Question

prove that √7 is an irrational number by condradiction method

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

assume root 7 is rational...
it can written as:
root 7 = a/b where a and b are lowest terms a nd b not equal to 0 
if a and b are in lowest terms that means they do not share any common factors.
our equation above can be written as 7= a^2/b^2 by squaring on both sides..
= 7b square=a square...
now multiply a square on both sides..
since both sides are equal , if 7 divides LHS it must also divide RHS..
so we can write 'a' as 7k,where k is some integer..k not equal to 0
=7b square = ( 7 k square)
= 7b square = 49k square
= b square = 7k square..
from same logic now b is also divided by 7 
thus we know that a nd b are divisible by 7 , which they do not share common factors of 7
we know that a nad b are in lowest terms..they do not share any common factor..
therefor we have a contradiction which means our assumption ,
since root 7 is not rational , so we conclude that root 7 is irrational..

hope this helps u=)