Edexcel IAL October 2021 P4 (WMA14/01) Q10 Proof by Contradiction
(a) A student’s attempt to answer the question
“Prove by contradiction that if \( n^3 \) is even, then \( n \) is even”
is shown below. Line 5 of the proof is missing.
Assume that there exists a number \( n \) such that \( n^3 \) is even, but \( n \) is odd.
If \( n \) is odd then \( n = 2p + 1 \) where \( p \in \mathbb{Z} \)
If \( n \) is odd then \( n = 2p + 1 \) where \( p \in \mathbb{Z} \)
So \( n^3 = (2p + 1)^3 \)
\( = 8p^3 + 12p^2 + 6p + 1 \)
\( = \)
This contradicts our initial assumption, so if \( n^3 \) is even, then \( n \) is even.
Complete this proof by filling in line 5.
(1)
(b) Hence, prove by contradiction that \( \sqrt[8]{2} \) is irrational.
(3)
(3)
Solution to Part (a): Complete the Proof
Key Concepts Used:
- Algebraic Definition of Odd/Even: An odd integer \( n \) is \( 2p + 1 \).
- Factorisation: Demonstrating oddness by factorising to \( 2(\text{integer}) + 1 \).
Step-by-Step Solution:
- The proof assumes \( n \) is odd, so \( n = 2p + 1 \).
- \( n^3 \) is expanded to \( 8p^3 + 12p^2 + 6p + 1 \).
- The term \( 2(4p^3 + 6p^2 + 3p) \) is obtained by factorising the first three terms.
- The final step (Line 5) must state the resulting conclusion about \( n^3 \). Since \( 4p^3 + 6p^2 + 3p \) is an integer, \( n^3 \) is of the form \( 2(\text{integer}) + 1 \).
Final Answer
Line 5 should be:
\( n^3 = 2(4p^3 + 6p^2 + 3p) + 1 \)
which is odd.
\( n^3 = 2(4p^3 + 6p^2 + 3p) + 1 \)
which is odd.
Solution to Part (b): Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume \(\sqrt[8]{2}\) is rational, i.e., \(\sqrt[8]{2}=\frac{a}{b}\).
- Definition of Rational Numbers: \(a\) and \(b\) are integers with no common factors (coprime).
- Integer Property of Cubes: Using the result from part (a) (if \(n^{3}\) is even, then \(n\) is even).
- Quadratic Discriminant: Use of \(b^{2}-4ac\) to show that the resulting quadratic equation has no real solutions for \(k\).
Step-by-Step Solution:
- Form the Assumption
Assume \(\sqrt[8]{2}\) is a rational number. This means \(\sqrt[8]{2}=\frac{a}{b}\), where \(a\) and \(b\) are positive integers with no common factors other than 1.
- Cube and Rearrange
Cube both sides and rearrange the equation:
\[ (\sqrt[8]{2})^{3} = \left(\frac{a}{b}\right)^{3} \implies 2 = \frac{a^{3}}{b^{3}} \]
\[ 2b^{3} = a^{3} \] - Deduce Property of \(a\)
The equation \(a^{3} = 2b^{3}\) shows that \(a^{3}\) is a multiple of 2, meaning \(a^{3}\) is even. Using the result from part (a), if \(a^{3}\) is even, then \(a\) must also be even.
- Substitute \(a\) and Deduce Property of \(b\)
Since \(a\) is even, let \(a = 2k\) for some integer \(k\). Substitute this back into the rearranged equation \(2b^{3} = a^{3}\):
\[ 2b^{3} = (2k)^{3} = 8k^{3} \]
Divide by 2:
\[ b^{3} = 4k^{3} \implies b^{3} = 2(2k^{3}) \] - Identify the Contradiction
The equation \(b^{3} = 2(2k^{3})\) shows \(b^{3}\) is a multiple of 2, meaning \(b^{3}\) is even. Therefore, \(b\) is also even. Since \(a\) is even (\(a = 2k\)) and \(b\) is even, they share a common factor of 2. This contradicts the initial assumption that \(a\) and \(b\) have no common factors (other than 1).
Final Answer
Since the assumption that \(\sqrt[8]{2}\) is rational leads to the contradiction that \(a\) and \(b\) have a common factor of 2, the assumption is false. Therefore, \(\sqrt[8]{2}\) is irrational.