Edexcel IAL June/October 2020 P4 (WMA14/01) Q1 Proof by Contradiction
Given that n is an integer, use algebra, to prove by contradiction, that if \( n^3 \) is even then n is even.
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Solution to Question: Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume \( n \) is odd.
- Algebraic Definition of Odd/Even: An odd integer \( n \) can be expressed as \( n = 2k + 1 \) for some integer \( k \).
- Factorisation: Showing \( n^3 \) can be written as \( 2(\text{integer}) + 1 \), which is the definition of an odd number.
Step-by-Step Solution:
- Form the Assumption
Assume the negation of the conclusion is true. Assume that \( n \) is odd, and \( n^3 \) is even.
- Define \( n \)
Since \( n \) is assumed to be odd, let \( n \) be written as \( n = 2k + 1 \), where \( k \) is an integer (\( k \in \mathbb{Z} \)).
- Cube \( n \)
Calculate \( n^3 \) using the definition of \( n \): \( n^3 = (2k + 1)^3 \).
Expanding it by considering:
\[ (a + b)^3 = (a + b)(a + b)^2 \]
So, \( n^3 \) becomes:
\[ n^3 = (2k + 1)(2k + 1)^2 \]
\[ n^3 = (2k + 1)(4k^2 + 4k + 1) \]
\[ n^3 = 8k^3 + 12k^2 + 6k + 1 \] - Factorise to show oddness
Factor out a 2 from the first three terms:
\[ n^3 = 2(4k^3 + 6k^2 + 3k) + 1 \] - Identify the Contradiction
Since \( k \) is an integer, the expression \( (4k^3 + 6k^2 + 3k) \) is also an integer. Let \( M = 4k^3 + 6k^2 + 3k \). Therefore, \( n^3 = 2M + 1 \), which is the definition of an odd integer. This contradicts the initial condition that \( n^3 \) is even.
Final Answer
Since the assumption that \( n \) is odd leads to the contradiction that \( n^3 \) is odd, the initial statement must be true: if \( n^3 \) is even, then \( n \) is even.