Edexcel IAL January 2023 P4 (WMA14/01) Q9 Proof by Contradiction
A student was asked to prove, for \( p \in \mathbb{N} \), that
“If \( p^3 \) is a multiple of 3, then \( p \) must be a multiple of 3”
The start of the student’s proof by contradiction is shown in the box below
Assumption:
There exists a number \( p, p \in \mathbb{N} \), such that \( p^3 \) is a multiple of 3, and \( p \) is NOT a multiple of 3
Let \( p = 3k + 1, k \in \mathbb{N} \).
Consider \( p^3 = (3k + 1)^3 = 27k^3 + 27k^2 + 9k + 1 \)
\( = 3(9k^3 + 9k^2 + 3k) + 1 \) which is not a multiple of 3
There exists a number \( p, p \in \mathbb{N} \), such that \( p^3 \) is a multiple of 3, and \( p \) is NOT a multiple of 3
Let \( p = 3k + 1, k \in \mathbb{N} \).
Consider \( p^3 = (3k + 1)^3 = 27k^3 + 27k^2 + 9k + 1 \)
\( = 3(9k^3 + 9k^2 + 3k) + 1 \) which is not a multiple of 3
(a) Show the calculations and statements that are required to complete the proof.
(3)
(3)
(b) Hence prove, by contradiction, that \( \sqrt[3]{3} \) is an irrational number.
(5)
(5)
Solution to Part (a): Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume \( p \) is not a multiple of 3.
- Integer Cases: Show that if \( p \) is not a multiple of 3, it must be of the form \( 3k + 1 \) or \( 3k + 2 \).
Step-by-Step Solution:
- Form the Assumption
Assume \( p^3 \) is a multiple of 3, but \( p \) is not a multiple of 3.
- Case 1: \( p = 3k + 1 \) (Already shown in the box)
\( p^3 = 3(9k^3 + 9k^2 + 3k) + 1 \)
which is not a multiple of 3. - Case 2: \( p = 3k + 2 \) (The required continuation)
\( p^3 = (3k + 2)^3 \)
\( p^3 = (3k + 2)(3k + 2)^2 \)
\( p^3 = (3k + 2)(9k^2 + 6k + 4) \)
\( p^3 = 27k^3 + 36k^2 + 12k + 18k^2 + 24k + 8 \)
\( p^3 = 27k^3 + 54k^2 + 36k + 9 – 1 \) - Factorize to show not a multiple of 3
Factor out 3 from the first three terms, and split \( 8 = 9 – 1 \),
\( p^3 = 27k^3 + 54k^2 + 36k + 9 – 1 \)
\( p^3 = 3(9k^3 + 18k^2 + 12k + 3) – 1 \)
The expression of \( p^3 \) is in the form \( 3(\text{integer}) – 1 \); hence, \( p^3 \) is one less than a multiple of 3 - Identify the Contradiction
Since \( 3(9k^3 + 18k^2 + 12k + 3) – 1 \) is not a multiple of 3, \( p^3 \) is not a multiple of 3. This contradicts the premise that \( p^3 \) is a multiple of 3.
Final Answer
Since both possible cases lead to the contradiction that \( p^3 \) is not a multiple of 3, the assumption is false. Therefore, if \( p^3 \) is a multiple of 3, then \( p \) must be a multiple of 3.
Solution to Part (b): Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume \( \sqrt[3]{3} = \frac{a}{b} \) where \( a, b \) are coprime integers.
- Definition of Rational Numbers: \( a, b \) are integers with no common factors.
- Integer Property of Cubes: Using the result from part (a) (if \( p^3 \) is a multiple of 3, then \( p \) is a multiple of 3).
Step-by-Step Solution:
- Form the Assumption
Assume \( \sqrt[3]{3} \) is a rational number. This means \( \sqrt[3]{3} = \frac{a}{b} \), where \( a \) and \( b \) are positive integers with no common factors (coprime).
- Cube and Rearrange
Cube both sides and rearrange the equation:
\( (\sqrt[3]{3})^3 = \left( \frac{a}{b} \right)^3 \implies 3 = \frac{a^3}{b^3} \)
\( 3b^3 = a^3 \) - Deduce Property of \( a \)
The equation \( a^3 = 3b^3 \) shows that \( a^3 \) is a multiple of 3. Using the result from part (a), if \( a^3 \) is a multiple of 3, then \( a \) must also be a multiple of 3.
- Substitute \( a \) and Deduce Property of \( b \)
Since \( a \) is a multiple of 3, let \( a = 3k \) for some integer \( k \). Substitute this back into the rearranged equation \( 3b^3 = a^3 \):
\( 3b^3 = (3k)^3 = 27k^3 \)
Divide by 3:
\( b^3 = 9k^3 \Rightarrow b^3 = 3(3k^3) \) - Identify the Contradiction
The equation \( b^3 = 3(3k^3) \) shows \( b^3 \) is a multiple of 3. Therefore, \( b \) is also a multiple of 3. Since \( a \) is a multiple of 3 and \( b \) is a multiple of 3, they share a common factor of 3. This contradicts the initial assumption that \( a \) and \( b \) have no common factors (other than 1).
Final Answer
Since the assumption that \( \sqrt[3]{3} \) is rational leads to the contradiction that \( a \) and \( b \) share a common factor of 3, the assumption is false. Therefore, \( \sqrt[3]{3} \) is irrational.