Example 3 - Chapter 9 Class 9 - Propositions and their Converses (Ganita Manjari)

Transcript

Example 3 In this example, n is any positive integer. Proposition P: If n is a perfect square, then it has an odd number of factors. Converse Q: If n has an odd number of factors, then it is a perfect square. You may recall that we came across these statements in the previous grade. Which of them are true? Let’s check if both statements are true or not Checking if Proposition P is true P: If n is a perfect square, then it has an odd number of factors. Perfect squares are numbers like 12, 22, 32, 42, 52, 62, 72, …. 1, 4, 9, 16, 25, 36, 49, … Finding factors of these perfect squares Factors of 16 1 × 16 = 16 2 × 8 = 16 4 × 4 = 16 Thus, factors of 16 are 1, 2, 4, 8, 16 i.e. 5 factors It has odd factors because the number 4 is multiplied by itself Every pair contributes two distinct factors except (4, 4) which contributes only one. Similarly, factors of 25 are 1, 5, 25 (3 factors) And factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36 (9 factors) Generally Generally, if 𝑛=𝑘^2, then 𝑘 pairs with itself. Every other factor has a different partner. Hence: " Number of factors "=𝟐𝒎+𝟏 which is odd. Thus, every perfect square has an odd number of factors ∴ Proposition P is true Checking if Converse Q is true Q: If n has an odd number of factors, then it is a perfect square. If every factor had a different partner, the total number of factors would be even. An odd total therefore requires a factor that pairs with itself. If that factor is 𝑓, then: 𝒏=𝒇 × 𝒇=𝒇^𝟐 Thus, 𝒏 is a perfect square. ∴ Converse Q is true