| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Contradiction proof of irrationality |
| Difficulty | Moderate -0.3 This is a standard proof by contradiction following the classic template for proving √p is irrational. The key lemma is provided, reducing it to a routine application of the contradiction method with minimal novelty. Slightly easier than average due to the given hint and standard structure, though it requires understanding proof technique rather than just computation. |
| Spec | 1.01d Proof by contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Assume \(\sqrt{7}\) is rational so that \(\sqrt{7} = \frac{a}{b} \Rightarrow 7 = \frac{a^2}{b^2}\) (where \(a\) and \(b\) have no factors in common) | M1 | Starts proof by contradiction and squares both sides; condone omission of statement about common factors |
| \(7 = \frac{a^2}{b^2} \Rightarrow 7b^2 = a^2\); so \(a^2\) is a multiple of 7 which means \(a\) is a multiple of 7 | A1 | Reaches \(7b^2 = a^2\) and deduces \(a^2\) is a multiple of 7 which means \(a\) is a multiple of 7 |
| \(a = 7k \Rightarrow 7b^2 = 49k^2 \Rightarrow b^2 = 7k^2\) | M1 | Sets \(a=7k\) and proceeds to \(b^2 = 7k^2\) |
| So \(b^2\) is a multiple of 7 which means \(b\) is a multiple of 7. As \(a\) and \(b\) are both multiples of 7, this contradicts the fact that \(a\) and \(b\) have no factors in common. Hence \(\sqrt{7}\) must be irrational. | A1 | Fully correct proof including: initial statement that \(a,b\) have no common factors; correct direction of implications; \(b^2\) multiple of 7 \(\Rightarrow\) \(b\) multiple of 7; minimal conclusion |
## Question 7:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Assume $\sqrt{7}$ is rational so that $\sqrt{7} = \frac{a}{b} \Rightarrow 7 = \frac{a^2}{b^2}$ (where $a$ and $b$ have no factors in common) | M1 | Starts proof by contradiction and squares both sides; condone omission of statement about common factors |
| $7 = \frac{a^2}{b^2} \Rightarrow 7b^2 = a^2$; so $a^2$ is a multiple of 7 which means $a$ is a multiple of 7 | A1 | Reaches $7b^2 = a^2$ and deduces $a^2$ is a multiple of 7 which means $a$ is a multiple of 7 |
| $a = 7k \Rightarrow 7b^2 = 49k^2 \Rightarrow b^2 = 7k^2$ | M1 | Sets $a=7k$ and proceeds to $b^2 = 7k^2$ |
| So $b^2$ is a multiple of 7 which means $b$ is a multiple of 7. As $a$ and $b$ are both multiples of 7, this contradicts the fact that $a$ and $b$ have no factors in common. Hence $\sqrt{7}$ must be irrational. | A1 | Fully correct proof including: initial statement that $a,b$ have no common factors; correct direction of implications; $b^2$ multiple of 7 $\Rightarrow$ $b$ multiple of 7; minimal conclusion |
---\begin{enumerate}
\item Use proof by contradiction to prove that $\sqrt { 7 }$ is irrational.\\
(You may assume that if $k$ is an integer and $k ^ { 2 }$ is a multiple of 7 then $k$ is a multiple of 7 )
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2023 Q7 [4]}}