2b + 1

xy = 2 (2ab + a + b)+ 1

Since (2ab + a + b) is an

integer, this means that xy

is also an odd integer.


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* Property of STI

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Types of Proofs

Prove that the sum of two

rational numbers is a rationalnumber.

Proof:

x is a rational number

means

x = a/b

y is a rational number

means

y = c/d

where a, b, c , and d are

integers.

x +y = a/b + c /d = (ad + cb)/(bd )

Since (ad + cb) and (bd )

are both integers, this

means that x + y is a

rational number.


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* Property of STI

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Types of Proofs

v Proof by Contradiction

• Proof b y c on trad ic t io n starts

by assuming the statement tobe proven is false and then

deriving consequences or

outcomes.

In the process, if acontradiction to what is known

to be true is reached, then it

implies that the initial

assumption that the statement

is false is incorrect.

Therefore, the statement must

be true.


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* Property of STI

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Types of Proofs

Examples:

Prove that if 3 x + 2 is odd,

then x is odd, where x is an

integer.

Proof:

Assume that 3 x + 2 is odd

but x is even.

x is even means

x = 2awhere a is an integer.

3 x + 2 = 3(2a) + 2

= 6a + 2

= 2(3a + 1)Since (3a +1) is an integer,

that means 3 x + 2 is even

(a contradiction).

Therefore the statement istrue.


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* Property of STI

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Types of Proofs

Prove that if x 2 is even, then x

is even.

Proof:

Assume that x 2 is even, but

x is odd.

x is odd means

x = 2a + 1

where a is an integer.

x 2 = (2a + 1)2

= 4a2 + 4a + 1

= 2(2a2 +2a) + 1

Since (2a2 + 2a) is an

integer, that means x 2 is

odd (a contradiction).

Therefore the statement is

true.


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* Property of STI

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Types of Proofs

Prove that the square root of

two is irrational.

Proof:

Assume that the square

root of two is rational.

If the square root of 2 is

rational, then it means that

where a and b are

integers.

Take note that a and bmust not have common

factors so that the fraction

will be in its simplest form.

This implies that either a or

b must be odd (or both).

b

a=2


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* Property of STI

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Types of Proofs

2 = a2/b2

2b2 = a2

This implies a2

is even. And if a2 is even, then a is

also even.

Since a is even, this

means that

a = 2 x

where x is an integer.

2b2 = a2

= (2 x )2

= 4 x 2

b2 = 2 x 2


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* Property of STI

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Types of Proofs

This again implies that b2

is even which makes beven also.

Since both a and b are

even, this produces a

contradiction to thestatement earlier that at

least one of them must be

odd.

This proves that the square

root of two is irrational.


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* Property of STI

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Types of Proofs

v Proof by Induction

• Proo f by induc t ion is useful

in proving that all elements of

an infinite set, such as the set

of natural numbers, possess a

certain property.

• Proof of induction starts off by

first proving that the property

in question is true for one of

the elements of the set,

usually the first one.

• The second step is to prove

that if the property is true for

one element of the set, it is

also true for the next element.

• Once these two proofs have

been established, then it can

be concluded that the property

is true for all members of the

set.


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* Property of STI

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Types of Proofs

• The first step is called the

bas is s tep while the second

step is called the i nduc t ion

step .

• Formal Definition:

Let P be the property to be

proven true for an infinite set.

Basis Step:

Prove that P (1) is true.

Induction Step:

Prove that if P (k ) is true,

then P (k + 1) is also true

for every positive integer k .


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* Property of STI

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Types of Proofs

Examples:

Prove that 1 + 2 + … + n =

½ n(n+ 1) for all n ≥ 1 (the sum

of the first n integers).

Basis Step: Prove P (1) is true.

Let n = 1:

1 = ½ (1) (1 + 1)

1 = 1

Induction Step: Prove that if

P (k ) is true, then P (k + 1) is

also true for all k .

Assume first that P (k ) is true.

That is,

( )12

1...21 +=+++ k k k


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* Property of STI

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Types of Proofs

Now prove that P (k + 1) istrue. That is,

( ) ( )( )212

11...21 ++=++++ k k k

( ) ( )( )212

11...21 ++=+++++ k k k k

( ) ( ) ( )( )212

111

2

1++=+++ k k k k k

( ) ( )[ ] ( )( )212

1121

2

1++=+++ k k k k k

[ ] ( )( )212

122

2

1 2++=+++ k k k k k

[ ] ( )( )212

123

2

1 2++=++ k k k k


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* Property of STI

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Types of Proofs

This proves that if P (k ) is true,

the P (k + 1) is also true for all

k .

Therefore, the property 1 + 2 +

… + n = ½ n(n+ 1) for all n ≥ 1

is true.

[ ] ( )( )212

123

2

1 2++=++ k k k k

( ) ( )( )212

1)2(1

2

1++=++ k k k k


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* Property of STI

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Types of Proofs

Prove that the sum of the first

n odd positive integers is n2 .


Let n = 1:

1 = (1)2

1 = 1





That is,

2)12(...531 k k =−++++


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* Property of STI

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Types of Proofs

Now prove that P (k + 1) istrue. That is,

This proves that if P (k ) is true,

then P (k + 1) is also true for allk .

Therefore, the statement that

the sum of the first n odd

positive integers is n2 is true.

( ) ( )2112...531 +=+++++ k k

( ) ( ) ( )211212...531 +=++−++++ k k k

( ) ( )22 112 +=++ k k k

( ) ( )22 11 +=+ k k


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* Property of STI

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Types of Proofs

Prove that n3 – n is divisible by

3 whenever n is a positive

integer.


Let n = 1:

13 - 1 = 0

0 is divisible by 3.





That is:

is divisible by 3.

k k −3


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* Property of STI

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Types of Proofs

Now prove that P (k + 1) is

true. That is:

is divisible by 3.

( ) ( )11 3 +−+ k k

( ) ( )1133 23 +−+++ k k k k

1133 23 −−+++ k k k k

k k k k −++ 33 23

k k k k 33 23 ++−

k k k k ++− 23 3


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* Property of STI

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Types of Proofs

Take note that it was

established earlier that k 3 – k

is divisible by 3. The term 3(k 2+ k ) is also divisible by 3. This

makes the entire term divisible

by 3.

This proves that if P (k ) is true,then P (k + 1) is also true for all

k .


n3 – n is divisible by 3

whenever n is a positive

integer is true.


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* Property of STI

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Types of Proofs

Prove that 7n – 1 is divisible by

6 for all n ≥ 1.


Let n = 1:

71 - 1 = 6

6 is divisible by 6.





That is:

is divisible by 6.

17 −k


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* Property of STI

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Types of Proofs


true. That is:

is divisible by 6.

17 )1( −+k

17 )1( −+k

177 −⋅ k

6777 +−⋅ k

6177 +−k


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* Property of STI

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Types of Proofs

Take note that it was

established earlier that 7 k – 1is divisible by 6. Therefore,

the term 7(7k – 1) is also

divisible by 6.

This proves that if P (k ) is true,then P (k + 1) is also true for all

k .


7n – 1 is divisible by 6 for all n

≥ 1 is true.


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* Property of STI

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Types of Proofs

Prove that 20 + 21 + 22 + … +

2n = 2(n+1) – 1 for all n ≥ 0.


Let n = 0:

20 = 2(0+1) – 1

1 = 2 – 1

1 = 1





That is,

( )122...222

1210−=++++

+k k


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true. That is,

( ) ( )122...222

111210−=++++

+++ k k

( ) ( )1222...222

21210−=+++++

++ k k k

( ) ( ) ( ) 12212 211 −=+− +++ k k k

( ) ( ) ( ) 12122 211 −=−+ +++ k k k

( ) ( ) 12122 21 −=−⋅ ++ k k

( ) ( ) 1212 22 −=− ++ k k

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MELJUN CORTES Automata Lecture Turing Machines 2