Question 2 - End-of-Chapter Exercises - Chapter 3 Class 9 - The World of Numbers (Ganita Manjari I)

Transcript

Question 2 Prove that √5 is an irrational number. We have to prove √5 is irrational Let us assume the opposite, i.e., √𝟓 is rational Hence, √5 can be written in the form 𝑎/𝑏 where a and b (b≠ 0) are co-prime (no common factor other than 1) Hence, √𝟓 = 𝒂/𝒃 √5 b = a Squaring both sides (√5b)2 = a2 5b2 = a2 𝒂^𝟐/𝟓 = b2 Hence, 5 divides a2 So, 5 shall divide a also Hence, we can say 𝑎/5 = c where c is some integer So, a = 5c By theorem: If p is a prime number, and p divides a2, then p divides a , where a is a positive number Now we know that 5b2 = a2 Putting a = 5c 5b2 = (5c)2 5b2 = 25c2 b2 = 1/5 × 25c2 b2 = 5c2 𝒃^𝟐/𝟓 = c2 Hence, 5 divides b2 So, 5 divides b also By theorem: If p is a prime number, and p divides a2, then p divides a , where a is a positive number By (1) and (2) 5 divides both a & b Hence, 5 is a factor of a and b So, a & b have a factor 5 Therefore, a & b are not co-prime. Hence, our assumption is wrong ∴ By contradiction, √𝟓 is irrational