Suppose we consider, √6 is a rational number .

Then we can express it in the form of a/b

∴√6 = a/b, where a and b are positive integer and they are co-prime, i.e.HCF(a,b) = 1

∴√6 = a/b

=>b√6 = a

=>(b√6)² = a² [squaring both sides]

=>6b² = a²………..(1)

Here, a² is divided by 6

∴ a is also divided by 6. [we know that if p divides a², then p divides a]

∴ 6|a

=>a = 6c [c∈ℤ]

=>a² = (6c)²

=>6b² = 36c² [from (1)]

=>b² = 6c²

Here, b² is divided by 6,

∴ b is also divided by 6.

∴ 6|a and 6|b

we observe that a and b have at least 6 as a common factor. But this contradicts that “a and b are co-prime.”

It means that our consideration of “√6 is a rational number” is not true.

Hence, √6 is an irrational number.

I would use the proof by contradiction method for this.

So let's assume that the square root of 6 is rational.

By definition, that means there are two integers a and b with no common divisors where:

a/b = square root of 6.

So let's multiply both sides by themselves:

(a/b)(a/b) = (square root of 6)(square root of 6)a2/b2= 6a2= 6b2

But this last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least) is even. So a2must be even. But any odd number times itself is odd, so if a2is even, then a is even.

Since a is even, there is some integer c that is half of a, or in other words:

2c = a.

Now let's replace a with 2c:

a2= 6b2(2c)2= (2)(3)b22c2= 3b2

But now we can argue the same thing for b, because the LHS is even, so the RHS must be even and that means b is even.

Now this is the contradiction: if a is even and b is even, then they have a common divisor (2). Then our initial assumption must be false, so the square root of 6 cannot be rational.