Ex 9.1, 10

Transcript

Ex 9.1, 10 Frame the converse for each of the propositions in Questions 1–12. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement. In Questions 8–12, n is a positive integer If n is the square of a prime number, then it has exactly 3 factors. Let’s write Proposition and Converse first Proposition: If n is the square of a prime number, then it has exactly 3 factors Converse: If n has exactly three factors, then it is the square of a prime number Checking if Proposition is true Proposition: If n is the square of a prime number, then it has exactly 3 factors Let’s prove this Let 𝑛=𝑝^2, where 𝑝 is a prime number Since 𝑝 is a prime number, it has factors 1, 𝑝 Thus, factors of 𝑛 are: 𝟏,□( ) 𝒑,□( ) 𝒑^𝟐 So, 𝑛 has exactly 3 factors. ∴ Proposition is true Checking if Converse is true Converse: If n has exactly three factors, then it is the square of a prime number Let’s prove this Let the three factors of n be: 1,□( ) 𝑎,□( ) 𝑛 where □( ) 1<𝑎<𝑛. Factors occur in pairs whose product is 𝑛 Factors 1 and 𝑛 form one pair, i.e. 1 × 𝑛 = 𝑛 Remaining factor 𝑎 must pair with itself, i.e. 𝑎 × 𝑎=𝑛 Therefore: 𝑎 × 𝑎=𝑛 𝒏=𝒂^𝟐 Now, 𝒂 must be prime. If 𝑎 were composite, it would have another factor between 1 and 𝑎. That would also be a factor of 𝑛, giving more than three factors. Hence, 𝑛 is the square of a prime number ∴ Converse is true Checking if Proposition is true Proposition: If n is the square of a prime number, then it has exactly 3 factors Let’s prove this Let 𝑛=𝑝^2, where 𝑝 is a prime number Since 𝑝 is a prime number, it has factors 1, 𝑝 Thus, factors of 𝑛 are: 𝟏,□( ) 𝒑,□( ) 𝒑^𝟐 So, 𝑛 has exactly 3 factors. ∴ Proposition is true Checking if Converse is true Converse: If n has exactly three factors, then it is the square of a prime number Let’s prove this Let the three factors of n be: 1,□( ) 𝑎,□( ) 𝑛 where □( ) 1<𝑎<𝑛. Factors occur in pairs whose product is 𝑛 Factors 1 and 𝑛 form one pair, i.e. 1 × 𝑛 = 𝑛 Remaining factor 𝑎 must pair with itself, i.e. 𝑎 × 𝑎=𝑛 Therefore: 𝑎 × 𝑎=𝑛 𝒏=𝒂^𝟐 Now, 𝒂 must be prime. If 𝑎 were composite, it would have another factor between 1 and 𝑎. That would also be a factor of 𝑛, giving more than three factors. Hence, 𝑛 is the square of a prime number ∴ Converse is true