Download Assignment #1 : Solutions

Discrete Mathematics
Summer 03
Assignment #2: Solutions
Section 1.4
2. P =1, Q = 0
P=1
1
0
Q=0
0
1
1
0
R=1
1
o
17. (p ^ ~q) v (~p ^ r)
p
q
o
r
o
p^~q
(p ^ ~q) v (~p ^ r)
~p^r
21. S = (p ^ q ^ ~r) v (~p ^ q ^ r) v (~p ^ ~q ^ r)
p
q
o
r
o
o
(p ^ q ^ ~r)
(~p ^ q ^ r)
(~p ^ ~q ^ r)
S
Discrete Mathematics
Summer 03
Section 1.5
3. 28710 = 28*1 + 2 4*1 + 23*1+ 22*1 + 21*1+ 20*1
= 1000111112
8. 1101102 = 20*0 + 21*1 + 22*1+ 23*0 + 24*1+ 25*1
= 0 + 2 + 4 + 0 + 16 + 32
= 54
12. 10012 + 11112 = 110002
23. 10810 = 011011002 switch
31. 79 + (-43)
79 = 010011112
-43 = 001010112
switch
01001111
+ 11010101
100100100 remove first 1
100100112 bin add 1
110101002
bin add 1
100101002
110101012
001001002 = 3610
42. 1011011111000101 = 1011 0111 1100 0101 = B 7 C 5
Section 2.1
11.d)  logicians x, x is not lazy.
12.b)  a real number x such that x is rational.
32. Statement:  n  Z, if n is prime then n is odd or n = 2.
Negation:  an integer n such that n is prime and both n is not odd and n2.
34. Statement:  animals x, if x is a cat then x has whiskers and x has claws.
Negation:  an animal x such that x is a cat and x does not have whiskers or x does not have
claws.
Section 2.2
11. Statement: Everybody trusts somebody.
Formally: (a)  people x,  a person y such that x trusts y.
Negation: Somebody trusts nobody.
Formally: (b)  a person x, such that  people y, x does not trust y.
17. Statement: There’s a prime number between every integer and its double.
Formally: (a)  integers x,  a prime number y such that x ≤ y and y ≤ 2*x
Negation: (b)  an integer x, such that  prime numbers y, x > y or y > 2*x
32. Being divisible by 6 is a sufficient condition for being divisible by 3.
 x, if x is divisible by 6 then x is divisible by 3.
Discrete Mathematics
Summer 03
40. b) ?isabove(w1, g)
“is w1 above g?”  NO.
c) ?color(w2, blue)
“is w2 colored blue?”  NO.
e) ?isabove(X, b1)
“for which blocks X is the predicate X is above b1 true?”  X = g.
g) ?isabove(g, X)
“for which blocks X is the predicate g is above X true?”  X = b1, X = w1.
41. Statement:  > 0,  an integer N such that  integers n, if n>N then L-<an<L+
Negation:  >0 such that  integers N,  an integer n such that n>N and an  L- or an  L+
42. b) ! integer x such that 1/x is an integer.
False, because there’s more than one such: 1 and –1.
Section 2.3.
1. e)
Given:
 real numbers a and b:
(a+b)2 = a2 + 2ab + b2
 a = log(t1) and b = log(t2) are particular real numbers
Conclude by universal instantiation:
(log(t1) + log(t2))2 = (log(t1))2 +2log(t1)log(t2) + (log(t2))2
16. invalid (converse error)
20. a)
All dogs are carnivorous.
Felix is not a dog.
Felix is not carnivorous.
non-carnivorous
carnivorous
dogs
x Felix
This diagram shows a case of an x for which the premises are true and the
conclusion is false.