Question

show that root 7 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

We will prove whether √7 is irrational by contradiction method. Let √7 be rational It can be expressed as √7 = a/b ( where a, b are integers and co-primes.
√7=a/b

7= a²/b² 
7b² = a²
7divides a²
By the Fundamental theorem of Arithmetic 
so, 7 divides a .

a = 7k (for some integer) 

a² = 49k² 
7b² = 49k² 
b² = 7k² 

7divides b²
7 divides b. 

Now 7 divides both a & b this contradicts the fact that they are co primes. 
this happened due to faulty assumption that √7 is rational. Hence, √7 is irrational. 

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