Prove that √2√5 is irrational

To prove that √2√5 is irrational, we can start by simplifying the expression. The product √2√5 can be rewritten as √(2 * 5), which equals √10. Thus, we need to show that √10 is irrational.

An irrational number is defined as a number that cannot be expressed as a fraction of two integers. To prove that √10 is irrational, we can use proof by contradiction.

Assume, for the sake of contradiction, that √10 is rational. This means that we can express √10 as the fraction a/b, where a and b are integers with no common factors (other than 1), and b ≠ 0:

√10 = a/b

Squaring both sides gives us:

10 = a²/b²

This implies:

a² = 10b²

Since the right side of this equation (10b²) is obviously even (because it’s 10 multiplied by any integer), we can conclude that a² must also be even. And if a² is even, then a must also be even (because the square of an odd number is odd).

Let’s say a = 2k for some integer k. Substituting this back into our equation gives us:

(2k)² = 10b²

Which simplifies to:

4k² = 10b²

Dividing both sides by 2 results in:

2k² = 5b²

This means that 5b² must also be even, which implies that b² is even. Consequently, b must also be even. Thus, both a and b are even, which contradicts our original assumption that a and b have no common factors other than 1.

Since our assumption that √10 is rational leads to a contradiction, we conclude that √10 is indeed irrational. Therefore, √2√5 is also irrational.