Transcript DS Lecture 6

Discrete Structures (CSC 102)
Lecture 6
Previous Lectures Summary
•Different forms of arguments
•Modus Ponens and Modus Tollens
•Additional Valid Arguments
•Valid Argument with False Conclusion
•Invalid argument with a true Conclusion
•Converse and Inverse error
•Contradictions and valid arguments
Predicates and Quantified statements I
Today’s Lecture
• Predicates
• Set Notation
• Universal and Existential Statement
• Translating between formal and informal language
• Universal conditional Statements
• Equivalent Form of Universal and Existential
statements
• Implicit Qualification
• Negations of Universal and Existential statements
Predicates
A predicate is a sentence which contains finite number of
variables and becomes a statement when specific values
are substituted for the variables.
The domain of a predicate variable is the set of all values that
may be substituted in place of the variable
Truth Set
If P(x) is a predicate and x has domain D, the truth set of
P(x) is the set of all elements of D that make P(x) true
when substituted for x. The truth set of P(x) is denoted by
{x  D | P( x)}
read as “the set of all x in D such that P(x)”.
Notation
For any two predicates P(x) and Q(x), the notation
P( x)  Q( x) means that every element in the truth set of
P(x) is in the truth set of Q(x). The notation P( x)  Q( x)
means that P and Q have identical truth sets.
Consider the predicate:
x  0,
xR
The truth set of the above predicate is
x  R


x0
Cont…
Example
Let P(x) = x is a factor of 8, Q(x)= x is a factor of 4
and R(x)= x < 5 and x  3 . The domain of x is
assumed to be Z . Use symbols  ,  to indicate
true relationships among P(x), Q(x) and R(x).
a. The truth set of P(x) is {1,2,4,8}, Q(x) is {1,2,4}.
Since every element in the truth set of Q(x) is in the
truth set of P(x), So Q( x)  P( x)
b. The truth Set of R(x) is {1,2,4}, which is identical to
Q( x)  R( x) .
the truth set of Q(x). Hence
Cont…
Let Q(x, y) be the statement
x+y=x−y
where the domain for x and y is the set of all real numbers.
Determine the truth value of:
(a) Q(5,−2).
(b) Q(4.7, 0).
(c) Determine the set of all pairs of numbers, x and y, such that
Q(x, y) is true.
Solution:
(a) Q(5,−2) says that 5 + (−2) = 5 − (−2), or 3= 7, which is false.
(b) Q(4.7, 0) says that 4.7+ 0 = 4.7 − 0, which is true.
(c) x + y = x − y if and only if x + 2y = x, which is true if and only if
y = 0. Therefore, x can be any real number and y must be zero.
Universal and Existential Statements
Let Q(x) be a predicate and D the domain of x. A
universal statement is of the form “ x  D, Q( x) ”. It is
true if and only if Q(x) is true for all x in D and it is
false if and only if Q(x) is false for at least one x in D. A
value for x for which Q(x) is false is called a
counterexample to the universal statement.
Example: Let D={1,2,3,4,5} and consider the
statement x  D, x 2  x. Show that this statement is true.
Solution: Check that " x 2  x ". is true for each individual
x in D.
52  5
12  1
32  3
22  2
42  4
62  6
Cont…..
Hence x  D, x 2  x. is true.
The technique used in first statement while showing the
truthness of the universal statement is called method of
exhaustion.
Consider the statement x  R, x 2  x. Find the counter
example to show that this statement is not true.
Counter example . Take x=1/2, then x is in R and
2
1 1
   
2 2
Hence x  R, x  x. is false.
2
Existential Quantifier
Let Q(x) be a predicate and D the domain of
x. An existential statement is of the form.
x D such that Q (x)
It is true if and only if Q(x) is true for at least
one x in D. It is false if and only if Q(x) is
false for all x in D.
The symbol  denotes “there exist” and is
called the existential quantifier.
Truth and falsity of Existential statements
Suppose P(x) is the predicate “x < |x|.” Determine the
truth value of ∃ x s.t. P(x) where the domain for x is:
(a) the three numbers 1, 2, 3.
(b) the six numbers −2,−1, 0, 1, 2, 3.
Solution
(a) P(1), P(2), and P(3) are all false because in each
case x = |x|. Therefore, ∃ x such that P(x) is false for
this domain.
(b) If we begin checking the six values of x, we find
P(−2) is true. It states that −2 < |−2|. We need to check
no further; having one case that makes the predicate
true is enough to guarantee that ∃ x s.t. P(x) is true.
Truth and falsity of Existential statements
Consider the statement m  Z s.t. m 2  m. Show
that this statement is true.
2
2
m
 m is true for at
Sol: observe that 1  1. Thus
least one integer m . Hence m  Z s.t. m 2  m is
true.
Let E={5,6,7,8,9,10} and consider the statement
m  E s.t. m 2  m
Show that this statement is false.
Sol: the statement is not true for every value of
the E. Thus m  E s.t. m 2  m is false.
Translating from formal to informal language
Rewrite the following statements in a variety of
equivalent but more informal ways. Do not use the
, 
symbol
a)
x  R, x 2  0.
b)
x  R, x 2  1.
c)
m  Z s.t. m  m
2
Solution: a) we can write the statement in many ways
like “ All real numbers have non negative squares”,
“No real number has a negative square”,
“ x has a non negative square, for each value of x”.
Cont….
b). Similarly we can translate the second statement in
these ways.
“ All real numbers have squares not equal to -1”,
“No real number have square equal to -1”.
c). “There is an integer whose square is equal to itself”,
“we can find at least one integer equal to its own
square”
Cont…
Write the following statement in English, using the
predicates
F(x): “x is a Freshman”
T (x, y): “x is taking y”
where x represents students and y represents courses:
∃x (F(x) ∧ T (x, Discrete Math))
Solution
The statement ∃ x (F(x)∧T (x, Discrete)) says that there
is a student x with two properties: x is a freshman and x
is taking Discrete. In English, “Some Freshman is taking
Discrete Math.”
Translating from informal Language to Formal language
“Every freshman at the College is taking CSC 102.”
Solution: There are various ways to answer this question,
depending on the domain.
• If we take as our domain all freshmen at the College
and use the predicate T (x) : “x is taking CSC 102”,
then the statement can be written as ∀x, T(x).
• We are making a conditional statement:
“If the student is a freshman, then the student is taking
CSC 101;”
∀x, (F(x) → T (x)).
Note that we cannot say ∀ x (F(x) ∧ T (x)), because this
says that every student is a freshman, which is not
something we can assume here.
Cont…..
“Every freshman at the College is taking some Computer
Science course.”
Sol: If we take as our domain for people all freshmen at
the College and our domain for courses, all Computer
Science courses.
Then we can use the predicate
T (x, y): “x is taking y”
The statement can be written as
∀x ∃y T(x, y).
Universal Conditional Statements
A reasonable argument can be made that the most
important form of statement in mathematics is the
universal conditional statement:
∀ x, if P(x) then Q(x)
Example: “Everyone who visited France stayed in
Paris.”
Sol: However, if we take all people as the universe ,
then we need to introduce the predicate F(x) for “x
visited France.” and P(x) is the predicate “x stayed in
Paris.” In this case, the proposition can be written as
∀ x, (F(x) → P(x)).
• We can write the following statements in a
variety of informal ways.
x  R, if x  2 then x 2  4
Sol:
• if a real number is greater then 2, then the
square is greater than 4.
• Whenever a real number is greater then 2,
its square is greater than 4.
• The squares of real number, greater than 2,
are greater than 4.
Exercise
Rewrite the following statements in the form
∀
,if
then
.
a) If a real number is an integer, then it is a rational
number.
a) All bytes have eight bits.
b) No fire trucks are green.
Sol: a). x  R, ifx  Z , thenx  Q.
b). ∀ x, if x is a byte, then x has eight bits.
c). ∀ x, if x is a fire truck, then x is not green.
Equivalent Forms of Universal and Existential statements
Observe that the two statements “∀ real
numbers x, if x is an integer then x is
rational” and “∀ integers x, x is rational”
mean the same thing.
In fact, a statement of the form
x  U , if P(x) then Q(x).
Can always be rewritten in the form
x  D, Q ( x )
Can be rewritten as
∀x, if x is in D then Q(x).
Contd.
The following statements are equivalent
∀ polygons P, if P is square, then P is a rectangle.
And
∀ squares P, P is a rectangle
The existential statements
∃ x belongs to U such that P(x) and Q(x).
And
∃ x belongs to D such that Q(x)
Are also equivalent provided D is taken to consist of all elements in
U that make P(x) true.
Equivalence form for existential statement
The following statements are equivalent:
∃ a number n such that n is prime and n is even
And
∃ a prime number n such that n is even.
Implicit Quantifications
• Consider “ If a number is an integer, then it is a
rational number”
The clue to indicate its universal quantifications comes
from the presence of the indefinite article “a”.
Existential quantification can also be implicit.
for instance, “ the number 24 can be written as a sum of
sum of two integers”
“∃ even integers m and n such that 24=m + n.”
Negations of Quantified Statement
The negation of the statement of the form
∀ x in D, Q(x)
is logically equivalent to a statement of the
form
∃ x in D such that ~Q(x)
Symbolically:
~ (x  D, Q( x))  x  D ~ Q( x).
Note: the negation of universal statement is
logically equivalent to existential statement.
Cont….
The negation of the statement of the form
∃ x in D such that Q(x)
is logically equivalent to a statement of the
form
∀ x in D, ~Q(x)
Symbolically:
~ (x  D  Q( x))  x  D, ~ Q( x).
Note: the negation of existential statement is
logically equivalent to universal statement.
Examples
Negate “Some integer x is positive and all integers
y are negative.”
Solution: Using all integers as the universe for x and y,
the statement is ∃ x s.t. (x > 0) ∧ ∀ y, (y < 0). The
negation is
~{∃x (x > 0) ∧ ∀ y (y < 0)}≡ ~∃x s.t. (x > 0) ∨ ~∀y, (y < 0):
by De Morgan’s law
≡ ∀ x, ~(x > 0) ∨ ∃ y s.t. ~(y < 0) properties of negation
≡ ∀ x, (x ≤ 0) ∨ ∃ y s.t. (y ≥ 0).
Therefore, the negation is “Every integer x is non
positive or there is an integer y that is nonnegative.”
Cont….
Negate “There is a student who came late to class
and there is a student who is absent from class.”
Solution: In symbols, if L(x) : “x came late to class” and
A(x) : “x is absent from class,” this statement can be
written as ∃ x st L(x) ∧ ∃ y st A(y).
Note that we must use a second variable y. By one of
De Morgan’s laws the negation can be written as
~(∃ x st L(x)) ∨ ~(∃ y st A(x)) ≡ ∀x, ~L(x) ∨ ∀ y, ~A(x).
In English this is “No student came late to class or no
student is absent from class.”
Lecture Summary
• Predicates
• Set Notation
• Universal and Existential Statement
• Translating between formal and informal language
• Universal conditional Statements
• Equivalent Form
• Implicit Qualification
• Negations