Question

To prove that root7 is irrational

Answer 1

let us assume that √7 be rational.

then it must in the form of p / q  [q ≠ 0] [p and q are co-prime]

√7 = p / q

=> √7 x q = p

squaring on both sides

=> 7q2= p2  ------> (1)

p2 is divisible by 7

p is divisible by 7

p = 7c  [c is a positive integer] [squaring on both sides ]

p2 = 49 c2 --------- > (2)

subsitute p2 in equ (1) we get

7q2 = 49 c2

q2 = 7c2

=> q is divisble by 7

thus q and p have a common factor 7.

there is a contradiction

as our assumsion p & q are co prime but it has a common factor.

so that √7 is an irrational.

Answer 2

Lets assume that √7 is rational number. ie √7=p/q.suppose p/q have common factor thenwe divide by the common factor to get √7 = a/b were a and b are co-prime number.that is a and b have no common factor.√7 =a/b co- prime number√7= a/ba=√7bsquaringa²=7b² .......1a² is divisible by 7a=7csubstituting values in 1(7c)²=7b²49c²=7b²7c²=b²b²=7c²b² is divisible by 7that is a and b have atleast one common factor 7. This is contridite to the fact that a and b have no common factor.This is happen because of our wrong assumption.√7 is irrational number.