Exercise 1.1
1. Which of the
following sentences are the statements? Find the truth values of those
sentences which are statements.
a) Kathmandu is the capital of Nepal.
Given sentence is statement; its truth value is T.
b) Nepal Exports oil.
Given sentence is statement; its truth value is F.
c) 3+5=8
Given sentence is statement; its truth value is T.
d) Where do you live?
Given sentence is interrogative sentence, so it is not a statement.
e) Understand Logic.
Given sentence is imperative sentence, so it is not a statement.
f) Oh! How beautiful the scene is?
Given
sentence is exclamatory sentence, so it is not a statement.
2. Let p: demand is
increasing and q: supply is decreasing. Express each of the following
statements into words.
a) ~p
Demand is not increasing.
b) ~q
Supply is not decreasing.
c) p ^ q
Demand is increasing and supply is decreasing.
d) p v q
Demand is increasing or supply is decreasing.
e) ~p ^ q
Demand is not increasing and supply is decreasing.
f) p v ~q
Demand is increasing or supply is not decreasing.
g) ~p ^
~q
Neither demand is increasing nor supply is decreasing.
h) ~(p ^ q)
It
is false that demand is increasing and supply is decreasing.
3. Express each of
the following statements into symbolic form.
a) p: Temperature is
increasing;
q: volume is
expanding
Temperature is
increasing and length is expanding.
Solution: p ^ q
b) p: Pressure is
decreasing;
q: volume is
increasing
Pressure is
increasing or volume is increasing.
Solution: p v q
c) p: demand is
increasing;
q: price is
increasing
Neither demand is
increasing nor demand is decreasing.
Solution: ~p ^
~q.
d) p: Pradeep is
bold;
q: Sandeep is
handsome.
It is false that
Pradeep is bold or Sandeep is handsome.
Solution: ~(p v q)
4) Construct truth
tables for the following compound statements:
(a) ~p ^ q
Solution:
|
p |
q |
~p |
~p ^ q |
|
T T F F |
T F T F |
F F T T |
F F T F |
(b) ~p v (~q)
Solution:
|
p |
q |
~p |
~q |
~p v (~q) |
|
T T F F |
T F T F |
F F T T |
F T T F |
F T T T |
c) ~p^q
Solution:
|
p |
q |
p^q |
~p^q |
|
T T F F |
T F T F |
T F F F |
F T T |
(d) ~[p v (~q)]
Solution:
|
p |
q |
~q |
p v (~q) |
~[p v (~q)] |
|
T T F F |
T F T F |
F T F T |
T T F T |
F F T F |
e) (p⇒q)^(q⇒p)
Solution:
|
p |
q |
p⇒q |
q⇒
p |
(p⇒q)^(q⇒p) |
|
T T F F |
T F T F |
T F T |
T T F T |
T F F T |
f) (~p^q) ⇒ (pvq)
Solution:
|
p |
q |
~p |
~p ^ q |
p v q |
(~p^q) ⇒ (pvq) |
|
T T F F |
T F T F |
F F T T |
F F T F |
T T T F |
T T T T |
5. Let p, q, r and s
be four simple statements. If p is true, q is false, r is true and s is false,
find the truth values of the following compound statements.
a) p^q
Solution:
|
p |
q |
p^q |
|
T |
F |
F |
b) p v (~q)
|
p |
q |
~q |
p v (~q) |
|
T |
F |
T |
T |
c) ~p ^ ~q
|
p |
q |
~p |
~q |
~p ^ ~q |
|
T |
F |
F |
T |
F |
d) q v (p ^ s)
|
P |
q |
s |
p ^ s |
q v (p ^ s) |
|
T |
F |
F |
F |
F |
e) ~(~p)
|
p |
~p |
~(~p) |
|
T |
F |
T |
f) (pvq)^(rvs)
|
p |
q |
r |
s |
p v q |
r v s |
(pvq)^(rvs) |
|
T |
F |
T |
F |
T |
T |
T |
6) If p and q are any
two statements, prove that:
a) p ^ ~p=c, where c=
contradiction
Solution:
We prove the given relations using truth table.
|
p |
~p |
p ^ ~p |
c |
|
T T F F |
F F T T |
F F F F |
F F F F |
So, from the above truth table, p ^ ~p ≡ c.
(b) pvq≡ qvp
Solution:
We know the given relations using truth table,
|
p |
q |
pvq |
qvp |
|
T T F F |
T T T F |
T T T F |
T T T F |
From the above truth table, p v q ≡ p v p.
c) ~[pv(~q)] ≡ ~p^q
Solution:
We prove the given relations using truth table.
|
p |
q |
~p |
~q |
pv(~q) |
~[pv(~q)] |
~p^q |
|
T T F F |
T F T F |
F F T T |
F T F T |
T T F T |
F F T F |
F F T F |
From the above truth table, ~[pv(~q)]≡~p ^ q.
d) ~[(~p)^q] ≡pv(~q)
We prove the given relations using truth table.
|
p |
q |
~p |
~q |
~p ^ q |
~[(~p)^q] |
pv(~q) |
|
T T F F |
T F T F |
F F T T |
F T F T |
F F T F |
T T F T |
T T F T |
From the above truth table, ~p[(~p ^ q)] ≡≡ p v
(~q).
7) Find the negation
of each of the following statements:
a) Light travels in a
straight line.
Solution: Light
does not travel in a straight line.
b) River can be used
to produce electricity.
Solution: River cannot
be used to produce electricity.
c) x>o
Solution: x ≤ 0
d) Some students are
weak in mathematics.
Solution: No
student is weak in mathematics.
e) All teacher are laborious.
Solution: Some
teachers are not laborious.
8) Find the truth
value and the negation of each of the following statements:
a) 3+2=5 or 6 is the
multiple of 5.
Solution:
Let p is a statement of 3 + 2 = 5 and q is a statement of 6
is a multiple of 5 i.e.
|
p |
q |
Pvq |
|
T |
F |
T |
Negation: 3 + 2 ≠ 5 and 6 is not a multiple of 5.
b) 8 is the prime
number and 4 is even.
Solution:
Let p: 8 is a prime number, q:4 is even.
|
p |
q |
p^q |
|
F |
T |
F |
Negation: 8 is not a prime number or 4 is even.
c) If 3>0 then
4+6=10
Solution:
Let p:3> 0, q: 4 + 6 = 10.
|
p |
q |
pà q |
|
T |
T |
T |
Negation: 3 ≤ 0 and 4 + 6 = 10.
d) If 2 is odd or 3
is natural number then 2+3=8
Solution:
Let p: 2 is odd, q: 3 is a natural number, r: 2 + 3 =
8
|
p |
q |
pvq |
r |
(pvq)
⇒r |
|
F |
T |
T |
F |
F |
Negation: 2 is odd or 3 is a natural number and 2 + 3 ≠ 8.
e) If 2×3=5⇒3>1
then 6 is even.
p: 2 ×
3 = 5, q: 3 > 1 and 6 is even.
|
p |
q |
p⇒q |
r |
(p⇒q)
⇒r |
|
F |
T |
T |
T |
T |
Negation: 2 * 3 = 5 à 3 > 1 and 6 is odd.
g) A triangle ABC is
right angled at B if and only if AB2+ BC2=AC2
Let p: A triangle ABC is a right angled at B.
q: AB2 + BC2 = AC2
|
P |
q |
p⇒q |
q⇒p |
p⇒q |
|
T |
T |
T |
T |
T |
Negation: A triangle is right angle at B and AB2 +
BC2 ≠ AC2.
9) Some statements
are given below:
Find the:
(i)
Antecedent
and Consequent
(ii)
Converse,
inverse and contrapositive
(iii)
Negation
of each of the given statements
Solution for (i)
a) If 3 is natural
number then 1/3 is rational number.
Antecedent: 3 is
a natural number.
Consequent: 1313 is
a rational number.
b) x2 = 4
whenever x=2.
Antecedent: x = 2
Consequent: x2
= 4
c) If the battery is
low then mobile doesn’t work well.
Antecedent: The
battery is low.
Consequent: The
mobile will not work well.
Solution for (ii):
a) If 3 is natural
number then 1/3 is rational number.
Converse: If 1/3
is a rational number then 3 is a natural number.
Inverse: If 3 is
not a natural number then 1/3 is not a natural number.
Contra positive:
If 1/3 is not a rational number then 3 is not a natural number.
b) x2 = 4
whenever x=2.
Converse: If x2 =
4 then x = 2.
Inverse: If x ≠ 2
then x2 ≠ 4.
Contra positive:
If x2 ≠ 4 then x ≠ 2.
c) If the battery is
low then mobile doesn’t work well.
Converse: If the
mobile does not work well, then the battery is low.
Inverse: If the
battery is not low then the mobile works well.
Contra positive:
If the mobile works well then the battery is not low.
Solution for (iii)
a) If 3 is natural
number then 1/3 is rational number.
3 is a natural number and 1/3 is not a rational number.
b) x2 = 4
whenever x=2.
x = 2 and x2 = 4.
c) If the battery is
low then mobile doesn’t work well.
The battery is low and the mobile works well.
10) If p and q be the
statements, prove that:
(a) p v(~p v q) is
tautology
Solution:
We prove the above relation by using truth table as follows:
|
p |
q |
~p |
~pvq |
p v(~p v q) |
|
T T F F |
T F T F |
F F T T |
T F T |
T T T T |
Here, the compound statement p v (~pvq) is true whatever be
the truth of its component. Therefore, p v (~p v q) is a tautology,
(b) (p^q) ⇒ (pvq)is a tautology
Solution:
We prove the above relation by using truth table as follows:
|
p |
q |
p ^ q |
pvq |
(p^q) ⇒
(ppv q) |
|
T T F F |
T F T F |
T F F F |
T T T F |
T T T T |
Here, the compound statement(p^q)à(pvq) is true whatever be
the truth of its component. Therefore, (p^q) ⇒ (pvq)is a tautology.
c) ~(pvq) ^ q is
contradiction
Solution:
We prove the above relation by using truth table as follows:
|
p |
q |
p v q |
~(pvq) |
~(pvq) ^ q |
|
T T F F |
T F T F |
T T T F |
F F F T |
F F F F |
Here, the statement ~(pvq)^ q is false whatever the truth
values of its component. Therefore, the statement ~(pvq)^ q is a contradiction.
d) (p^q)^~(pvq) is
contradiction
Solution:
We prove the above relation by using truth table as follows:
|
p |
q |
p^q |
pvq |
~(pvq) |
(p^q)^~(pvq) |
|
T T F F |
T F T F |
T F F F |
T T T F |
F F F T |
F F F F |
Here, the statement (p^q)^~(pvq) is false whatever the truth
values of its component. Therefore, the statement (p^q)^~(pvq)is a
contradiction.