About this chapter
Pure 4 opens with proof by contradiction: assume the statement is false, follow the algebra, and reach something impossible. The method is short to describe but easy to lose marks on, because every step, including the opening assumption and the final conclusion, has to be written out in words.
This chapter covers writing the assumption correctly, parity and divisibility arguments, irrational numbers such as √6 and log 3 to base 2, integer equations, inequalities, curves that never meet, curves with no stationary points, and the classic proof that there are infinitely many primes. One question gives a student's flawed proof to correct.
The ten questions
- Q1 (5 marks): find and correct the errors in a student's proof that √6 is irrational.
- Q2 (5 marks): prove that a curve made of an exponential, a linear and a sine term has no stationary points.
- Q3 (5 marks): x + y and x − y have the same parity, so x² − y² = 18 has no positive integer solutions.
- Q4 (5 marks): prove x²/(x − 1) ≥ 4 for x greater than 1, and show it fails for other x.
- Q5 (6 marks): prove the log of 3 to base 2 is irrational, then use it for the log of 9 to base 8.
- Q6 (10 marks): cubes of non-multiples of 3, then the cube root of 3 is irrational, and why the same method fails for the cube root of 8.
- Q7 (10 marks): an identity with sec² and cosec², a contradiction proof of an inequality, and an exact equation.
- Q8 (8 marks): none of 2 to n divides n! + 1, so there are infinitely many primes, and a student's claim disproved.
- Q9 (9 marks): prove two curves never meet, then find when a changed curve does meet them.
- Q10 (10 marks): a rational plus an irrational is irrational, then √2 + √3 and √3 − √2.
Key skills tested
The method. Assume the statement is false, use correct algebra to reach something impossible, and finish with: this is a contradiction, so the statement is true.
Writing the assumption. "For all x" becomes "there is some x for which it is false", and "there are no integers" becomes "assume there are integers".
Parity and divisibility. Even numbers are 2k and odd numbers 2k + 1. An odd number squared is odd, so if n² is even, n is even. A number that is not a multiple of 3 is 3k + 1 or 3k + 2.
Irrational numbers. Write √6 = a/b, where a and b are integers with no common factors, then show that a and b do share a factor.
Integer equations. Factorise, for example x² − y² = (x−y)(x+y), then use factor pairs or odd and even. A power of 2 is even, but a power of 3 is odd.
Inequalities. Assume it fails for some x, multiply only by something known to be positive, then complete the square: (x−2)² < 0 is impossible.
No real solutions. Assume the curves meet and eliminate y. A quadratic in x² with a negative discriminant has no real solutions.
Calculus and ranges. A stationary point means dy/dx = 0. Use facts such as an exponential being positive and cos x lying between −1 and 1.
Infinitely many primes. Assume p is the largest prime. p! + 1 is not divisible by 2, 3, and so on up to p, so it has a prime factor bigger than p.
Worked example
Question 2 from this chapter. The curve C has equation y = e to the power 3x, plus 6x, minus 2 sin 3x. Use proof by contradiction to prove that C has no stationary points.
Assume that C has a stationary point, so dy/dx = 0 for some value of x.
Differentiating, dy/dx = 3 e to the power 3x, plus 6, minus 6 cos 3x. At the stationary point this is zero, so 3 e to the power 3x = 6(cos 3x − 1). Since cos 3x is at most 1, the right-hand side is zero or negative.
But an exponential is always positive, so the left-hand side is positive. This is a contradiction, so C has no stationary points.
The mark scheme needs both facts, that the exponential is positive and that cos 3x is at most 1. A sketch alone does not prove it.
Seen on real papers
Every question in our booklets is original. These are the recent papers where each question type has appeared.
- January 2024, Q8: proof by contradiction that a curve has no stationary points, as in Q2.
Where marks are lost
- No reason why the equation fails. In January 2024 many simply said the equation had no solutions, with no explanation, and gained no further credit.
- Methods that are not proofs. A graph on its own, testing values of x, or using small-angle approximations could not earn the second method mark.
- A contradiction but no conclusion. Some reached the contradiction but never said that the curve therefore has no stationary points, or concluded without saying there was a contradiction.
- The wrong assumption. Assuming dy/dx = 0 for all x, instead of for some x, earns nothing for the opening line.
Common questions
What must the first line of a proof by contradiction say?
The negation of the statement in words, such as "Assume there is some x for which..." or "Assume √6 is rational".
Why include "no common factors"?
Because the contradiction in an irrationality proof is showing that a and b do share a factor. Without that condition there is nothing to contradict.
Can a diagram prove the result?
No. The January 2024 examiners were clear that a diagram alone was not sufficient; the argument has to be algebraic or in words.
Where this chapter leads
- Pure 4 Chapter 2: Partial Fractions: proof by contradiction used again on a rational function.
- Pure 4 Chapter 5: Differentiation: proving an implicit curve has no stationary points.