Discrete Mathematics
Summer 03
Midterm Solutions
1. In each of these tables, the last column contains the truth-value of the given expression.
If this column contains all T, then the expression is a tautology. If it contains all F, then
it is a contradiction. Otherwise, it is neither.

(p~q)  (q  ~p)
NEITHER
p
F
F
T
T

q
F
T
F
T
~p
T
T
F
F
~q
T
F
T
F
p~q
F
F
T
F
q~p
T
T
T
F
(p~q)  (q  ~p)
F
F
T
F
p V q V r V ~p
TAUTOLOGY
p
F
F
F
F
T
T
T
T

q
F
F
T
T
F
F
T
T
r
F
T
F
T
F
T
F
T
~p
T
T
T
T
F
F
F
F
p V q V r V ~p
T
T
T
T
T
T
T
T
(pq) V (~pV(p~q))
TAUTOLOGY
p
F
F
T
T

q
F
T
F
T
pq
F
F
F
T
~p
T
T
F
F
~q
T
F
T
F
p~q
F
F
T
F
(~pV(p~q)
T
T
T
F
(pq) V (~pV(p~q))
T
T
T
T
(pq) (p~q) ~p
NEITHER
p
F
q
F
~p
T
~q
T
pq
T
p~q
T
(pq) (p~q) ~p
T
Discrete Mathematics
F
T
T
T
F
T
T
F
F
F
T
F
Summer 03
T
F
T
T
T
F
T
F
F
2. “ If my car is in the repair shop, then I cannot get to class”.
Let A = “my car is in the repair shop”
B = “I cannot get to class”
(1)
(2)
(3)
(4)
(5)
statement: A  B
negation: A ^ ~B
contrapositive: ~B  ~A
inverse: ~A  ~B
converse: B  A
(1) and (3) are equivalent.
(4) and (5) are equivalent.
3. Critical rows – where the premises are all – are shown in dark gray. The argument is
valid if there are no critical rows where the conclusion is false.

pq
Hence : ~p V q
p
F
F
T
T
q
F
T
F
T
Premise
pq
T
F
F
T
Conclusion
~pVq
T
T
F
T
Valid.

pq
qr
~p
Hence: ~r
p
F
F
F
F
T
T
Q
F
F
T
T
F
F
r
F
T
F
T
F
T
pq
T
T
T
T
F
F
Premises
qr
T
T
F
T
T
T
~p
T
T
T
T
F
F
Conclusion
~r
T
F
T
F
T
F
Discrete Mathematics
T
T
T
T
F
T
Summer 03
T
T
F
T
F
F
T
F
Invalid. 2nd and 3rd critical rows have false conclusion.

r~r
Hence: ~r
Premise
r~r
F
T
r
T
F
Conclusion
~r
F
T
Valid.

pVq
r>~r
Hence: p  q
p
F
F
F
F
T
T
T
T
q
F
F
T
T
F
F
T
T
R
F
T
F
T
F
T
F
T
~r
T
F
T
F
T
F
T
F
PVq
F
F
T
T
T
T
T
T
Premises
r~r
F
F
F
F
F
F
F
F
Conclusion
pq
F
F
F
F
F
F
T
T
Valid. (There are no critical rows in which the conclusion is false because there are
no critical rows at all)
4. (1) x y, x>y
(2) x y, x>y
The first statement says that for any x, there exists a smaller y.
The second statement says that there exists an x which is larger than all y.
Note that the second statement can never be true, because it implies that x > x.

If the variables range over the integers, then the first is true and the second is false.
Discrete Mathematics
Summer 03

If the variables range over the positive integers, then neither is true. The first
statement is false, because when x = 1, there is no smaller positive integer y.
Negations:
(1) x y, x <= y
(2) x y, x<= y
5. De Morgan’s laws:
(1) ~(pq) = ~p V ~q
(2) ~(pVq)= ~p  ~q
6.
 Prove: Product of 2 odd integers is odd.
Let a , b be any two odd integers. This means that a = 2k+1 and b = 2m+1 for some
integers m, k.
Then the product a*b = (2k+1)*(2m+1) = 4km+2k+2m+1
= 2(2km+k+m) +1
Hence, a*b is odd.

Prove by induction: Every positive integer is either even or odd.
Base Case: The smallest positive integer is 1. 1 is odd since 1 = 2*0+1.
Inductive Hypothesis: Any positive integer k is either even or odd.
Show that the integer k+1 is either even or odd.
Inductive Step:
According to the inductive hypothesis, the integer k is either even or odd.
Case1: k is even.
I.e., k = 2m for some integer m.
Hence k+1 = 2m+1.
Therefore, by definition of odd, k+1 is odd.
Case2: k is odd.
I.e., k=2m+1 for some integer m.
Hence k+1 = 2m+2.
Therefore, by definition of even, k+1 is even.

Prove using the quotient remainder theoren: Every positive integer is either even or odd.
Discrete Mathematics
Summer 03
The theorem that for any integer n and positive integer d, there are unique integers q,r
such that
n = qd+r and 0r<d.
Let d=2. Then this case of the theorem states that any integer n can be written
as n=2q+r, where r=0 or 1. If r=0, then n=2q, and hence n is even. If r=1, then n=2q+1,
and so n is odd. Hence n is either odd or even.
7. Statement: There is no smallest positive rational number.
Proof:
For a contradiction, suppose the opposite, i.e., that there exists a smallest positive
rational number, call it R. Since R is rational, we can write it as a/b, for positive integers
a,b. Let R’ = a/(2b). R’ is also a ratio of integers, and hence rational. It is easy to show
that R’ < R. But this contradicts our assertion that R is the smallest rational number.

8. Statement: For all integers a, x, and y, if a | xy then a | x or a | y.
This statement is false. A counter example is a=12, x=3, y=4.
12 | 3*4, but neither 12 | 3 nor 12 | 4.
9. 504 = 23*32*7.
10.
 gcd(48, 20) = gcd(20, 8); since 48 = 2*20+8.
gcd(20, 8) = gcd( 8, 4); since 20 = 2*8 + 4.
gcd(8, 4) = gcd(4, 0); since 8 = 2*4+0.
Therefore, gcd(48, 20) = gcd(4, 0) = 4.

gcd(p2q3, p3q2) = p2q2.
lcm(p2q3, p3q2) = p3q3.
Extra Credit:
lcm(a,b)*gcd(a,b) = a*b.
Discrete Mathematics
Summer 03