Introduction
To prove that the sum of the square roots of 2 and 3 is irrational, we will utilize a proof by contradiction.
Assumption
- Assume that √2 + √3 is rational.
- This means it can be expressed as a fraction: a/b, where a and b are integers, and b ≠ 0.
Isolate One Root
- Rearranging gives us √2 = (a/b) - √3.
- Squaring both sides results in: 2 = (a/b)² - 2(a/b)√3 + 3.
Reorganize the Equation
- This simplifies to: 2 = (a² + 3b²) / b² - 2a√3 / b.
- Rearranging leads to: 2a√3 / b = (a² + 3b²) / b² - 2.
Isolate the Square Root
- Now, we have: √3 = [b(2 - (a² + 3b²) / b²)] / (2a).
- This implies √3 is expressed as a fraction involving integers, indicating that √3 is rational.
Contradiction
- However, it is a well-established fact that √3 is irrational.
- Therefore, our initial assumption that √2 + √3 is rational must be false.
Conclusion
- Hence, we conclude that √2 + √3 is indeed irrational.