Edexcel IAL October 2023 P4 (WMA14/01) Q4 Proof by Contradiction
a. Prove by contradiction that for all positive numbers \( k \),
\( k + \frac{9}{k} \geq 6 \)
b. Show that the result in part (a) is not true for all real numbers.
(4)
Solution to Part (a): Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume the negation, \( k + \frac{9}{k} < 6 \).
- Algebraic Manipulation: Multiplying by \( k \) (since \( k > 0 \)).
- Property of Squares: Showing the resulting inequality contradicts the property that a squared real number is non-negative.
Step-by-Step Solution:
- Form the Assumption
Assume the negation of the statement is true. That is, assume \( k + \frac{9}{k} < 6 \) for all positive numbers \( k \).
- Rearrange the Inequality
Since \( k \) is a positive number (\( k > 0 \)), we can multiply the inequality by \( k \) without reversing the inequality sign:
\( k^2 + 9 < 6k \) - Form a Quadratic Inequality
Move \( 6k \) to the left side:
\( k^2 – 6k + 9 < 0 \) - Factorise the Expression
The left-hand side is a perfect square:
\( (k – 3)(k – 3) < 0 \)
\( (k – 3)^2 < 0 \) - Identify the Contradiction
The statement \( (k – 3)^2 < 0 \) implies that the square of a real number is negative. This contradicts the fundamental property that the square of any real number must be non-negative, i.e., \( (k – 3)^2 \geq 0 \).
Final Answer
Since the assumption leads to the contradiction \( (k – 3)^2 < 0 \), the original statement must be true: \( k + \frac{9}{k} \geq 6 \) for all positive numbers \( k \).
Solution to Part (b): Proof by Contradiction
Key Concepts Used:
- Counter-Example: Finding a specific value of \( k \) (a non-positive real number) that violates the inequality.
Step-by-Step Solution:
- Choose a counter-example
The proof in part (a) relied on \( k > 0 \). Choose a negative real number, for instance, \( k = -10 \).
- Evaluate the expression
Substitute \( k = -10 \) into \( k + \frac{9}{k} \):
\( -10 + \frac{9}{-10} = -10 – 0.9 = -10.9 \) - Compare with the statement
The result \( -10.9 \) is clearly less than 6.
Final Answer
If \( k = -10 \), then \( k + \frac{9}{k} = -10.9 \). Since \( -10.9 < 6 \), the inequality \( k + \frac{9}{k} \geq 6 \) is not true for all real numbers.