Prove that √2 is an irrational number

Prove that √2 

The number $\sqrt{2}$ is irrational. The proof can be as follows:

Let's assume that $\sqrt{2}$ is rational. Then we have:

$\sqrt{2}=\frac{p}{q}$ where $p,q\in Z$

If we raise this equation to the second power we get:

$2=\frac{p^2}{q^2}$

$2q^2=p^2$

But this equality is not possible for integer $p$ and $q$. If we analyze the prime factorizarion of both sides we can see that number $2$ appears even number of times on the right hand side and odd number of times on the left side.

This is a contradiction which leads to the conclusion that the assumption is false, so $\sqrt{2}$ is an irrational number .

                                  OR

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 = a^2/b^2

2b^2 = a^2 ….. Eq.1

Here, 2 divides a^2 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 )…

a^2 = 4c^2…Eq.2

Substituting Eq. 1 in Eq. 2,

2 b^2 = 4c^2

b^2 = 2c^2

2c^2 = b^2

2 divides b^2 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.
For more explanation watch it
https://youtu.be/he-Zfpe87aw

Regards

Shikha kaushal