Question

prove that √7 is irrational​

Answer 1

Answer:

The number which is does not have any square root is irrational

Step-by-step explanation:

hope it helps!!

Answer 2

Step-by-step explanation:

Given:-

√7

To find:-

Prove that √7 is irrational

Solution:-

Given number =√7

Let assume that √ is a rational number.

It must be in the form of p/q ,where p and q are integers and q≠0

Let √7 = a/b ( where a and b are co-primes)

On squaring both sides then

=> (√7)^2 = (a/b)^2

=> 7 = a^2/b^2

=> 7b^2 = a^2--------(1)

=> b^2 = a^2/7

=> 7 divides a^2

=>7 divides a also

(Since if p is a prime and a positive integer then if p divides a^2 then p divides a also)

=> a is a multiple of 7

=> 7 is a factor of a ------(2)

Put a = 7c in (1) then

=> 7 b^2 = (7c)^2

=> 7 b^2 = 49c^2

=>b^2 = 49c^2/7

=>b^2 = 7c^2

=> c^2 = b^2/7

=> 7 divides b^2

=> 7 divides b also

(Since if p is a prime and a positive integer then if p divides a^2 then p divides a also)

=> b is a multiple of 7

=>7 is a factor of b ------(3)

from (2)&(3)

7 is a common factor of a and b

But a and b are co-primes which have only one factor .i.e.1

This contradicts to our assumption.

=>√7 is not a rational number.

√7 is an irrational number.

Hence , Proved.

Used formulae:-

• if p is a prime and a positive integer then if p divides a^2 then p divides a also

Used Method:-

• Method of Contradiction (Indirect method)