Standard +0.3 Part (i) is a standard proof by contradiction requiring basic parity arguments. Part (ii) tests understanding of inequality manipulation when dividing by negatives—a common error but straightforward to identify and correct. This is easier than average as it requires recognizing a single well-known algebraic mistake rather than complex problem-solving.
7. (i) Given that \(a\) and \(b\) are integers such that
$$a + b \text { is odd }$$
Use algebra to prove by contradiction that at least one of \(a\) and \(b\) is odd.
(ii) A student is trying to prove that
$$( p + q ) ^ { 2 } < 13 p ^ { 2 } + q ^ { 2 } \quad \text { where } p < 0$$
The student writes:
$$\begin{gathered}
\qquad \begin{array} { c }
p ^ { 2 } + 2 p q + q ^ { 2 } < 13 p ^ { 2 } + q ^ { 2 } \\
2 p q < 12 p ^ { 2 } \\
\text { so as } p < 0 \quad 2 q < 12 p \\
q < 6 p
\end{array}
\end{gathered}$$
a. Identify the error made in the proof.
b. Write out the correct solution.
M1 For using the "correct"/allowable language in setting up the contradiction. Expect to see a minimum of "assume" or "let" or "there is" or other similar words and "\(a + b\) is odd" and "neither \(a\) nor \(b\) is odd". "There exists integers \(a\) and \(b\) such that \(a + b\) is odd then neither \(a\) nor \(b\) is odd"
A1
Sets \(a = 2k\) and \(b = 2m\) and then attempts \(a + b = 2k + 2m = \ldots\). Obtains \(a + b = 2k + 2m = 2(k+m)\). States that \(a + b\) is even, giving a contradiction that \(a + b\) is odd. "if \(a + b\) is odd that at least one of \(a\) and \(b\) is odd"
(ii) (a)
The error is that as \(p < 0 \Rightarrow 2q > 12p\)
B1
(ii) (b)
\(q > 6p\)
B1
**(i)** | | M1 For using the "correct"/allowable language in setting up the contradiction. Expect to see a minimum of "assume" or "let" or "there is" or other similar words and "$a + b$ is odd" and "neither $a$ nor $b$ is odd". "There exists integers $a$ and $b$ such that $a + b$ is odd then neither $a$ nor $b$ is odd" |
| --- | --- | --- |
| | A1 | Sets $a = 2k$ and $b = 2m$ and then attempts $a + b = 2k + 2m = \ldots$. Obtains $a + b = 2k + 2m = 2(k+m)$. States that $a + b$ is even, giving a contradiction that $a + b$ is odd. "if $a + b$ is odd that at least one of $a$ and $b$ is odd" |
**(ii) (a)** | The error is that as $p < 0 \Rightarrow 2q > 12p$ | B1 | |
**(ii) (b)** | $q > 6p$ | B1 | |
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7. (i) Given that $a$ and $b$ are integers such that
$$a + b \text { is odd }$$
Use algebra to prove by contradiction that at least one of $a$ and $b$ is odd.\\
(ii) A student is trying to prove that
$$( p + q ) ^ { 2 } < 13 p ^ { 2 } + q ^ { 2 } \quad \text { where } p < 0$$
The student writes:
$$\begin{gathered}
\qquad \begin{array} { c }
p ^ { 2 } + 2 p q + q ^ { 2 } < 13 p ^ { 2 } + q ^ { 2 } \\
2 p q < 12 p ^ { 2 } \\
\text { so as } p < 0 \quad 2 q < 12 p \\
q < 6 p
\end{array}
\end{gathered}$$
a. Identify the error made in the proof.\\
b. Write out the correct solution.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q7 [5]}}