Question

Prove that √2 is irrational no.​

Answer 1

Answer:-

Lets first assume √2 to be rational.

√2 = a/b, where b is not equal to 0.

Here, a and b are co-primes whose HCF is 1.

√2 = a/b ( squaring both sides )… 2 = a²/b²

2b²= a² ….. Eq.1

Here, 2 divides a² also a ( bcz, If a prime number divides the square of a positive integer, then it divides the integer itself )

Now, let a = 2c ( squaring both sides )

a²= 4c².........…Eq.2

Substituting Eq. 1 in Eq. 2,

2 b² = 4c²

b²= 2c²

2c²= b²

2 divides b² as well as b.

Conclusion:

Here, a & b are divisible by 2 also. But our assumption that their HCF is 1 is being contradicted.

Therefore, our assumption that √2 is rational is wrong. Thus, it is irrational.

Answer 2

Answer:

• To prove that the square root of 2 is irrational is to first assume that its negation is true. Therefore, we assume that the opposite is true, that is, the square root of 2 is rational. ... So, 2 = a b \sqrt 2 = {\Large{{a \over b}}} 2 =ba where a and b are integers but b ≠ 0 b \ne 0 b=0.