Prove by contradiction that \( \sqrt{2} \) is an irrational number. - AQA - A-Level Maths Pure - Question 10 - 2018 - Paper 3
Question 10
Prove by contradiction that \( \sqrt{2} \) is an irrational number.
Worked Solution & Example Answer:Prove by contradiction that \( \sqrt{2} \) is an irrational number. - AQA - A-Level Maths Pure - Question 10 - 2018 - Paper 3
Step 1
Assume that \( \sqrt{2} \) is rational
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Answer
We start by assuming that ( \sqrt{2} ) is a rational number. Thus, we can express it as ( \sqrt{2} = \frac{a}{b} ), where ( a ) and ( b ) are integers with no common factors, and ( b \neq 0 ).
Step 2
Manipulate the equation
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Answer
Squaring both sides, we have:
2=ba⇒2=b2a2⇒2b2=a2.
Step 3
Deduce that \( a \) is even
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Answer
From ( 2b^2 = a^2 ), we can deduce that ( a^2 ) is even since it is equal to ( 2b^2 ), which is clearly even. Consequently, ( a ) must also be even.
Step 4
Express \( a \) as even
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Answer
Since ( a ) is even, we can write ( a = 2k ) for some integer ( k ).
Step 5
Substitute back into the equation
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Answer
Substituting ( a = 2k ) back into the equation gives:
2b2=(2k)2=4k2⇒b2=2k2.
Step 6
Deduce that \( b \) is even
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Answer
From ( b^2 = 2k^2 ), we can conclude that ( b^2 ) is even, which means ( b ) is also even.
Step 7
Explain the contradiction
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Answer
Since both ( a ) and ( b ) are even, this implies that they have a common factor of 2, contradicting our initial assumption that ( a ) and ( b ) have no common factors.
Step 8
Conclude that \( \sqrt{2} \) is irrational
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Answer
Thus, we accept that the assumption that ( \sqrt{2} ) is rational must be incorrect. Therefore, we conclude that ( \sqrt{2} ) is an irrational number.
