Pearson Edexcel IAL Mathematics P4.1 Proof Question Bank

Practise contradiction arguments for integers, inequalities and curves, using algebra to expose an impossible conclusion.

Exam points

assume the negation of a statement and derive a contradiction from parity or integrality
use algebraic identities or curve equations to show an intersection or value is impossible

Covered syllabus points

P4.1.1 - Proof by contradiction

[Maximum number: 4]

Prove by contradiction that for all positive numbers k

k+\frac{9}{k} \geqslant 6

Assume that there exists a positive number k such that $k+\frac9k<6$ .
B1
$k+\frac9k<6\Rightarrow k^2+9<6k\Rightarrow k^2-6k+9<0$
M1
$(k-3)^2<0$
A1
But numbers squared are $\ge 0$ , hence $k+\frac9k\ge 6$ .
A1*

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Covered syllabus points

P4.1 - Proof question 1