1.3.7

Proof by Contradiction (A2 Only)

Test yourself

What is Proof by Contradiction?

Proof by contradiction is when we assume that the conjecture is false and use a series of logical steps to obtain a statement that is mathematically impossible. This proves that the original conjecture is always true.

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How do you prove by contradiction?

  • To prove a conjecture by contradiction, assume that the conjecture is always false.
  • Next, we use known theorems to make logical steps from the assumption.
  • You will reach a point where, by inspection, the logical step either:
    • Contradicts the starting assumption.
    • Violates a known theorem.
  • This proves that the original conjecture is always true.
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Example

  • Prove that there are no integers xx and yy that satisfy:
    • x2y2=10x^2 - y^2 = 10
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Assumption

  • If the conjecture is false, then that means the following equation is true:
    • x2y2=10x^2-y^2 = 10
  • Where x,yx,y \in ℤ (are integers).
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Logical steps

  • Looking at the left hand side, we see that it is the difference of two squares:
    • x2y2(x+y)(xy)=10x^2-y^2 \equiv (x+y)(x-y) = 10
  • We don't have any more information about xx and yy except that they are integers, so we need to look at what happens for the different cases of xx and yy.
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$$y$$ are oddIllustrative background for Both $$x$$ and 
$$y$$ are odd ?? "content

Both xx and yy are odd

  • It is a known theorem that odd + odd = even and odd − odd = even.
    • So x+y=2mx + y = 2m and xy=2nx-y = 2n, where mm and nn are integers.
    • This means (x+y)(xy)=4mn(x+y)(x-y) = 4mn.
  • By inspection, we see that the left-hand side of our initial assumption is a multiple of 4.
  • This is a contradiction as the right-hand side is equal to 10, which is not a multiple of 4.
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Illustrative background for Both $$x$$ and 
$$y$$ are evenIllustrative background for Both $$x$$ and 
$$y$$ are even ?? "content

Both xx and yy are even

  • It is a known theorem that even + even = even and even − even = even.
    • So x+y=2px + y = 2p and xy=2qx-y = 2q, where pp and qq are integers.
    • This means (x+y)(xy)=4pq(x+y)(x-y) = 4pq.
  • By inspection, we see that the left-hand side is a multiple of 4.
  • This is a contradiction, as 10 is not a multiple of 4.
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 $$y$$ is oddIllustrative background for $$x$$ is even,
 $$y$$ is odd ?? "content

xx is even, yy is odd

  • It is a known theorem that even + odd = odd and even − odd = odd.
    • So x+y=2k+1x+y=2k+1 and xy=2l+1x-y = 2l+1, where kk and ll are integers.
    • This means (x+y)(xy)=(2k+1)(2l+1)=4kl+2l+2k+1(x+y)(x-y) = (2k+1)(2l+1) = 4kl + 2l + 2k + 1.
  • By inspection, we see that the left-hand side is equal to even + even + even + odd, which is an odd number.
  • This is a contradiction as 10 is an even number.
    • The same applies for odd xx and even yy.
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Final statement

  • There are no integers xx and yy that satisfy:
    • x2y2=10x^2-y^2 = 10

Jump to other topics

1Proof

2Algebra & Functions

2.1Powers & Roots

2.2Quadratic Equations

2.3Inequalities

2.4Polynomials

2.5Graphs

2.6Functions

2.7Transformation of Graphs

2.8Partial Fractions (A2 Only)

3Coordinate Geometry

4Sequences & Series

5Trigonometry

6Exponentials & Logarithms

7Differentiation

8Integration

9Numerical Methods

10Vectors

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