Prove that 2 + √5 is an irrational number.

To show that \(2 + \sqrt{5}\) is an irrational number, we can use a proof by contradiction. Let's assume the opposite: that \(2 + \sqrt{5}\) is a rational number.

Definition of Rational Numbers

A rational number can be expressed as the fraction of two integers, where the denominator is not zero. So, we can write:

2 + √5 = a/b

where \(a\) and \(b\) are integers and \(b \neq 0\).

Isolating the Square Root

Next, we can rearrange this equation to isolate \(\sqrt{5}\):

√5 = (a/b) - 2

Now, we can express \(2\) as a fraction:

√5 = (a - 2b)/b

Analyzing the Expression

The right side of the equation, \((a - 2b)/b\), is a fraction of two integers, which means it is rational. Therefore, we have:

√5 is rational.

Contradiction with Known Facts

However, it is a well-established fact that \(\sqrt{5}\) is an irrational number. This creates a contradiction because we cannot have both \(\sqrt{5}\) being rational and irrational at the same time.

Final Thoughts

Since our initial assumption that \(2 + \sqrt{5}\) is rational leads to a contradiction, we conclude that:

2 + √5 is indeed an irrational number.