prove that √3 is an irrational number. hence show that 7+2√3 is also an irrational number
Answers
Answer:Let us assume that √3 is a rational number. That is, we can find integers a and b (≠ 0) such that √3 = (a/b) Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. √3b = a ⇒ 3b2=a2 (Squaring on both sides) → (1) Therefore, a2 is divisible by 3 Hence ‘a’ is also divisible by 3. So, we can write a = 3c for some integer c. Equation (1) becomes, 3b2 =(3c)2 ⇒ 3b2 = 9c2 ∴ b2 = 3c2 This means that b2 is divisible by 3, and so b is also divisible by 3. Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that √3 is rational. So, we conclude that √3 is irrational. Now √2 = 1.4142... √3 = 1.7321... 1.45, 1.5, 1.55, 1.6, 1.65, 1.7 lies between √2 and √3. Hence the rational numbers between √2 and √3 are: 145/100, 15/10, 155/100, 16/10, 165/100 and 17/100