Standard +0.3 This is a guided proof by contradiction with scaffolding provided. Part (a) requires only factoring out 2 from a given expansion (routine algebra), and part (b) follows a standard template similar to the classic √2 irrationality proof but with the cube root. The structure is heavily scaffolded and the techniques are standard A-level proof methods, making it easier than average despite being a proof question.
10. (a) A student's attempt to answer the question
"Prove by contradiction that if \(n ^ { 3 }\) is even, then \(n\) is even" is shown below. Line 5 of the proof is missing.
Assume that there exists a number \(n\) such that \(n ^ { 3 }\) is even, but \(n\) is odd.
If \(n\) is odd then \(n = 2 p + 1\) where \(p \in \mathbb { Z }\)
So \(n ^ { 3 } = ( 2 p + 1 ) ^ { 3 }\)
$$\begin{aligned}
& = 8 p ^ { 3 } + 12 p ^ { 2 } + 6 p + 1 \\
& =
\end{aligned}$$
This contradicts our initial assumption, so if \(n ^ { 3 }\) is even, then \(n\) is even.
Complete this proof by filling in line 5.
(b) Hence, prove by contradiction that \(\sqrt [ 3 ] { 2 }\) is irrational.
Requires correct algebra and statement that expression is odd. Allow even+even+even+1=odd
Part (b):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Assume \(\sqrt[3]{2} = \frac{p}{q}\) (in simplest form), cube to get \(2 = \frac{p^3}{q^3}\)
M1
Sets up contradiction AND cubes. Condone omission of "simplest form" for this mark
\(p^3 = 2q^3 \Rightarrow p^3\) is even, therefore \(p\) is even
A1
States both \(p^3\) and \(p\) are even
Write \(p = 2m\) so \((2m)^3 = 2q^3\)
M1
Writes \(p=2m\) and attempts to find \(q^3\)
\(q^3 = 4m^3 \Rightarrow q^3\) is even, therefore \(q\) is even
A1
States both \(q^3\) and \(q\) are even
This contradicts \(\frac{p}{q}\) being in simplest form, so \(\sqrt[3]{2}\) is irrational *
A1*
Completely correct proof with conclusion. \(\frac{p}{q}\) must be stated in simplest form in assumption
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# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes $2(4p^3+6p^2+3p)+1$ which is odd | B1 | Requires correct algebra and statement that expression is odd. Allow even+even+even+1=odd |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume $\sqrt[3]{2} = \frac{p}{q}$ (in simplest form), cube to get $2 = \frac{p^3}{q^3}$ | M1 | Sets up contradiction AND cubes. Condone omission of "simplest form" for this mark |
| $p^3 = 2q^3 \Rightarrow p^3$ is even, therefore $p$ is even | A1 | States both $p^3$ and $p$ are even |
| Write $p = 2m$ so $(2m)^3 = 2q^3$ | M1 | Writes $p=2m$ and attempts to find $q^3$ |
| $q^3 = 4m^3 \Rightarrow q^3$ is even, therefore $q$ is even | A1 | States both $q^3$ and $q$ are even |
| This contradicts $\frac{p}{q}$ being in simplest form, so $\sqrt[3]{2}$ is irrational * | A1* | Completely correct proof with conclusion. $\frac{p}{q}$ must be stated in simplest form in assumption |
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10. (a) A student's attempt to answer the question\\
"Prove by contradiction that if $n ^ { 3 }$ is even, then $n$ is even" is shown below. Line 5 of the proof is missing.
Assume that there exists a number $n$ such that $n ^ { 3 }$ is even, but $n$ is odd.
If $n$ is odd then $n = 2 p + 1$ where $p \in \mathbb { Z }$\\
So $n ^ { 3 } = ( 2 p + 1 ) ^ { 3 }$
$$\begin{aligned}
& = 8 p ^ { 3 } + 12 p ^ { 2 } + 6 p + 1 \\
& =
\end{aligned}$$
This contradicts our initial assumption, so if $n ^ { 3 }$ is even, then $n$ is even.
Complete this proof by filling in line 5.\\
(b) Hence, prove by contradiction that $\sqrt [ 3 ] { 2 }$ is irrational.
\hfill \mbox{\textit{Edexcel P4 2021 Q10 [6]}}