Prove that √6 is an irrational number.

To prove that √6 is an irrational number, we'll use a proof by contradiction.

Suppose √6 is a rational number, which means it can be expressed as a fraction in the form a/b, where a and b are integers with no common factors other than 1 and b ≠ 0.

Let's assume that (√6) = a/b, where a/b is in its simplest form.

Squaring both sides of the equation, we get:
6 = (a^2)/(b^2)

Cross-multiplying, we have:
6b^2 = a^2

From this equation, we can see that a^2 must be divisible by 6. This implies that a must also be divisible by 6, since the square of an integer is divisible by that integer.

Let's express a as 6c, where c is an integer. Substituting this into the equation, we have:
6b^2 = (6c)^2
6b^2 = 36c^2
b^2 = 6c^2

Now, we see that b^2 is also divisible by 6, which means b must be divisible by 6.

We have shown that both a and b are divisible by 6, contradicting our assumption that a/b is in its simplest form. Therefore, our initial assumption that √6 is a rational number must be false.

Hence, we conclude that √6 is an irrational number.