Edexcel IAL June 2022 P4 (WMA14/01) Q9 Proof by Contradiction
Use proof by contradiction to show that, when n is an integer,
\( n^{2} – 2 \)
is never divisible by 4
(4)
Solution to Question: Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume \( n^{2} – 2 = 4k \).
- Integer Cases: Considering the two possible cases for an integer \( n \); \( n \) is even or \( n \) is odd.
- Factorisation: Showing the resulting expression is not a multiple of 4 in either case.
Step-by-Step Solution:
- Form the Assumption
Assume that \( n^{2} – 2 \) is divisible by 4 for some integer \( n \). This means \( n^{2} – 2 = 4k \) for some integer \( k \).
- Rearrange
\( n^{2} = 4k + 2 = 2(2k + 1) \). Since \( n^{2} = 2(\text{odd integer}) \), \( n^{2} \) is even, which implies \( n \) is even. Let \( n = 2m \).
- Case 1: \( n \) is even (Let \( n = 2m \))
Substitute \( n = 2m \) back into the assumption \( n^{2} – 2 = 4k \):
\( (2m)^{2} – 2 = 4k \)
\( 4m^{2} – 2 = 4k \)
\( 2 = 4m^{2} – 4k \)
\( 2 = 4(m^{2} – k) \)This implies 2 is a multiple of 4, which is false. \( 4(m^{2} – k) \) is an integer multiple of 4, so it must be 0,4,8, …, none of which equal 2. This is a contradiction.
- Case 2: \( n \) is odd (Let \( n = 2m + 1 \))
Substitute \( n = 2m + 1 \) back into the assumption \( n^2 – 2 = 4k \):
\( (2m + 1)^2 – 2 = 4k \)
\( 4m^2 + 4m + 1 – 2 = 4k \)
\( 4m^2 + 4m – 1 = 4k \)
\( -1 = 4k – 4m^2 – 4m \)
\( -1 = 4(k – m^2 – m) \)This implies \(-1\) is a multiple of 4, which is false. This is also a contradiction.
Final Answer
Since the assumption leads to a contradiction regardless of whether \( n \) is even or odd, the original statement is true: \( n^2 – 2 \) is never divisible by 4.