Prove that √3 is an Irrational Number

Answer: Assume, on the other hand, that 3 is a rational number.

The symbol x/y can denote it.

Here x and y are co-primes and y≠ 0.

⇒ √3 = x/y

⇒ 3 = x²/y² 

⇒ 3y² = x²……..(1)

Now, we have x = 3s

Where s is some integer.

⇒ x² = 9s²…..(2)

Now, from equations (1) and (2)

⇒ 3y² = 9s²

⇒ y² = 3s²

Now, we have two cases to solve further.

Case1: 

Assume that s is an even number. Then s2 is even, and 3s2 is even, implying that y2 is even, and thus y is even, which is impossible to do. If both y and s are even, then gcd(y,s)>=2 contradicts itself.

Case2: 

Assume that s is out of the ordinary. Then s2 and 3s2 are odd, implying that y2 is odd and q is odd. Because y and s are both odd, we can write

y = 2m−1 and a = 2n−1 for some m,n∈N.

y² = 3s²

(2m−1)² = 3(2n−1²

4m²−4m+1 = 3(4n²−4n+1)

4m²−4m+1 = 12n²−12n+3

4m²−4m = 12n²−12n+2

2m²−2m = 6n²−6n+1

2(m²−m) = 2(3n²−3n)+1

We can see that the left-hand side of this equation is even, yet the right-hand side is odd, indicating a contradiction. As a result, there is no rational number r for which r² = 3.

As a result, √ 3 is an irrational number.