INFS 501 Fall 2002 Solutions – HW6
Homework 6: Solutions
Problem 1
(1)
qr
r
Therefore q
premise
premise
by modus tollens
(2)
pq
q
Therefore p
premise
by (1)
by disjunctive syllogism
(3)
q  u  s
q
Therefore u  s
premise
by (1)
by modus ponens
(4)
us
Therefore s
by (3)
by conjunctive simplification
(5)
p
s
Therefore p  s
by (2)
by (4)
by conjuctive addition
pst
ps
Therefore t
premise
by (5)
by modus ponens
(6)
Problem 2
(1)
q  s
s
Therefore q
premise
premise
by disjunctive syllogism
(2)
pq
q
Therefore p
premise
by (1)
by modus tollens
(3)
rs
s
Therefore r
premise
premise
by disjunctive syllogism
(4)
p
r
Therefore p  r
by (2)
by (3)
by conjunctive addition
p  r u
p  r
Therefore u
premise
by (4)
by modus ponens
(5)
(6)
s  t
premise
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INFS 501 Fall 2002 Solutions – HW6
s
Therefore t
premise
by modus ponens
(7)
wt
t
Therefore w
premise
by (6)
by disjunctive syllogism
(8)
by (5)
by (7)
by conjuctive addition
u
w
Therefore uw
Problem 3
The statement is not existential.
An informal negation of the statement is “There are some easy questions on the exam” or “Some of the
questions on the exam are easy”.
A formal version of the statement is “ questions on the exam x, x is not easy” or “x, if x is a question on
the exam then x is not easy”
In informal language, these can be written as “None of the questions on the exam are easy”
Problem 4
Invalid.
Let D be the set of all discrete mathematics students, T the set of all thoughtful people, and V the set of all
people who can tell a valid from an invalid argument.
Any one of the following diagrams could represent the given premises.
Only in drawing (1) is the conclusion true. Hence it is possible for the premises to be true while the
conclusion is false, and so the argument is invalid.
Problem 5
1.
Yes: 4rs = 2(2rs) and 2rs is an integer because r and s are integers and products of integers are integers.
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INFS 501 Fall 2002 Solutions – HW6
2.
Yes: 6r+4s2+3 = 2(3r+2s2+1) + 1 and 3r+2s2+1 is an integer because r and s are integers and products and
sums of integers are integers.
3.
Yes: r2+2rs+s2 = (r+s)2 and r + s is an integer that is greater than or equal to 2 because both r and s are
positive integers, and so each is greater than or equal to 1.
Problem 6
For example, let n = 25.
Then n is a perfect square because 25 = 52 and since 25 = 9+16 = 32+42, n can be written as a sum of two other
perfect squares.
Problem 7
This incorrect proof begs the question.
The second sentence of the proof states a conclusion that follows from the assumption that m x n is even.
The next to last sentence of the proof states this conclusion as if it were known to be true. But it is not known
to be true. In fact, it is the main task of the proof to derive this conclusion, not from the assumption that it is
true but from the hypothesis of the theorem.
Problem 8
Proof: Let m and n be any odd integers. By definition of odd, m=2r+1 and n=2s+1 for some integers r and s.
By substitution, m-n = (2r+1)-(2s+1)=2(r-s).
Since(r-s) is an integer (being a difference of integers), then (m-n) equals twice some integer, and so m-n is
even by definition of even.
Problem 9
Counterexample: Let m=3. Then m2 – 4 = 9 – 4 = 5 which is not composite.
Problem 10
If m and n are perfect squares, then m=a2 and n = b2 for some integers a and b.
We may take a and b to be nonnegative because for any real number x, x 2=(-x)2 and if x is negative then –x is
nonnegative.
By substitution,
m + n + 2 X (square root of mn)
= a2 + b2 + 2 X (square root of a2b2)
= a2+b2+2ab
since a and b are nonnegative
=(a + b)2
But a + b is an integer (since a and b are), and so m + n + 2 X (square root of mn) is a perfect square.
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