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Prove that root 2 is an irrational no |
| Answer» Let √2 is an irrational numberLet √2=p/q (where p and q are integers and q is not equal to 0) -1Let a/b is the lowest form of p/q So,p/q=a/b (where a and b are co- prime numbers)Putting in 1√2=a/bSquarring both side(√2)^2 = (a/b)^22=a^2/b^2a^2 = 5b^2So,2 is the factor of a^2also 2 is the factor of aLet a=2m( where m is some integer) (2m)^2= 2b^24b^2=2b^2 4b^2/2 = b^2 2b^2 = b^2b^2=2b^2 So,2 is the factor of b^2also 2 is the factor of bHere,2 is the factor of both a and b .This contradict our supposition that √2 is an rational number is wrong. Hence ,√2 is an irrational number | |