In the previous section, we saw Implications. In this section, we will see validating statements.
• Truth value of a statement tells us whether a statement is true or false.
• If a statement is true, then its truth value is true.
♦ We write it as T.
• If a statement is false, then its truth value is false.
♦ We write it as F.
• Validating statements is the process of finding the truth value of a statement.
• Validating simple statements is easy. But for compound statements, we have to follow certain rules.
• The rules depend on the connectives/implications present in the compound statement.
Rule 1: Statements with “and”.
• If the connective in a compound statement is “and”, then the truth value of each of the component statements must be T. Then only we get T for the compound state.
• So when the component statements are p and q, we can use the following two steps:
Step 1: Show that p is true.
Step 2: Show that q is true.
Rule 2: Statements with “or”.
• If the connective in a compound statement is “or”, then the truth value of at least one component statement must be T. Then only we get T for the compound state.
• So when the component statements are p and q, we can use the following three cases:
Case 1: p is false but q is true.
Case 2: p is true but q is false.
Case 3: Both p and q are true.
• If we get any of the above three cases, then the truth value of the compound statement is T.
Rule 3: Statements with “if-then”.
• When the compound statement is in the form “if p then q”, we know that:
Whenever p is true, q is also true.
• So we can think of a method to validate such statements:
♦ First, we assume that p is true.
♦ Based on this assumption, we check q.
♦ If q is true, then the compound statement can be given the truth value T
• This method is known as direct method.
• When the compound statement is in the form “if p then q”, we know that:
if ~q then ~p is applicable.
• So we can think of another method to validate such statements:
♦ First, we assume that ~q is true.
♦ Based on this assumption, we check ~p.
♦ If ~p is true, then the compound statement can be given the truth value T
• This method is known as contrapositive method.
• So for validating “if p then q” statements, we have two methods:
(i) Direct method (ii) Contrapositive method.
Rule 4: Statements with “if and only if”.
• When the compound statement is in the form “p if and only if q”, we know that:
♦ Whenever p is true, q is also true.
♦ Whenever q is true, p is also true.
• So we can think of a method to validate such statements:
In this method, we need to show two items:
(i) If p is true, then q is true.
(ii) If q is true, then p is true.
Rule 5
This rule is used to prove that, a statement is true. In this rule, we make use of contradiction. The steps are:
(i) We assume that p is false.
This is same as assuming ~p to be true.
(ii) Then we arrive at some result which contradicts the assumption.
(iii) Due to the contradiction, we conclude that, p is true.
Rule 6
This rule is used to prove that, a statement is false. In this rule, we simply present an example where the statement is not valid. Such an example is called counter example. When the counter example is presented, the statement is immediately written off as false.
◼ We may be able to present numerous examples where a statement is valid. Whatever be the number of such examples we present, we cannot conclude that the statement is true.
◼ We may be able to present numerous counter examples where a statement is not valid. Just one counter example is sufficient to conclude that the
statement is false.
Let us see some solved examples:
Solved example 14.10
Check whether the following statement is true or not.
If x, y ∈ Z are such that x and y are odd, then xy is odd.
Solution:
1. Consider the statement:
r: If x, y ∈ Z are such that x and y are odd, then xy is odd.
2. The component statements are:
p: x, y ∈ Z are such that x and y are odd.
q: xy is odd.
3. This is a statement with "if-then". So we will apply Rule 3. We will use the direct method.
♦ First, we assume that p is true.
♦ Based on this assumption, we check q.
4. Assuming p to be true:
• Let us assume that, both x and y are indeed elements of the set Z and they are indeed odd.
• We know that, Z is the set of integers. So we can write:
We assume that both x and y are indeed odd integers.
5. Since they are odd integers, we can write:
♦ x = 2m + 1
♦ y = 2n + 1
✰ Where m and n are some integers.
6. Now we calculate the product xy. We get:
$\begin{array}{ll}
{}&{xy}
& {~=~}& {(2m+1)(2n+1)} &{} \\
{}&{}
& {~=~}& {4mn + 2m + 2n + 1} &{} \\
{}&{}
& {~=~}& {2(2mn + m + n) + 1} &{} \\
\end{array}$
7. Consider the result in (6):
• m and n are integers
⇒ (2mn + m + n) can be odd or even.
⇒ 2(2mn + m + n) will be even.
⇒ 2(2mn + m + n) + 1 will be odd.
8. Based on (6) and (7), we can write:
The product xy is odd.
• So the statement q is true.
9. We obtained the truth value of q as T.
• We obtained this by assuming that the truth value of p is T.
• So the truth value of “if p then q” is T.
• We can write:
The truth value of the given compound statement r is T.
Alternatively, using the contrapositive method:
1. We have: if ~q then ~p
• The contrapositive statements are:
~q: xy is even.
~p: One among x, y is even.
2. For applying the contrapositive method,
♦ First, we assume that ~q is true.
♦ Based on this assumption, we check ~p.
3. Assuming ~q to be true:
~q: xy is even.
4. When xy is even, x or y has to be even.
(supposing x = 2n, we get xy = 2ny, which is even)
• xy will never become even if both x and y are odd.
• So ~p is true.
5. We obtained the truth value of ~p as T.
• We obtained this by assuming that the truth value of ~q is T.
• So the truth value of “if ~q then ~p” is T.
• We can write:
The truth value of the given compound statement r is T.
Solved example 14.11
Check whether the following statement is true or false by proving its contrapositive.
If x, y ∈ Z such that xy is odd, then both x and y are odd.
Solution:
1. Consider the statement:
r: If x, y ∈ Z such that xy is odd, then both x and y are odd.
2. The component statements are:
p: x, y ∈ Z such that xy is odd.
q: both x and y are odd.
3. This is a statement with "if-then". So we will apply the contrapositive method for validation.
• We have: if ~q then ~p
• The contrapositive statements are:
~q: One among x, y is even.
~p: xy is even.
2. For applying the contrapositive method,
♦ First, we assume that ~q is true.
♦ Based on this assumption, we check ~p.
3. Assuming ~q to be true:
~q: One among x, y is even.
4. When one among x, y is even, xy will be even.
(supposing x = 2n, we get xy = 2ny, which is even)
• xy will never become even if both x and y are odd.
• So ~p is true.
5. We obtained the truth value of ~p as T.
• We obtained this by assuming that the truth value of ~q is T.
• So the truth value of “if ~q then ~p” is T.
• We can write:
The truth value of the given compound statement r is T.
Solved example 14.12
Prove that the following statement is true.
p: √7 is irrational
Solution:
1. Assume that √7 is rational.
2. If √7 is rational, we can write: $\sqrt{7} = \frac{a}{b}$
Where the ratio $\frac{a}{b}$ is in the simplest form.
•
That is, a and b do not have any common factors.
3. Squaring both sides, we get: $7 = \frac{a^2}{b^2}$
4. Based on this we can write: $b^2 = \frac{a^2}{7}$
•
b is an integer. So b2 will be an integer.
5. Let us analyze the result in (4):
$\text{An integer}(b^2)~=~\frac{\text{An integer}(a) \times \text{An integer}(a)}{7}$
•
So $\frac{a}{7}$ must be an integer. Let it be c.
6. We can write: $\frac{a}{7}= c$
•
Thus we get a = 7c
•
Squaring both sides, we get: a2 = 49c2.
7. But from (3) we have a2 = 7b2.
8. Equating the results in (6) and (7), we get:
49c2 = 7b2.
•
From this we get: 7c2 = b2.
9. Based on this we can write: $c^2 = \frac{b^2}{7}$
•
c is an integer. So c2 will be an integer.
10. Let us analyze the result in (9):
$\text{An integer}(c^2)~=~\frac{\text{An integer}(b) \times \text{An integer}(b)}{7}$
•
So $\frac{b}{7}$ must be an integer. Let it be d.
11. Let us compare two results:
♦ In (5), we saw that, a is divided by 7 to give an integer.
♦ In (10), we saw that, b is divided by 7 to give an integer.
•
That means, a and b has a common factor 7. This contradicts our assumption in (2) that, a and b have no common factors.
12. So it is clear that the assumption that √7 is rational is wrong.
•
We can write:
p: √7 is irrational
is true.
Solved example 14.13
By giving a counter example, prove that the following statement is false.
If n is an odd integer, then n is prime.
Solution:
1. Consider the statement:
r: If n is an odd integer, then n is prime.
2. The component statements are:
p: n is an odd integer.
q: n is prime.
3. This is a statement with "if-then".
•
We have to prove "if p then q" is false.
4. For statement p, we take n = 9 as an example. This is acceptable because 9 is an odd number.
5. Now we consider statement q.
•
We know that 9 is not a prime number. So when n = 9, the statement q becomes false.
6. So we presented a counter example which proved "if p then q" to be false.
•
That is., r: If n is an odd integer, then n is prime.
is false.
The link to some solved examples is given below:
In the next section, we will see some miscellaneous examples.
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