ICS-252 Discrete Structure II
Assistant Professor Department of Computer Science & Assistant Professor, Department of Computer Science & Engineering, University of Hail, KSA. [email protected]
ICS-252 Discrete Structure II Lecture 6
Outlines
- Proof Methods • Proof Methods • Proof Strategies
DMA=Discrete Mathematics and its DMA=Discrete Mathematics and its Applications (86-100) ICS ‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail.
Proof Methods Proof :
A proof is a valid argument that establishes the truth of mathematical statement. There are two types of proofs;
Formal Proof : In this type all steps are supplied and rules for each
step in the arguments are given. Useful theorems can be long and hard to follow hard to follow.
Informal Proof :
Proof of theorems designed for the human consumption are always informal proofs. More than one rule of inference may be used in each step. Steps may be skipped. Rules of inference are not explicitly stated.
Proof Methods : The following are the proofs methods; 1) Direct Proof: 1)
Direct Proof: It It is a way of showing the truth of a given statement is a way of showing the truth of a given statement
by a straightforward combination of established facts.
ICS ‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail. 3 Proof Methods Sum of two even integers is an even number.
Example: Proof: Let x and y are two even numbers. Since they are
even therefore we can write x= 2a, y = 2b for all integers a and b. x + y = 2a + 2b = 2 (a + b) From this it is clear that x + y has 2 as a factor and therefore is even. Hence, sum of two even integers is an even number.
It is also known as proof by contradiction.
2) Indirect Proof:
It is a form of proof that establishes the proof or validity of a proposition by showing that proposition is being false would imply a contradiction. A proposition must be either true or false and its falsity has been shown impossible, t f l d it f l it h b h i ibl then proposition must be true. ICS
‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail.
Proof Methods For all integers n, if 3n + 1 is even, then n is odd. Example: Solution: Suppose the contradiction that n is not odd. It means n is even.
We can write for all integers n, 3n + 1 is even then n is even. If n is even mean n is multiple of 2, therefore n = 2a, for integers a. Then 3n + 1 = 3(2a) +1 = 6a + 1 ………………………. (1) 6a is even because 2(3a). But 6a + 1 is odd. Therefore 3n + 1 is odd from equation (1). By assuming n is even, we shown that 3n + 1 is odd which is an contradiction to our assumption contradiction to our assumption. Therefore if n is odd then 3n +1 is even, which is impossible. It follows that the original statement if 3n + 1 is even, then n is odd is true. ICS
‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail. 5 Proof Methods
It is also known as proof by cases. It
3 -1) Exhaustive Proof:
is a special type of proof by cases where each case involves checking a single example.
Prove that , if n is a positive
Example 1: ( n 1 )
- 3 n
3 ≥
integer with n ≤ 4.
Proof by exhaustion need only to verify for
Solution:
n=1,2,3,4
3 3 n
1
n=1, (n + 1) = (2) = 8 and 3 = 3 = 3; It follows 8 > 3;
3 3 n
2
n = 2, (n + 1) = (3) = 27 and 3 = 3 = 9; It follows 27 > 9;
3 3 n
3
3
3
3
n = 3, (n + 1) 3 ( 1) = (4) (4) = 64 and 3 64 d 3 =3 3 = 27; It follows 64 > 27;
27 It f ll
64
27
3 3 n
4
n = 4, (n + 1) = (5) =125 and 3 = 3 = 81; It follows 125>81 Proof holds.
Proof Methods
A proof by cases must cover all cases
3-2) Proof by cases: that arises in a theorem.
2 Example 3: Prove that if n is an integer then n
≥ n We can prove for every integer by considering
Solution:
three cases when n=0, n three cases when n=0 n ≥ 1 and n ≤ -1 ≥ 1 and n ≤ -1
2
2
n=0, because 0 =0 therefore n
Case (i) ≥ n holds.
n
Case (ii)
≥ 1, multiply both sides of inequality by positive
2
n, we get n. n ≥ n.1, implies n ≥ n hold for n ≥ 1.
2
2
n
Case (iii) ≤ -1, however n ≥ 0, implies n ≥ n.
2 Hence n ≥ n hold for all inequalities.
ICS ‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail. 7 Proof Methods
It is a theorem with a statement involving
4) Existence Proof:
the existential qualifier. It has two types; It proves the existence of
a) Constructive Existence Proof:
a mathematical object with certain properties by creating or providing a method to create this object. or providing a method to create this object Show that there is a positive integer that can be
Example 10:
written as the sum of cubes of positive integers in two different ways;
Solution: After doing some computation we found that
3
3
3
3
1729 = 10 + 9 = 12 + 1 It proves the
a) Non-constructive Existence Proof:
existence of a mathematical object with certain properties, but does not provide a means of constructing an example.
Proof Methods Example 11:
On the other hand if is irrational then we can let x= and y = ,
2
2
2
2
2
2
y rational.
Show that there exist irrational numbers x and y such that x
x = and y = with x
2
2
2
We know that is irrational. Consider the number if it is rational , we have two irrational numbers
Solution:
y is rational.
2
2 So that x
y
Uniqueness Proofs:
x y ≠ property, then x y.
Equivalently, we can show that if x and y both have the desired property then x = y
We show that if , then y does not have the desired property.
b) Uniqueness:
We show that an element x with the desired property exists.
a) Existence:
It has two fundamental properties;
Rather we have shown that either the pair have the desired property and we do not know which of these pairs work.
y rational. 9 ICS ‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail.
Proof Methods
2 = = =
2
2
2 2 .
2 2 ) 2 (
2
= We have not found irrational number x and y such that is x
Proof Methods
Show that if a and b are real numbers and ≠
Example 13: a Then there is a unique real number r such that ar + b = 0.
Solution: , note that the real number r = -b/a is a solution
First of ar + b = 0, It follows that a(-b/a) + b = -b + b = 0.
Consequently, a real number r exist for which ar + b = 0. This is the existence part of the theorem.
Second, suppose that s is a real number such that as + b = 0, then ar + b = as + b, It follows that ar = as. Dividing both side by a, which is nonzero, we have r = s. this means that if , then This is the uniqueness q
s s ≠ ≠ r r
- b + b ≠ ≠
as as part of the theorem. ICS ‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail. 11 Proof Strategies
Generally, if the statement is a conditional
Proof Strategies:
statement, you should first try a direct proof; if this fails, you can try an indirect proof if neither of these approaches works , you might try a proof by contradiction.
To begin a direct proof of To begin a direct proof of
Forward and Backward Reasoning: Forward and Backward Reasoning:
a conditional statement, you start with the premises. Using these premises, together with axioms and known theorems, you can construct a proof using a sequence of steps that leads to the conclusion. This type of reasoning is called
forward reasoning (see example in direct proof). But
forward reasoning is often difficult to use to prove more forward reasoning is often difficult to use to prove more complicated results. In such cases it is helpful to use backward reasoning.
Proof Strategies Example:
- ( x y ) /
2 > xy When x and y are distinctive positive real numbers.
Solution: We will proof by backward.
- ( ( x y ) ) / /
2 2 xy > ( x y ) / 2
- 2
4 > xy ( x y ) 2 > + 4 xy 2 x xy y xy
4 x − 2 +
2 > + +
2 xy y > 2 2 ( x ) − y >
It follows that x ICS ‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail. ≠ y. 13 Proof Strategies Counterexamples show that
Looking for Counterexamples:
certain statements are false. When confronted with a conjecture, you might first try to prove this conjecture, and if your attempts are unsuccessful, you might try to find a counterexample counterexample.
Show that the statement “ Every positive integer
Example 17:
is the sum of the square of three integers” is false by finding a counterexample.
We first look for a counterexample that “Every
Solution:
positive integer is the sum of three squares of integers” is false, If we find a particular integer that is not the sum of the false If we find a particular integer that is not the sum of the squares if three integers. To look for a counterexample, we try successive positive integers as a sum of three squares, we find that
Proof Strategies
We try successive positive integers as a sum of three squares, we find that 2 2 2
1 = 2 2 + +
1 2
2 2 = = 2 + + + +
1 2
1
1
1 2
3
1
1
1 = 2 2 + + 2 4 =
2 To show that there are not three squares that add up to 7, as
the sum of three squares we can use are those not exceeding 7, namely 0, 1, and 4. Because no three terms exceeding 7 namely 0 1 and 4 Because no three terms where each term is 0, 1 or 4 add up to 7. it follows that 7 is a counterexample. We conclude that the statement “ Every positive integer is the sum of the square of three integers” is false. ICS ‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail. 15 Proof Strategies
In tilings, checkerboard is a rectangle divided into
Tilings:
squares of same size by horizontal and vertical lines. The game of checkers is played on a board with 8 rows and 8 columns; this board is called the standard checkerboard shown in the figure. shown in the figure
Proof Strategies Dominoes:
A domino is a rectangular piece that is one square by two squares as shown in figure.
We say that a board is tiled by dominoes when all its squares are covered with no overlapping dominoes and no dominoes overhanging the board We now develop some dominoes overhanging the board. We now develop some results about tiling boards using dominoes.
17 ICS ‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail. Proof Strategies Example 18:
Can we tile the standard checkerboard using dominoes. How many ways we have to fill it? Solution: There are the following ways to tile the checkerboard. 1) Tile it by placing 32 dominoes horizontally. 2) Tile it by placing 32 dominoes vertically. 3) Tile it by placing some horizontally and some vertically dominoes.
This method is called constructive existence proof.
Home Work
Prove that the only consecutive integers not
Question 1:
exceeding 100 that are perfect powers are 8 and 9 (An
a
integer is a perfect power if it equals n , where a is an integer greater than 1).
Formulate a conjecture about the decimal digits Formulate a conjecture about the decimal digits
Question 2: Question 2:
that occur as the final digit of the square of an integer and prove your result.
Can we tile a board obtained by removing one of
Question 3:
the four corner squares of a standard checkerboard? Can we tile the board obtained by deleting the
Question 4:
upper left and lower right corner squares of a standard checkerboard.
ICS ‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail. 19 Home Work
Prove or disprove that you can use dominoes to
Question 5:
tile the standard checkerboard with two adjacent corners removed (that is, corners that are not opposite).
Question 6: Prove that you can use dominoes to tile a
rectangular checkerboard with an even number of squares. rectangular checkerboard with an even number of squares Show that by removing two white squares and two
Question 7:
black squares from an 8 x 8 checkerboard (colored as in the text) you can make it impossible to tile the remaining squares using dominoes.
Prove that there are 100 consecutive positive
Question 8:
integers that are not perfect squares. Is your proof constructive or non-constructive? Prove that there exists a pair of consecutive
Question 9:
integers such that one of these integers is a perfect square
Thank you for your Attention.
21 ICS ‐252 Dr. Salah Omer, Assistant Professor, CSSE, University of Hail.