A-Level AQA Maths: Pure 1.2 Proof by Contradiction: Prove by contradiction that \( \

Prove by contradiction that $ \sqrt{2} $ is an irrational number. - AQA - A-Level Maths: Pure - Question 10 - 2018 - Paper 3

Question 10

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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:

\sqrt{2} = \frac{a}{b} \Rightarrow 2 = \frac{a^2}{b^2} \Rightarrow 2b^2 = a^2.

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:

2b^2 = (2k)^2 = 4k^2 \Rightarrow b^2 = 2k^2.

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.

Prove by contradiction that \( \sqrt{2} \) is an irrational number. - AQA - A-Level Maths: Pure - Question 10 - 2018 - Paper 3

Worked Solution & Example Answer:Prove by contradiction that \( \sqrt{2} \) is an irrational number. - AQA - A-Level Maths: Pure - Question 10 - 2018 - Paper 3

Assume that \( \sqrt{2} \) is rational

Manipulate the equation

Deduce that \( a \) is even

Express \( a \) as even

Substitute back into the equation

Deduce that \( b \) is even

Explain the contradiction

Conclude that \( \sqrt{2} \) is irrational

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