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Discrete Mathematics
CS 2610
February 10, 2009
Agenda
Previously

Functions
And now


Finish functions
Start Boolean algebras (Sec. 11.1)
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But First
p  q  r, is NOT true when only one of p, q, or r
is true. Why not?
It is true for (p Λ ¬q Λ ¬r)
It is true for (¬p Λ q Λ ¬r)
It is true for (¬p Λ ¬q Λ r)
So what’s wrong? Raise your hand when you know.
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Injective Functions (one-to-one)
If function f : A  B is 1-to-1 then every b  B
has 0 or 1 pre-image.
Proof (bwoc): Say f is 1-to-1 and b  B has 2 or
more pre-images.
Then a1, a2 st a1  A and a2  A, and a1 ≠ a2.
So f(a1) = b and f(a2) = b, meaning f(a1) = f(a2).
This contradicts the definition of an injection
since when a1 ≠ a2 we know f(a1) ≠ f(a2).
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Combining Real Functions
Given f :RR and g :RR then

(f  g): RR, is defined as
(f  g)(x) = f(x)  g(x)

(f · g): RR is defined as
(f · g)(x) = f(x) · g(x)
Example:
Let f :RR be f(x) = 2x and and g :RR be g(x) = x3
(f+g)(x) = x3+2x
(f · g)(x) = 2x4
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Monotonic Real Functions
Let f: AB such that A,B  R
f is strictly increasing iff
 for all x, y  A x > y  f(x) > f(y)
f is strictly decreasing iff
 for all x, y  A, x > y  f(x) < f(y)
Example:
f: R+  R+, f(x) = x2 is strictly increasing
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Increasing Functions are Injective
Theorem: A strictly increasing function is always
injective
Proof:
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Floor and Ceiling Function
Definition: The floor function .:R→Z, x is the
largest integer which is less than or equal to x.
x reads the floor of x
Definition: The ceiling function . :R→Z, x is the
smallest integer which is greater than or equal to
x.
x reads the ceiling of x
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Example Ceiling and Floor Functions
Example:
-2.8 = -3
2.8 = 2
2.8 = 3
-2.8 = -2
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Ceiling and Floor Properties
Let n be an integer
(1a)
x = n if and only if n ≤ x < n+1
(1b)
x = n if and only if n-1 < x ≤ n
(1c)
x = n if and only if x-1 < n ≤ x
(1d)
x = n if and only if x ≤ n < x+1
(2)
x-1 < x ≤ x ≤ x < x+1
(3a)
-x = - x
(3b)
-x = - x
(4a)
x+n = x+n
(4b)
x+n = x+n
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Ceiling and Floor Functions
Let n be an integer, prove x+n = x+n
Proof

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

Let k = x
Then k ≤ x < k+1
So k+n ≤ x+n < k+1+n
I.e., k+n ≤ x+n < (k+n)+1
Since both k and n are integers, k+n is an integer
Thus, x+n = k+n = x+n (by our choice of k)
This concludes the proof

This also concludes Chapter 2!
11
Boolean Algebras (Chapter 11)
Boolean algebra provides the operations and
the rules for working with the set {0, 1}.
These are the rules that underlie electronic
and optical circuits, and the methods we
will discuss are fundamental to VLSI design.
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Boolean Algebra
The minimal Boolean algebra is the algebra formed
over the set of truth values {0, 1} by using the
operations functions +, ·, - (sum, product, and
complement).
The minimal Boolean algebra is equivalent to
propositional logic where





O corresponds to False
1 corresponds to True
 corresponds logical operator AND
+ corresponds logical operator OR
- corresponds logical operator NOT
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Boolean Algebra Tables
x x
0 1
1 0
x y
x + y xy
0 0
0
0
0 1
1
0
1 0
1
0
1 1
1
1
x,y are Boolean variables – they assume values 0 or 1
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Boolean n-Tuples
Let B = {0, 1}, the set of Boolean values.
Let Bn = { (x1,x2,…xn) | xi  B, i=1,..,n}
B1= { (x1) | x1  B,}
B2= { (x1, x2), | xi  B, i=1,2}
Bn= { ((x1,x2,…xn) | xi  B, i=1,..,n,}
.
For all nZ+, any function f:Bn→B is called a
Boolean function of degree n.
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Example Boolean Function
F(x,y,z) =B3B
x
y
z
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
1
1
0
1
1
1
1
0
0
1
1
1
1
F(x,y,z)=x(y+z)
B3 has 8 triplets
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Number of Boolean Functions
How many different Boolean functions of degree 1
are there?
How many different Boolean functions of degree 2
are there?
How many different functions of degree n are
there ?
 There are 22ⁿ distinct Boolean functions of
degree n.
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