Addition of Integers
How do we (humans) add two integers?
Example:
1 11
7583
+ 4932
carry
1 25 1 5
Binary expansions:
1 1
(1011)2
+ (1010)2
carry
(101 01 )2
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
1
Addition of Integers
Let a = (an-1an-2…a1a0)2, b = (bn-1bn-2…b1b0)2.
How can we algorithmically add these two binary
numbers?
First, add their rightmost bits:
a0 + b0 = c02 + s0,
where s0 is the rightmost bit in the binary
expansion of a + b, and c0 is the carry.
Then, add the next pair of bits and the carry:
a1 + b1 + c0 = c12 + s1,
where s1 is the next bit in the binary expansion of
a + b, and c1 is the carry.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
2
Addition of Integers
Continue this process until you obtain cn-1.
The leading bit of the sum is sn = cn-1.
The result is:
a + b = (snsn-1…s1s0)2
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
3
Addition of Integers
Example:
Add a = (1110)2 and b = (1011)2.
a0 + b0 = 0 + 1 = 02 + 1, so that c0 = 0 and s0 = 1.
a1 + b1 + c0 = 1 + 1 + 0 = 12 + 0, so c1 = 1 and s1 = 0.
a2 + b2 + c1 = 1 + 0 + 1 = 12 + 0, so c2 = 1 and s2 = 0.
a3 + b3 + c2 = 1 + 1 + 1 = 12 + 1, so c3 = 1 and s3 = 1.
s4 = c3 = 1.
Therefore, s = a + b = (11001)2.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
4
Addition of Integers
procedure add(a, b: positive integers)
c := 0
for j := 0 to n-1
begin
d := (aj + bj + c)/2
sj := aj + bj + c – 2d
c := d
end
sn := c
{the binary expansion of the sum is (snsn-1…s1s0)2}
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
5
Matrices
A matrix is a rectangular array of numbers.
A matrix with m rows and n columns is called an
mn matrix.
Example:
1 
 1
A  2.5  0.3
 8
0 
is a 32 matrix.
A matrix with the same number of rows and
columns is called square.
Two matrices are equal if they have the same
number of rows and columns and the corresponding
entries in every position are equal.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
6
Matrices
A general description of an mn matrix A = [aij]:
 a11
 a21
 .
A
 .
 .
am1
a12 ... a1n 
a22 ... a2 n 
.
. 

.
. 
.
. 
am 2 ... amn 
ai1,
ai 2 , ..., ain 
 a1 j 
a 
 2j
 . 
 . 
 . 
 amj 
j-th column
of A
i-th row of A
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
7
Matrix Addition
Let A = [aij] and B = [bij] be mn matrices.
The sum of A and B, denoted by A+B, is the mn
matrix that has aij + bij as its (i, j)th element.
In other words, A+B = [aij + bij].
Example:
 2 1  5 9  3 10
 4 8    3 6   1 14
  3 0  4 1  7 1 
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
8
Matrix Multiplication
Let A be an mk matrix and B be a kn matrix.
The product of A and B, denoted by AB, is the mn
matrix with (i, j)th entry equal to the sum of the
products of the corresponding elements from the
i-th row of A and the j-th column of B.
In other words, if AB = [cij], then
k
cij  ai1b1 j  ai 2b2 j  ...  aik bkj   ait btj
t 1
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
9
Matrix Multiplication
A more intuitive description of calculating C = AB:
0
3

2

1

A
0
0
  1 1
1
4
5
0
2 1 
B  0  1
3 4 
- Take the first column of B
- Turn it counterclockwise by 90 and superimpose
it on the first row of A
- Multiply corresponding entries in A and B and
add the products: 32 + 00 + 13 = 9
- Enter the result in the upper-left corner of C
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
10
Matrix Multiplication
- Now superimpose the first column of B on the
second, third, …, m-th row of A to obtain the
entries in the first column of C (same order).
- Then repeat this procedure with the second,
third, …, n-th column of B, to obtain to obtain
the remaining columns in C (same order).
- After completing this algorithm, the new matrix
C contains the product AB.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
11
Matrix Multiplication
Let us calculate the complete matrix C:
0
3

2

1

A
0
0
  1 1
1
4
5
0


C


February 25, 2002
2 1 
B  0  1
3 4 
9
8
15
-2
7
15
20
-2





Applied Discrete Mathematics
Week 5: Mathematical Reasoning
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Identity Matrices
The identity matrix of order n is the nn matrix
In = [ij], where ij = 1 if i = j and ij = 0 if i  j:
1 0 ... 0
0 1 ... 0
. .
.
A  . .
.
. .
.


0 0 ... 1
Multiplying an mn matrix A by an identity matrix
of appropriate size does not change this matrix:
AIn = ImA = A
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
13
Powers and Transposes of Matrices
The power function can be defined for square
matrices. If A is an nn matrix, we have:
A0 = In,
Ar = AAA…A (r times the letter A)
The transpose of an mn matrix A = [aij], denoted
by At, is the nm matrix obtained by interchanging
the rows and columns of A.
In other words, if At = [bij], the bij = aji for
i = 1, 2, …, n and j = 1, 2, …, m.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
14
Powers and Transposes of Matrices
Example:
2 1 
A  0  1
3 4 
2
0
3


A 
1  1 4
t
A square matrix A is called symmetric if A = At.
Thus A = [aij] is symmetric if aij = aji for all
i = 1, 2, …, n and j = 1, 2, …, m.
3
5 1
A  1 2  9
3  9 4 
A is symmetric, B is not.
February 25, 2002
1 3 1
B  1 3 1
1 3 1
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
15
Zero-One Matrices
A matrix with entries that are either 0 or 1 is
called a zero-one matrix. Zero-one matrices are
often used like a “table” to represent discrete
structures.
We can define Boolean operations on the entries in
zero-one matrices:
a
b
ab
a
b
ab
0
0
0
0
0
0
0
1
0
0
1
1
1
0
0
1
0
1
1
1
1
1
1
1
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
16
Zero-One Matrices
Let A = [aij] and B = [bij] be mn zero-one
matrices.
Then the join of A and B is the zero-one matrix
with (i, j)th entry aij  bij. The join of A and B is
denoted by A  B.
The meet of A and B is the zero-one matrix with
(i, j)th entry aij  bij. The meet of A and B is
denoted by A  B.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
17
Zero-One Matrices
Example:
1 1
A  0 1 
1 0
Join:
1  0 1  1  1 1
A  B  0  1 1  1   1 1
1  0 0  0 1 0
Meet:
1  0 1  1  0 1
A  B  0  1 1  1   0 1
1  0 0  0 0 0
February 25, 2002
0 1 
B  1 1
0 0
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
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Zero-One Matrices
Let A = [aij] be an mk zero-one matrix and
B = [bij] be a kn zero-one matrix.
Then the Boolean product of A and B, denoted by
AB, is the mn matrix with (i, j)th entry [cij],
where
cij = (ai1  b1j)  (ai2  b2i)  …  (aik  bkj).
Note that the actual Boolean product symbol has a
dot in its center.
Basically, Boolean multiplication works like the
multiplication of matrices, but with computing 
instead of the product and  instead of the sum.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
19
Zero-One Matrices
Example:
A  1 0
1 1
B  0 1
0 1
(1  0)  (0  0) (1  1)  (0  1) 0 1

A B  

 (1  0)  (1  0) (1  1)  (1  1)  0 1
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
20
Zero-One Matrices
Let A be a square zero-one matrix and r be a
positive integer.
The r-th Boolean power of A is the Boolean
product of r factors of A. The r-th Boolean power
of A is denoted by A[r].
A[0] = In,
A[r] = AA…A
February 25, 2002
(r times the letter A)
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
21
Let’s proceed to…
Mathematical
Reasoning
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
22
Mathematical Reasoning
We need mathematical reasoning to
• determine whether a mathematical argument is
correct or incorrect and
• construct mathematical arguments.
Mathematical reasoning is not only important for
conducting proofs and program verification, but
also for artificial intelligence systems (drawing
inferences).
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
23
Terminology
An axiom is a basic assumption about
mathematical structured that needs no proof.
We can use a proof to demonstrate that a
particular statement is true. A proof consists of a
sequence of statements that form an argument.
The steps that connect the statements in such a
sequence are the rules of inference.
Cases of incorrect reasoning are called fallacies.
A theorem is a statement that can be shown to be
true.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
24
Terminology
A lemma is a simple theorem used as an
intermediate result in the proof of another
theorem.
A corollary is a proposition that follows directly
from a theorem that has been proved.
A conjecture is a statement whose truth value is
unknown. Once it is proven, it becomes a theorem.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
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Rules of Inference
Rules of inference provide the justification of
the steps used in a proof.
One important rule is called modus ponens or the
law of detachment. It is based on the tautology
(p(pq))  q. We write it in the following way:
p
pq
____
q
The two hypotheses p and p  q are
written in a column, and the conclusion
below a bar, where  means “therefore”.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
26
Rules of Inference
The general form of a rule of inference is:
p1
p2
.
.
.
pn
____
q
The rule states that if p1 and p2 and …
and pn are all true, then q is true as well.
These rules of inference can be used in
any mathematical argument and do not
require any proof.
February 25, 2002
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
27
Rules of Inference
p
_____
Addition
 pq
q
Modus
pq
_____ tollens
 p
pq
_____
Simplification
p
pq
Hypothetical
qr
_____ syllogism
 pr
p
q
_____ Conjunction
 pq
February 25, 2002
pq
Disjunctive
p
_____ syllogism
q
Applied Discrete Mathematics
Week 5: Mathematical Reasoning
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