Edexcel IAL June 2023 P4 (WMA14/01) Q7 Proof by Contradiction
Use proof by contradiction to prove that \( \sqrt{7} \) is irrational.
(You may assume that if \( k \) is an integer and \( k^2 \) is a multiple of 7 then \( k \) is a multiple of 7)
(4)
Solution to Question: Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume \( \sqrt{7} = \frac{a}{b} \) where \( a, b \) are coprime integers.
- Definition of Rational Numbers: Any number that can be expressed as a fraction \( a/b \), where \( a \) and \( b \) are integers (\( b \neq 0 \)).
- Integer Property of Squares: Using the given assumption that if \( k^2 \) is a multiple of 7, then \( k \) is a multiple of 7.
Step-by-Step Solution:
- Form the Assumption
Assume \( \sqrt{7} \) is a rational number. This means \( \sqrt{7} = \frac{a}{b} \), where \( a \) and \( b \) are positive integers with no common factors (coprime).
- Square and Rearrange
Square both sides and rearrange the equation:
\( (\sqrt{7})^2 = \left(\frac{a}{b}\right)^2 \implies 7 = \frac{a^2}{b^2} \)
\( 7b^2 = a^2 \) - Deduce Property of \( a \)
The equation \( a^2 = 7b^2 \) shows that \( a^2 \) is a multiple of 7. Using the given assumption, if \( a^2 \) is a multiple of 7, then \( a \) must also be a multiple of 7.
- Substitute \( a \) and Deduce Property of \( b \)
Since \( a \) is a multiple of 7, let \( a = 7k \) for some integer \( k \). Substitute this back into the rearranged equation \( 7b^2 = a^2 \):
\( 7b^2 = (7k)^2 = 49k^2 \)
Divide by 7:
\( b^2 = 7k^2 \) - Identify the Contradiction
The equation \( b^2 = 7k^2 \) shows \( b^2 \) is a multiple of 7. Therefore, \( b \) is also a multiple of 7. Since \( a \) is a multiple of 7 and \( b \) is a multiple of 7, they share a common factor of 7. This contradicts the initial assumption that \( a \) and \( b \) have no common factors (other than 1).
Final Answer
Since the assumption that \( \sqrt{7} \) is rational leads to the contradiction that \( a \) and \( b \) share a common factor of 7, the assumption is false. Therefore, \( \sqrt{7} \) is irrational.