Standard +0.8 This is a proof by contradiction requiring students to adapt the classic √2 irrationality proof to the cube root case. While the structure mirrors a standard proof, it demands careful handling of prime factorization arguments with cubes and independent construction of a rigorous argument, making it moderately challenging but still within reach of well-prepared students.
Question 10:
10 | Begins proof by contradiction,
assumes that 3 2 is rational OE | AO3.1a | M1 | Assume 3 2is rational
a
3 2 = ,
b
aandb have no common factors
⇒ 3 2b=a
⇒2b3 =a3
∴ a is even
let a=2dthen 2b3 =8d3
⇒b3 =4d3
∴ b is even
Hence, a and b have a common
factor of 2.
This is a contradiction.
∴the assumption that3 2is rational
must be incorrect and it is proved
that 3 2is an irrational number
Uses language and notation
correctly to state initial
assumptions | AO2.5 | B1
Manipulates fraction including
cubing. | AO1.1a | M1
Deduces a is even | AO2.2a | R1
Deduces b is even | AO2.2a | R1
Explains why there is a
contradiction | AO2.4 | E1
Completes rigorous argument to
show that 3 2is irrational | AO2.1 | R1
Total | 7
Q | Marking Instructions | AO | Marks | Typical Solution