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Validating Statements

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Validating Statements - Lesson Summary

Every mathematical statement is either true of false.

Ex:

2 is a positive number and a prime number. (True)

5 is a factor of 25 or 30. (True)

For every integer x, x2 is a negative number. (False)

If x is odd, then x2 is odd. (True)

x - 2 is negative if and only if x is less than 2. (True)

The truth value of a compound statement depends on the truth values of the component statements involved in it.


To validate the component statements, certain rules are followed.

Rule I: The first rule is used to validate compound statements with the connective word "and," or conjunctions. To validate whether a conjunction is true, we have to show that all of its component statements are true.

For mathematical statement s p and q, to show that the statement "p and q" is true, we need to show both p and q are true.

Ex: Validity of the statement — 2 is a positive number and a prime number.


The component statements:

p: 2 is a positive number.

q: 2 is a prime number.

Component statements p and q are true.

Hence, the given compound statement is true.

Rule 2: The second rule is used to validate compound statements with the connective word "or".

To validate whether a disjunction is true, we have to show that one of its component statements is true.

If p and q are mathematical statements, then in order to show that the statement "p or q" is true, we need to show that either statement p is true or statement q is true.

Ex: Check the validity of the statement, "Five is a factor of 21 or 30."

The component statements:

p: 5 is a factor of 21.

q: 5 is a factor of 30.

The statement p is false, while the statement q is true.
Hence, it is validated that the given compound statement is true.

Rule III: Rule three is used to validate statements with "if-then," or implications.

To prove the statement "if p, then q", we need to show any one of the following cases is true.


Case 1: By assuming that p is true, show that q must be true. (Direct method)
Case 2: By assuming that q is false, show that p must be false. (Contrapositive method)

Example: If x is odd, then x2 is odd.

The component statements:

p: x is odd.

q: x2 is odd.

Consider p: x is odd, is true.

x = 2n + 1, for some integer n.

x2 = (2n + 1)2

    = 4n2 + 4n + 1

    = 2(2n2 + 2n) + 1

    = 2l + 1, where l = 2n2 + 2n

Hence, x2 is odd.

Hence, the given statement is proved.

Rule IV: Rule four is used to prove bi-implications.


To prove the statement "p if and only if q," we need to show the following.

If p is true, then q is true.

If q is true, then p is true.

Example: x - 2 is negative if and only if x is less than 2.

The component statements are

If x - 2 is negative, then x is less than 2.

If x is less than 2, then x - 2 is negative.

Both the implications are true.

Therefore, the bi-implication 'x - 2 is negative if and only if x is less than 2.' is true.

Contradiction Method

Some mathematical statements cannot be proved directly.
Such statements can be proved by the contradiction method.

Contradiction Method: To prove that a statement p is true, first, assume that p is not true or negation p is true. Then, we arrive at some result that contradicts the assumption. This concludes that p is true.

Consider the example of a situation where the statement is not valid. Such examples are called counter examples.

Counter example: The example of a situation where a statement is not valid.

Put x = 0,

(3 + x)(3 - x) ≠ 9 + x2

⇒ (3 + 0)(3 - 0) ≠ 9 + 0

⇒ 9 ≠ 9

On simplification, 9 ≠ 9, which is a contradiction.

Hence, the given statement is false.

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