Hi faizulla

A classic proof in mathematics!

To prove that √2 is an irrational number, we'll use a simple yet elegant proof by contradiction.

Assumption:

Let's assume that √2 is a rational number. This means it can be expressed as a fraction of two integers, a and b, where b is non-zero.

√2 = a/b

Squaring Both Sides:

Squaring both sides of the equation gives:

2 = a^2/b^2

Multiplying Both Sides by b^2:

Multiplying both sides by b^2 gives:

2b^2 = a^2

Even and Odd Numbers:

Now, we know that:

- If a number is even, its square is also even.

- If a number is odd, its square is also odd.

Since 2b^2 is even (2 is even), a^2 must also be even. This implies that a is even.

Expressing a as 2k:

Since a is even, we can express it as:

a = 2k

Substituting a = 2k:

Substituting a = 2k into the equation 2b^2 = a^2 gives:

2b^2 = (2k)^2

Simplifying:

Simplifying the equation gives:

2b^2 = 4k^2

Dividing Both Sides by 2:

Dividing both sides by 2 gives:

b^2 = 2k^2

Contradiction:

Now, we have:

b^2 = 2k^2

This implies that b^2 is even, which means b is also even.

Conclusion:

We started with the assumption that √2 is rational, which led us to the conclusion that both a and b are even. However, this creates a contradiction, as we can always divide both a and b by 2, resulting in a smaller fraction that still equals √2.

This contradiction proves that our initial assumption - √2 is rational - is false. Therefore, √2 is an irrational number.

Thanks.