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Prove Root 2 +root 3 whole square is irrational |
| Answer» (√2+√3)^2=2+3+2√5 =5+2√5Let us first prove that √5 is an irrational,Let us assume that √5 is a rational numberSo, √5=p/q,where q≠0,HCF(p,q)=0 (√5)^2=(p/q)^2,squaring on both the sides 5=p^2/q^2 5q^2=p^2 ......eqn 1q^2 is a multiple of 5;p^2 is also a multiple of 5;Therefore,even p is a multiple of 5;Let p=5r where r is a positive integer, 5q^2=(5r)^2......by eqn 1 5q^2=25r^2 q^2=5r^2r^2 is a multiple of 5q^2 is also a multiple of 5Therefore,even q is a multiple of 5 By this we come to know that the HCF is not 1.This contradiction arised due to our wrong assumption that √5 is a rational number.Therefore,√5 is an irrational number.Now,let us assume that (5+2√5) is a rational number, So, (5+2√5)=a/b,where b≠0,HCF(a,b)=1 2√5=a-5b/b √5=a-5b/2b Irrational≠integer/integer=rationalThis contradiction arised due to our wrong assumption that (5+2√5) is a rational number.Therefore,(5+2√5) is a irrational number.i.e;(√2+√3)^2 is a irrational number .Hence,proved | |