Prove that 3 + 2√5 is irrational.

To show that the number \(3 + 2\sqrt{5}\) is irrational, we can start by assuming the opposite: that it is rational. A rational number can be expressed as the fraction of two integers, say \(a/b\), where \(a\) and \(b\) are integers and \(b \neq 0\).

Assumption of Rationality

Let’s assume:

• Let \(x = 3 + 2\sqrt{5}\) be rational.
• Then we can write \(x = \frac{a}{b}\) for some integers \(a\) and \(b\).

Isolating the Square Root

Next, we can isolate the square root term:

• Subtract 3 from both sides: \(x - 3 = 2\sqrt{5}\).
• Now divide by 2: \(\frac{x - 3}{2} = \sqrt{5}\).

Squaring Both Sides

To eliminate the square root, we square both sides:

• \(\left(\frac{x - 3}{2}\right)^2 = 5\).
• This simplifies to \((x - 3)^2 = 20\).

Expanding the Equation

Expanding the left side gives:

• \(x^2 - 6x + 9 = 20\).
• Rearranging leads to \(x^2 - 6x - 11 = 0\).

Analyzing the Roots

Now, we can use the quadratic formula to find the roots of this equation:

• The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Here, \(a = 1\), \(b = -6\), and \(c = -11\).

Calculating the Discriminant

The discriminant (\(D\)) is calculated as follows:

• \(D = (-6)^2 - 4 \cdot 1 \cdot (-11) = 36 + 44 = 80\).

Determining Rationality

Since the discriminant \(D = 80\) is not a perfect square, the roots of the equation are irrational. Therefore, \(x\) cannot be expressed as a fraction of two integers.

Conclusion

This contradiction shows that our initial assumption that \(3 + 2\sqrt{5}\) is rational must be false. Thus, we conclude that:

3 + 2√5 is indeed irrational.