CMSC 203 : Section 0201 Hw2 Solution
February 24, 2007
1. Exercise set 2.1
Problem 18 c,d
Solution
(10 points)
c. ∀s ∈ C(s), C(s) → ¬E(s)
d. ∃s ∈ C(s), C(s) ∧ M (s)
2. Exercise set 2.2
Problem 3 b,d
Solution
(10 points)
b. ∃ computer c, c has no CPU
d. ∀ bands b, b has won less than 10 grammy awards
Problem 12
Solution
(10 points)
Statement: The product of any irrational number and any rational number is is irrational
Proposed negation: The product of any irrational number and any rational number is
rational
The proposed negation is incorrect. The correct negation would be:
There exists at least one irrational and a rational number such that the product of those
numbers is not irrational (rational)
Problem 15
Solution
(10 points)
Given that D = {-48,-14,-8,0,1,3,16,23,26,32,36}
a. ∀x ∈ D, if x is odd then x > 0
This is true
b. ∀x ∈ D, if x < 0 then x is even
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This is true
c. ∀x ∈ D, if x is even then x ≤ 0
This is false. Consider x = 16, 26, 32, 36 for these x the statement fails
d. ∀x ∈ D, if the ones digit of x is 2, then tens digit is 3 or 4
This is true
e. ∀x ∈ D, if the ones digit of x is 6, then tens digit is 1 or 2
This is false. Consider x = 36 for which the given statement fails
Problem 21
Solution
(5 points)
∀n ∈ Z, if n is a prime number then n is odd or n = 2
Negation is:
∃n ∈ Z, n is prime and n is not odd and n 6= 2
Problem 43
Solution
(10 points)
Having a large income is not a necessary condition for a person to be happy.
Negation would be:
∃p ∈ P , p does not have a large income and p is happy
3. Exercise set 2.3
Problem 12 b,e,f
Solution
(10 points)
b. ∀s ∈ S, V(s, Star Wars)
All students at your school have seen movie Star Wars
e. ∃s ∈ S, ∃t ∈ S, and ∃m ∈ M such that s 6= t and V (s, m) ∧ V (t, m)
There exists at least two distinct students s and t such that both have seen same movie
m∈M
f. ∃s ∈ S and ∃t ∈ S such that s 6= t and ∀m ∈ M , V (s, m) → V (t, m)
There exists atleast two distinct students s and t such that for any movie m ∈ M if s has
seen movie m then t has also seen movie m
Problem 13 b
Solution
(10 points)
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b. ∃x ∈ D such that ∀y ∈ E, x + y = −y
The negation would be
∀x ∈ D such that ∃y ∈ E, x + y 6= −y
Given that D = E = {-2,-1,0,1,2}
It can be observed that the negation is true
Consider y = -2 then the equation becomes:
x + (−2) 6= −(−2)
x 6= 4
This holds for all values of x ∈ D
Problem 29
Solution
(10 points)
∃x ∈ R such that ∀y ∈ R− , x > y
a. Interchanging the symbols ∃ and ∀
∀x ∈ R such that ∃y ∈ R− , x > y
b. It can be seen that only original statement is true.
∃x ∈ R such that ∀y ∈ R− , x > y
This is true for any positive real number x. Any positive real number will always be greater
than all negative real numbers R−
∀x ∈ R such that ∃y ∈ R− , x > y
This is false. Since x contains all negative numbers as well. So x > y does not hold for all
x.
Problem 40 c,d,f,g,h,i
Solution
(15 points)
c. ∃x ∈ R such that ∀y ∈ R, x = y + 1
This statement is false. It states that there exists a real number x such that for all real
numbers y, x = y + 1. However this can not be true.
d. ∀x ∈ R+ , ∃y ∈ R+ such that xy = 1
This is true since since for any positive real number x there exists a positive real number
y = 1/x such that xy = 1
f. ∀x ∈ Z + and ∀y ∈ Z + , ∃z ∈ Z + such that z = x − y
The statement states that for any positive integers x and y there exists a positive integer
z such that z = x − y. However this is not always true. e.g. Consider x = 2 and y = 6
(any values greater than x)
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g. ∀x ∈ Z and ∀y ∈ Z, ∃z ∈ Z such that z = x − y
This statement is true. It states that for any integers x and y there exists and integer z
such that z = x − y this is true for any integer x and y.
h. ∃u ∈ R+ such that v ∈ R+ , uv < v
This statement is true. The statement states that there exists a positive real number u
such that for any positive real number v, uv < v is satisfied. This is true for any u less
than 1 e.g. if u = 1/2, uv < v for any v ∈ R+
i. ∀v ∈ R+ , ∃u ∈ R+ such that uv < v
This statement is true. The statement holds for any u ∈ R+ which is less than 1.
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