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Wollongong College Australia
A College of the University of Wollongong
Diploma in Information Technology
Mid-Session Test #1
WUCT121 Discrete Mathematics
Autumn Session 2008 SOLUTIONS
Time Allowed: 50 minutes
DIRECTIONS TO CANDIDATES
Total No. of Questions: 4
All questions are to be attempted.
You may answer the questions in any sequence.
Questions are NOT of equal value.
All notation is as used in lectures.
Working ( including all necessary reasoning ) is to be shown for all solutions.
No examination aids or materials are allowed.
This exam represents 10% of the total subject marks.
ITC Education Ltd trading as
Wollongong College Australia
CRICOS 02723D
ABN 14105312329
WCA-WUCT121-EXMST1
Page 1 of 5
Question 1.
(a)
(16 marks)
Write in logical notation using simple statements and connectives:
If people marry in haste then they can regret it at their leisure,
but if they marry with prudence then they are more likely to avoid jurisprudence.
Statement form is
( P fi Q ) Ÿ (R fi S )
where
P : people marry in haste
Q : they can regret it at their leisure
R : they marry with prudence
S : they are more likely to avoid jurisprudence
[7]
(b)
(i) Let x and y be non-zero real numbers. Write the following statement in simple English:
x
((x > 0 ) Ÿ ( y > 0 )) ⁄ ((x < 0 ) Ÿ ( y < 0 )) T > 0
y
“The statement that any two non-zero real numbers are both positive or are both negative is
not logically equivalent to the statement that their ratio is positive”.
[5]
(ii)
Determine whether the statement in (i) is true or false. If it is a true statement,
briefly explain why. If it is a false statement, demonstrate why it is false.
It is a false statement because a ratio of two non-zero real numbers is positive only when
both the numerator and the denominator have the same sign – if they are of opposite sign the
ratio is negative. Consequently, the LHS statement and the RHS statement have the same
truth-value for any non-zero values of x and y, and so are logically equivalent, contrary to
the given statement.
[4]
Question 2.
(a)
(17 marks)
Using a truth table, determine whether the following statement is a tautology, a
contradiction, or is contingent:
~((p ⁄ ~q ) Ÿ (q ⁄ ~p)) ¤ ( ~p ¤ ~q)
p
T
T
F
F
q
T
F
T
F
~
9
F
T
T
F
( (
p ⁄
2
T
T
F
T
~ q
1
F
T
F
T
) Ÿ ( q ⁄
5
4
T
T
F
F
F
T
T
T
~ p
3
F
F
T
T
)
)
¤
10
F
F
F
F
(
~
6
F
F
T
T
p ¤
8
T
F
F
T
~ q )
7
F
T
F
T
[10]
WCA-WUCT121-EXMST1
Page 2 of 5
(b)
Using the “quicker” method, prove that the following statement is a tautology:
((p Ÿ ( q ⁄ r ) ) Ÿ ( p fi ~q)) fi r
p
q
( (
p
Ÿ ( q ⁄ r ) Ÿ
2
Step #1
Step #2
Step #3
Step #4
Step #5
1
(
p
5
fi ~ q ) ) fi
4
3
r
6
F
T
F
T
T
T
T
T
T
T
F
Step #5 reveals that the assumption in Step #1 that the entire statement can be false leads to
a contradiction i.e. that the variable q has to be both True and False simultaneously.
Consequently the statement must be a tautology.
[7]
Question 3.
(15 marks)
Using the Rules of Substitution and Substitution of Equivalence and the following three
tautologies,
1. ~(p ⁄ q) ¤ (~p Ÿ ~q)
2. p fi q ¤ ~p ⁄ q
3. (p ¤ q) ¤ ((p fi q) Ÿ (q fi p))
formally prove that the following statement is also a tautology:
~(p Ÿ (q Ÿ r) ) fi ((~r ⁄ ~q) ⁄ ~p)
Using tautology (1) and the Rule of Substitution, substituting (q Ÿ r) for q in tautology (1)
shows that ~(p Ÿ (q Ÿ r)) ¤ (~p ⁄~(r Ÿ q)) --- (A) is a tautology.
Using the commutativity of ⁄ in the RHS of A produces the tautology ~(p Ÿ (q Ÿ r)) ¤ (~(r Ÿ
q) ⁄ ~ p) --- (B)
Applying the Rule of Substitution to tautology (1), substituting r for p shows that ~(r Ÿ q) is
logically equivalent to ((~r ⁄ ~q) --- (C).
Applying the Rule of Substitution of Equivalence to statement (B), substituting (C), results in
the tautology ~(p Ÿ (q Ÿ r)) ¤ ((~r ⁄ ~q) ⁄ ~ p) --- (D).
Finally, applying tautology (3) to (D) shows that ~(p Ÿ (q Ÿ r) ) fi ((~r ⁄ ~q) ⁄ ~p) as
required.
Evidence of some understanding of the Rule of Substitution and the Rule of Substitution of
Equivalence is what is being sought here.]
[15]
WCA-WUCT121-EXMST1
Page 3 of 5
Question 4.
(a)
(32 marks)
Write in predicate calculus notation using quantifiers and variables:
The sum of any two integers is also an integer.
Long Format
"x ( x Π٠fi "y ( y Π٠fi (x + y)Π٠))
"x ( x Π٠fi "y ( y Π٠fi ( $u ( u Π٠٠( x + y) = u ))))
Short Format
" x, y Œ Ÿ ( x + y Œ Ÿ)
" x, y Œ Ÿ, $u Œ Ÿ ( x + y = u)
(b)
[5]
Write in simple English without using quantifiers or variables:
$x Œ —, "y Œ —, x < y.
There is a smallest real number.
(c)
[3]
Write down the negation of the statement in part (a) as follows:
(i)
in predicate calculus notation using quantifiers and variables
Q4 (c)(i) “Long” Symbolic Format
~"x ( x Œ Ÿfi " y ( y Œ Ÿ fi (x + y)Œ Ÿ )) ∫$ x ( x Œ Ÿ Ÿ $ y ( y Œ Ÿ Ÿ (x + y)∉ Ÿ ))
~"x ( x Œ Ÿ fi " y ( y Œ Ÿ fi ( $ u ( u Œ Ÿ Ÿ ( x + y) = u )))) ∫
$ x ( x Œ Ÿ Ÿ $ y ( y Œ Ÿ Ÿ (" u ( u Œ Ÿ Ÿ ( x + y) π u ))))
Q4 (c)(i) “Short” Symbolic Format
~" x, y Œ Ÿ ( x + y Œ Ÿ)∫ $ x, y Œ Ÿ ( x + y œ Ÿ)
~" x, y Œ Ÿ $ u Œ Ÿ ( x + y = u) ∫ $ x, y Œ Ÿ, " u Œ Ÿ ( x + y π u)
(ii)
in simple English without using logical quantifiers and variables.
There are two integers whose sum is not an integer.
(d)
[6]
[3]
Determine whether the statement in (a) is true or false. If it is true, briefly explain why it
is true. If it is false, demonstrate why it is false.
The statement in part 4(a) is true, given that the set of integers Ÿ is closed under the
operation of addition.
[3]
(e)
Write down the negation of the statement in part (b) as follows:
(i)
in predicate calculus notation using quantifiers and variables,
~$ x (x Œ — Ÿ (" y, y Œ — fi x < y)) ∫ " x (x Œ — fi ($ y, y Œ — Ÿ x ≥ y))
WCA-WUCT121-EXMST1
Page 4 of 5
~$ x Œ — (" y Œ —, x < y) ∫ " xŒ — ($ yŒ —, x ≥ y))
(ii)
in simple English without using logical quantifiers and variables.
There is no smallest real number.
(f)
[6]
[3]
Determine whether the statement in (b) is true or false. If it is true, briefly explain why it
is true. If it is false, demonstrate why it is false.
The statement in 4(b) is false, for if x is claimed to be the smallest real number the fact that
x–1 is less than x results in a contradiction. Consequently, by Modus Tollens, there can be no
smallest rational number. However, the term ‘smallest’ is sometimes interpreted to mean,
‘having the least absolute value’. With this meaning of smallest, the statement in 4 (b) would
be true because 0 Œ — and 0 ≤ | x| for any real number x.
[3]
WCA-WUCT121-EXMST1
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