meljun cortes automata lecture types of proofs 2

Theory of Computation (With Automata Theory)

Types of Proofs *Property of STI of 15

TOPIC TITLE: Types of Proofs

Specific Objectives:

At the end of the topic session, the students are expected to:

Cognitive:

1. Explain what definitions are. 2. Define the different terminologies related to sets such as

subsets, infinite sets, empty sets, etc. 3. Differentiate theorems, lemmas, and corollaries. 4. Apply the different methods of proving the correctness of

mathematical statements.

Affective:

1. Listen to others with respect. 2. Participate in class discussions actively.

MATERIALS/EQUIPMENT:

o topic slides o OHP

TOPIC PREPARATION:

o Have the students review whatever topics were discussed during the previous semesters.

o Prepare the slides to be presented in class. o It is imperative for the instructor to incorporate various kinds of

teaching strategies while discussing the suggested topics. o Prepare additional examples on the topic to be presented.



Definitions Page 1 of 27

Mathematical Definitions and Theorems Definitions A mathematical definition describes or gives meaning to a mathematical concept or object. Definitions must be precise. There should not be any ambiguity in the statements, nor should there be any room for alternative interpretations. Definitions must clearly state what constitutes that object and what does not. For example:

Definition of a Prime Number: A prime number is a natural number that can only be divided by one and itself.

Once the definition has been given, mathematical statements can be made about them, usually stating whether or not they possess a certain property. For example:

There are an infinite number of prime numbers . The only even prime number is two.

[Definitions, Page 1 of 27]

Theorems Page 2 of 27

Theorems Mathematical statements on a concept or object may be true or not. Once the correctness of a statement has been established, it is then called a theorem. Normally, theorems are proven based on known facts and previously proven theorems. Theorems often pertain to concepts or objects that have major importance. Lemmas are similar to theorems except that they have minor significance and are often simply used in proving theorems. In other words, they are stepping stones in the process of proving a theorem since they form part of the proof of a larger, more important theorem . Corollaries are statements whose correctness can be deduced immediately from other theorems. [Theorems, Page 2 of 27]



Direct Proof Page 3 of 27

Types of Proof There are different ways by which the correctness of a statement can be established or proven. Here, we will look at 3—Direct Proof, Proof by Contradiction, and Proof by Induction. 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 combinations of theorems and lemmas in such a way that the correctness directly follows from the given assumptions by logical deduction. This type of proof is called the direct proof or direct argument.


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:

We start with the known fact that a number is said to be divisible by another number if dividing the first by the second will result in an integer. Hence, if x is divisible by y, then it can be expressed as x/y = a where a is an integer. Rewriting it gives

x = (a)y equation 1

Similarly, if y is divisible by z, it can also be expressed as y/z = b where b is an integer. Therefore,

y = (b )z equation 2

Substituting equation 2 into equation 1 gives

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

x = (ab )z

Note that since a and b are integers, then ab is also an integer (Known Fact: The product of 2 integers is an integer). And since x/z = ab , this means that x is indeed divisible by z. Therefore, division is transitive.




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

Proof:

Take note that 2 times any integer plus 1 will always give an odd number. So if x is an odd integer, it can be written in the form:

x = (2a + 1) where a is an integer. Similarly, since y is also an integer, it can also be written as:

y = (2b + 1) where b is an integer. Multiplying x and y gives:

xy = (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1

Rewriting the equation by factoring out 2 from 4ab + 2a + 2b gives

xy = 2 (2ab + a + b ) + 1 Since a and b are integers, then (2ab + a + b) is also an integer. This makes xy of the form 2 times an integer + 1. Therefore, xy is an odd integer.


Prove that the sum of two rational numbers is a rational number.

Proof:

A rational number can be expressed as a ratio of two integers. Hence, if x and y are rational numbers, then

x = a/b

y = c/d where a, b , c, and d are integers. Adding x and y gives

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

Since a, b , c, and d are integers, it follows that both the numerator (ad + cb) and the denominator (bd) are integers , This making x + y a ratio of two integers. Therefore, x + y is a rational number.

[Direct Proof, Pages 3–6 of 27]



Proof by Contradiction Page 7 of 27

Proof by Contradiction

Proof by contradiction is an indirect proof. It starts by assuming that the statement to be proven is false, and then deriving consequences or outcomes. In the process, if a contradiction to what is known to be true is reached, then it implies that the initial assumption that the statement is false is incorrect. This leads to the conclusion that the original statement must be true.


Examples: Prove that if 3x + 2 is odd, then x is odd, where x is an integer.

Proof:

The statement to be proven is x is an odd integer. Hence, when using proof by contradiction, this will be assumed to be false. The initial assumptions will then be that 3x + 2 is odd but x is even. Take note that an even number can be expressed as 2 time any integer. So if x is even, then it can be written as

x = 2a where a is an integer. Substituting x = 2a into 3x + 2 and rewriting the equation gives

3x + 2 = 3(2a) + 2 = 6a + 2 = 2(3a + 1)

Since a is an integer, (3a +1) is also an integer, and 3x + 2 is of the form 2 times an integer. This means that 3x + 2 is even. This contradicts the initial assumption that it is supposed to be odd. Therefore, since x cannot be even, it has to be odd. This proves the original statement.




Prove that if x2 is even, then x is even.

Proof:

The statement to be proven is x is even. Using the proof by contradiction, we assume that x2 is even but x is odd. If x is odd, then it can be written as

x = 2a + 1 where a is an integer. Squaring x = 2a +1 and rewriting gives

x2 = (2a + 1)2 = 4a2 + 4a + 1 = 2(2a2 +2a) + 1

Since a is an integer, (2a2 + 2a) is also an integer. So x2 is of the form 2 times an integer + 1. This makes x2 odd which contradicts the initial assumption that x2 is supposed to be even. Therefore the original statement is true.


Prove that the square root of two is irrational.

Proof:

For the proof by contradiction, assume that the square root of two is rational. If the square root of 2 is rational, then it means that

ba

=2

where a and b are integers. Take note that for the fraction to be in its simplest form, a and b must not have common factors. This implies that at least a or b must be odd.




Squaring both sides of the equation gives

2 = a2/b2

2b2 = a2 This implies that a2 is even since it is of the form 2 times an integer. If a2 is even, then a is also even (this was proven in an earlier example). Since a is even, it can be written in the form

a = 2x where x is an integer. Substituting a = 2x into the equation 2b2 = a2 and simplifying gives

2b2 = a2 = (2x)2 = 4x2

b2 = 2x2


The equation b2 = 2x2 again implies that b2 is even (2 times an integer) which also makes b even. Since both a and b are even, this contradicts the earlier statement that at least one of them must be odd. This proves the correctness of the statement that that the square root of two is irrational.

[Proof by Contradiction, Pages 7–12 of 27]



Proof by I nduction Page 13 of 27

Proof by Induction Proof by induction is useful in proving that all elements of an infinite set, such as the set of natural numbers, possess a certain property.

The first step in the proof of induction is 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 (arbitrary) element of the set, it is also true for the next element. Once these two proofs have been established, then it is concluded that the property is true for all members of the set. Analogy.

Prove that all students in the class are human beings. The first step will try to establish that the first student is human. Having done this, the second step is to prove that if a student is human, then the person sitting beside him is also human. So if the first step establishes that the first student is human, the second step will establish that the person next to him is also human. And if the second person is human, then so are the third, the fourth, etc. Eventually, this proves that all students in the class are human beings.

Proof by Induction Page 14 of 27

The first step is called the basis step while the second step is called the induction step. Formal Definition:

Let P be the property to be proven true for an infinite set. Basis Step:

Prove that P(1) is true.

The term P(1) means that the property is true for the first element of the set.

Induction Step:

Prove that if P(k) is true, then P(k + 1) is also true for every positive integer k.

The induction step establishes that if the property is true for an arbitrary element k, then it also holds true for the next element, which is k + 1. In other words, if it is true for the kth element, it is also true for the (k + 1)th element.




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 (prove that the statement is applicable for the first element of the series . In other words, let n =1 and show that the equation is true.

Let n = 1.

11

)11)(1(1 21

=

+=

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 (the statement is applicable to the kth element of the series). So let n = k and assume the resulting equation to be true:

( )1...21 21 +=+++ kkk

Take note that this equation is assumed to be true (the statement holds for the kth element).

Proof of Induction Page 16 of 27

Now prove that P(k + 1) is true. In other words, prove that the statement is true for the (k + 1)th element. Replacing n by k + 1 gives

( ) ( )( )211...21 21 ++=++++ kkk

On the left-hand side of the equation, the number before (k + 1) is k. Rewriting the equation to indicate this gives

( ) ( )( )211)1(

...21 21

21

++=+++=+++ kkk

kkk43421

as assumed earlier. Substituting and s implifying the left side gives

( ) ( ) ( )( )[ ] ( )( )

[ ] ( )( )[ ] ( )( )2123

2122

21)1(2)1(

2111

212

21

212

21

21

21

21

21

++=++

++=+++

++=+++

++=+++

kkkk

kkkkk

kkkkk

kkkkk




Factoring the left side gives

[ ] ( )( )( )( ) ( )( )2121

2123

21

21

212

21

++=++

++=++

kkkk

kkkk

Since both sides of the equation are the same, this proves that if P(k) is true, then 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.


Prove that the sum of the first n odd positive integers is n2. The first odd integer is 1, the second is 3, the third is 5, and so on and so forth. The nth odd integer is (2n – 1). So this statement is actually expressed as:

2)12(...531 nn =−++++ Basis Step: Prove P(1) is true.

Let n = 1:

11)1(1 2

==


Assume first that P(k) is true (the statement is applicable to the kth element of the series). So let n = k and assume the resulting equation to be true:

2)12(...531 kk =−++++ Take note that this equation is assumed to be true (the statement holds for the kth element).




Now prove that P(k + 1) is true. In other words, prove that the statement is true for the (k + 1)th element. So let n = k + 1. That is,

( ) ( )2112...531 +=+++++ kk On the left-hand side of the equation, the odd number before (2k + 1) is (2k – 1). Rewriting the equation to indicate this gives

( ) ( ) ( )2

2

11212...531 +=++=

−++++ kkk

k444 3444 21

as assumed earlier. Substituting and factoring gives

( ) ( )22 112 +=++ kkk

( ) ( )22 11 +=+ kk Since both sides of the equation are the same, this proves that if P(k) is true, then P(k + 1) is also true for all k. Therefore, the statement that the sum of the first n odd positive integers is n2 is true.


Prove that n3 – n is divisible by 3 whenever n is a positive integer. Basis Step: Prove P(1) is true.

Let n = 1.

13 - 1 = 0

0 is divisible by 3.

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 (the statement is applicable to the kth element of the series). So let n = k and assume that resulting statement to be true:

kk −3 is divisible by 3 Take note that this statement is assumed to be true (the statement holds for the kth element).





( ) ( )11 3 +−+ kk is divisible by 3 Expanding the term (k + 1)3 gives

( ) ( )1133 23 +−+++ kkkk

1133 23 −−+++ kkkk

kkkk −++ 33 23 Reordering and regrouping the terms give

kkkk 33 23 ++−

( ) ( )kkkk ++− 23 3


Take note that it was assumed earlier that k3 – k is divisible by 3. The second term, 3(k2 + k), is also divisible by 3 since it is of the form 3 times an integer. Since you have the sum of two integers that are both 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. Therefore, the statement that n3 – n is divisible by 3 whenever n is a positive integer is true.




Prove that 7n – 1 is divisible by 6 for all n ≥ 1. Basis Step: Prove P(1) is true.

Let n = 1:

71 - 1 = 6

6 is divisible by 6.


Assume first that P(k) is true (the statement is applicable to the kth element of the series). So let n = k and assume that resulting statement to be true.

17 −k is divisible by 6 Take note that this statement is assumed to be true (the statement holds for the kth element).



( ) 17 1 −+k is divisible by 6 By algebraic manipulation,

177 −⋅ k Replacing –1 by (–7 + 6) gives

6777 +−⋅ k Factoring out 7 from the first two terms gives

( ) 6177 +−k




Recall that it was assumed earlier that 7k – 1 is divisible by 6. Therefore, the term 7(7k – 1) is also divisible by 6 because if an integer is multiplied to a number that is divisible by 6, the product is also divisible by 6. This proves that if P(k) is true, then P(k + 1) is also true for all k. Therefore, the statement that 7n – 1 is divisible by 6 for all n ≥ 1 is true.


Prove that 20 + 21 + 22 + … + 2n = 2(n+1) – 1 for all n ≥ 0.

Notice that the first n in the series is 0 and not 1 (as in the previous examples). This means that the basis step is to prove that P(0) is true, and not P(1) since 0 is the first element of the series.

Basis Step: Prove P(0) is true. Let n = 0:

11121

122 )10(0

=−=

−= +


Assume first that P(k) is true (the statement is applicable to the kth element of the series). So let n = k and assume that resulting statement to be true:

( ) 122...222 1210 −=++++ +kk Take note that this statement is assumed to be true (the statement holds for the kth element).





( ) ( ) 122...222 111210 −=++++ +++ kk On the left-hand side of the equation, the number before 2(k + 1) is 2k. Rewriting the equation to indicate this gives

( ) ( ) 12212

2...222 21

1

210 −=+−=

++++ ++

+

kk

k

k444 3444 21

as assumed earlier. Substituting and manipulating the left side gives

( ) ( ) ( )

( )

( )

( ) ( ) 1212

12122

12122

12212

22

2)1(

2)1()1(

211

−=−

−=−⋅

−=−+

−=+−

++

++

+++

+++

kk

kk

kkk

kkk

Since both sides of the equation are the same, this proves that if P(k) is true, then P(k + 1) is also true for all k. Therefore, the statement 20 + 21 + 22 + … + 2n = 2(n+1) – 1 for all n ≥ 0 is true.

[Proof by Induction, Pages 13–27 of 27]

meljun cortes automata lecture types of proofs 2

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