Edexcel IAL Sample Assessment 2018 P4 (WMA14/01) Q6 Proof by Contradiction
Prove by contradiction that, if \(a, b\) are positive real numbers, then \(a + b > 2\sqrt{ab}\).
(4)
Selected Question: Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume the negation of the statement is true.
- Algebraic Manipulation: Factorising the resulting inequality to reach a contradiction.
- Property of Squares: Any real number squared is non-negative, \((x)^{2}\geq 0\).
Step-by-Step Solution:
- Form the Assumption
Assume the negation of the statement is true. That is, assume \(a+b<2\sqrt{ab}\) for positive real numbers \(a\) and \(b\).
- Rearrange the Inequality
Move the term \(2\sqrt{ab}\) to the left side of the inequality:
\[ a-2\sqrt{ab}+b<0 \] - Factorise the Expression
Recognise that the left-hand side is a perfect square, corresponding to the form \((x-y)^{2}=x^{2}-2xy+y^{2}\).
Here, \(x=\sqrt{a}\) and \(y=\sqrt{b}\):
\[ (\sqrt{a}-\sqrt{b})^{2}<0 \] - Identify the Contradiction
Since \(a\) and \(b\) are real numbers, \((\sqrt{a}-\sqrt{b})\) is also a real number. The square of any real number must be greater than or equal to zero, i.e., \((\sqrt{a}-\sqrt{b})^{2}\geq 0\).
- Conclusion
The statement \((\sqrt{a}-\sqrt{b})^{2}<0\) contradicts the fundamental property that the square of a real number cannot be negative. Therefore, the initial assumption is false, and the original statement is true.
Final Answer
Since the assumption leads to the contradiction \((\sqrt{a}-\sqrt{b})^{2}<0\), the original statement must be true: \(a+b\geq 2\sqrt{ab}\) for all positive real numbers \(a\) and \(b\).