Question

prove that√7 is irrational​

Answer 1

Answer:

• Lets assume that √7 is rational number. 

ie √7=p/q.

• suppose p/q have common factor then

we 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/b

→ a=√7b

★Squaring

→a²=7b² __________1

★a² is divisible by 7

→a=7c

★substituting values in 1

→ (7c)²=7b²

→ 49c²=7b²

→ 7c²=b²

→ b²=7c²

• b² is divisible by 7
• That 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.

Answer 2

SOLUTION

Let the assume that √7 be rational.

Let the assume that √7 be rational.then it must in the form of p/q {q=0} [p & q are co-prime].

√7= p/q

=) √7× q=p

Squaring both sides

$= > {7q}^{2} = {p}^{2} ..........(1)$

p^2 is divisible by 7

p is divisible by 7.

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

=)p^2 = 49c^2...............(2)

Substitute p^2 in equation (1) we get,

=) 7q^2 = 49c^2

=) q^2 = 7c^2

=) q is divisible by 7.

Thus q & p have a common factor 7.

There is a contradiction as our assumption p& q are co-prime but it has a common factor.

So that √7 is an irrational.

hope it helps ☺️