CS0441 Discrete Structures
Recitation 6
Xiang Xiao
Section 2.6 Q28
Find the Boolean product of A and B, where
1 0 0 1


A 0 1 0 1


1 1 1 1
1
0
B
1

1
0
1

1

0
Section 2.6 Q28
1 0
1 0 0 1 

0
1

A B  0 1 0 1  

 1 1
1 1 1 1 

1
0


(1  1)  (0  0)  (0  1)  (1  1)
 (0  1)  (1  0)  (0  1)  (1  1)

(1  1)  (1  0)  (1  1)  (1  1)
0  0  0  0 1
1  0  0  1
 0  0  0  1
0  1  0  0   1

 
0  1  1  0  1
1  0  1  1
(1  0)  (0  1)  (0  1)  (1  0) 
(0  0)  (1  1)  (0  1)  (1  0) 

(1  0)  (1  1)  (1  1)  (1  0) 
0
1

1 
Mathematical Induction
For which nonnegative integers n is n2 <= n!? Prove your answer.
Let P(n) be the statement that n2 <= n!. We will attempt to prove that P(n) is true for all
nonnegative integers n>= a.
Basic Step: k = 4, we have n2= 16, n! = 24, n2 <= n!
Inductive Step:
The inductive hypothesis is the statement P(k) is true for the positive integer k >= 4.
To complete the inductive step, we must show that if P(k) is true, then P(k+1) must be
true. So we prove if k2 <= k!, then (k+1)2 <= (k+1)!
We need to prove k2 + 2k + 1<= (k + 1)*k! = k*k! + k!
We already know k2 <= k! given the inductive hypothesis, now lets prove that
2k + 1 <= k*k! is true for all k >= 4. Divide both side by k, the left side is 2 + 1/k < 3 for all k
>= 4, the right side is k! >= 4! = 24 for all k >= 4. Therefore, 2k + 1 <= k*k!.
Given k2 <= k! and 2k + 1 <= k*k!, we have k2 + 2k + 1 <= k*k! + k!= (k+1)!
We have proved that if P(k) is true, then P(k+1) is the true. The inductive step is
complete.
That is, we have shown that for nonnegative integer k >= 4, k2 <= k!.
Section 5.1 Q28
Prove that n 2  7n  12 is nonnegative whenever n is an integer
with n>= 3.
Let P(n) be the statement that n2 – 7n + 12 is nonnegative. We will attempt to prove
that P(n) is true for all integers n>= 3.
Basic Step: k = 3, we have n2 – 7n + 12 = 0, P(3) is nonnegative
Inductive Step:
The inductive hypothesis is the statement P(k) is true for the positive integer k >= 3.
To complete the inductive step, we must show that if P(k) is true, then P(k+1) must
be true. So we must prove if k2 -7k + 12 >= 0, then (k+1)2 – 7(k+1) + 12 >= 0.
(k+1)2 – 7(k+1) + 12 = k2 + 2k + 1 -7k – 7 + 12 = (k2 -7k + 12) + (2k – 6)
We already know k2 -7k + 12 >= 0 given the inductive hypothesis.
When k >= 3, 2k – 6 >= 0.
Therefore, (k2 -7k + 12) + (2k + 6) >= 0, when k >= 3. P(k+1) is true.
We have proved that if P(k) is true, then P(k+1) is the true. The inductive step is
complete.
That is, we have shown that n2 -7n + 12 is nonnegative whenever n is an integer
with n >= 3.
Section 5.1 Q60
Use mathematical induction to show that ( p1  p2  p3...  pn ) is
equivalent to p1  p2  ...  pn whenever p n are propositions.
Let P(n) be the statement that ( p1  p2  p3...  pn ) is equivalent to p1  p2  ...  pn
Basic Step: k = 2, P(2) asserts that ( p1  p2 )  p1  p2 .This is one of De Morgan’s Laws
Inductive Step:
The inductive hypothesis is the statement P(k) is true for k propositions
To complete the inductive step, we must show that if P(k) is true, then P(k+1) must be
true. So we must prove if this equality holds for every collection of k propositions, then it
must hold for every collection of (k+1) propositions.
( p1  p2  p3...  pk  pk 1 )  (( p1  p2  p3...  pk ))  pk 1 )
 ( p1  p2  p3...  pk )  pk 1
 (p1  p2  p3...  pk )  pk 1
De Morgan's Law
P(k)
 p1  p2  p3...  pk  pk 1
We have proved that if P(k) is true, then P(k+1) is the true. The inductive step is complete.
That is, we have shown that ( p1  p2  p3...  pn ) is equivalent to p1  p2  ...  pn