Edexcel IAL Mathematics P4.1.1 proof by contradiction

Proof by contradiction questions start with the opposite assumption, then use algebra, parity or inequalities to reach an impossible result.

Exam points

Negate the statement first, then derive a contradiction from parity or integer form.
Use positive a,b assumptions or trig identities to expose an impossible inequality.
Close the proof by linking the contradiction back to the original claim.

[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*

All question bank results loaded

P4.1.1 - Proof by contradiction question 1