Prove that √5 is an irrational number.

To show that √5 is an irrational number, we can use a proof by contradiction. This method involves assuming the opposite of what we want to prove and then demonstrating that this assumption leads to a contradiction.

Assumption

Let’s assume that √5 is a rational number. By definition, a rational number can be expressed as a fraction of two integers, say a/b, where a and b are integers with no common factors (other than 1) and b is not zero. Thus, we can write:

√5 = a/b

Squaring Both Sides

Next, we square both sides of the equation:

5 = a²/b²

This implies:

a² = 5b²

Analyzing the Equation

The equation a² = 5b² indicates that a² is a multiple of 5. This means that a must also be a multiple of 5 (since the square of a number is divisible by a prime if and only if the number itself is divisible by that prime).

Letting a = 5k

We can express a as:

a = 5k for some integer k. Now, substituting this back into our equation gives:

(5k)² = 5b²

Which simplifies to:

25k² = 5b²

Dividing both sides by 5 results in:

5k² = b²

Conclusion from the Analysis

This means that b² is also a multiple of 5, which implies that b must also be a multiple of 5. Now we have shown that both a and b are multiples of 5, which contradicts our initial assumption that a and b have no common factors other than 1.

Final Statement

Since our assumption that √5 is rational leads to a contradiction, we conclude that √5 cannot be expressed as a fraction of two integers. Therefore, √5 is an irrational number.