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| 1. |
show that _/3 is irrational number. |
| Answer» We have to prove √3 is irrational Let us assume the opposite, i.e., √3 is rational Hence, √3 can be written in the form ?/? where a and b (b≠ 0) are co-prime (no common factor other than 1) Hence, √3 = ?/? √3 b = a Squaring both sides (√3b)2 = a2 3b2 = a2 ?^2/3 = b2 Hence, 3 divides a2 So, 3 shall divide a also Hence, we can say ?/3 = c where c is some integer So, a = 3c Now we know that 3b2 = a2 Putting a = 3c 3b2 = (3c)2 3b2 = 9c2 b2 = 1/3 × 9c2 b2 = 3c2 ?^2/3 = c2 Hence 3 divides b2 So, 3 divides b also By (1) and (2) 3 divides both a & b Hence 3 is a factor of a and b So, a & b have a factor 3 Therefore, a & b are not co-prime. Hence, our assumption is wrong ∴ By contradiction, √3 is irrational | |