| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2020 |
| Session | October |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Parity and evenness proofs |
| Difficulty | Moderate -0.5 This is a standard proof by contradiction using parity arguments. Students assume n is odd, write n=2k+1, expand n³=(2k+1)³, and show it's odd—contradicting that n³ is even. While it requires understanding proof structure and algebraic manipulation, it's a textbook example of this proof technique with straightforward steps, making it slightly easier than average. |
| Spec | 1.01d Proof by contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Assume there exists a number \(m\) such that when \(m^3\) is even, \(m\) is odd | B1 | For setting up the contradiction. Condone contrapositive. Accept "assume if \(m^3\) is even then \(m\) is odd" |
| If \(m\) is odd then \(m = 2p+1\) (where \(p\) is an integer) and \(m^3 = (2p+1)^3 = \ldots\) | M1 | Attempts to cube an odd number. Accept attempt at \((2p+1)^3\) or \((2p-1)^3\) |
| \(= 8p^3 + 12p^2 + 6p + 1\) | A1 | \((2p+1)^3 = 8p^3+12p^2+6p+1\) or simplified equivalent \(2(4p^3+6p^2+3p)+1\). For \((2p-1)^3\): \(8p^3-12p^2+6p-1\) |
| \(2\times(4p^3+6p^2+3p)+1\) is odd, contradiction, so if \(n^3\) is even then \(n\) is even | A1 | Fully correct proof with correct calculations, reason (even+1 = odd) and conclusion |
# Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume there exists a number $m$ such that when $m^3$ is even, $m$ is odd | B1 | For setting up the contradiction. Condone contrapositive. Accept "assume if $m^3$ is even then $m$ is odd" |
| If $m$ is odd then $m = 2p+1$ (where $p$ is an integer) and $m^3 = (2p+1)^3 = \ldots$ | M1 | Attempts to cube an odd number. Accept attempt at $(2p+1)^3$ or $(2p-1)^3$ |
| $= 8p^3 + 12p^2 + 6p + 1$ | A1 | $(2p+1)^3 = 8p^3+12p^2+6p+1$ or simplified equivalent $2(4p^3+6p^2+3p)+1$. For $(2p-1)^3$: $8p^3-12p^2+6p-1$ |
| $2\times(4p^3+6p^2+3p)+1$ is odd, contradiction, so if $n^3$ is even then $n$ is even | A1 | Fully correct proof with correct calculations, reason (even+1 = odd) and conclusion |
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\item Given that $n$ is an integer, use algebra, to prove by contradiction, that if $n ^ { 3 }$ is even then $n$ is even.\\
\end{enumerate}
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\hfill \mbox{\textit{Edexcel P4 2020 Q1 [4]}}