Edexcel IAL October 2021 P4 (WMA14/01) Q10 Proof by Contradiction
Prove by contradiction that there is no greatest odd integer.
(2)
Solution to Question: Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume a greatest odd integer exists.
- Algebraic Definition of Odd Integer: Representing the integer as \(2k + 1\).
- Closure Property of Integers: Showing that adding 2 to an odd integer creates a larger odd integer.
Step-by-Step Solution:
- Form the Assumption
Assume that there exists a greatest odd integer. Let this integer be \(n\).
- Define \(n\)
Since \(n\) is the greatest odd integer, it can be written as \(n = 2k + 1\), where \(k\) is some integer (\(k \in \mathbb{Z}\)).
- Construct a larger odd integer
Consider the integer \(n + 2\):
\[ n + 2 = (2k + 1) + 2 = 2k + 3 \] - Show the new integer is odd and greater than \(n\)
The integer \(n + 2\) can be written as \(2(k + 1) + 1\). Since \((k + 1)\) is an integer, \(n + 2\) is an odd integer. Since 2 is a positive number,
\[ n + 2 > n \] - Identify the Contradiction
The existence of an odd integer \(n + 2\) that is greater than \(n\) contradicts the initial assumption that \(n\) was the greatest odd integer.
Final Answer
Since the assumption leads to a contradiction, it must be false. Therefore, there is no greatest odd integer.