Proof
Proof
- A-level
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Write down the converse of the statement "if $n$ is divisible by $4$, then $n$ is even". Say whether the converse is true, and justify your answer.
3 marks
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Prove by deduction that the product of two consecutive integers is even.
3 marks
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Prove by exhaustion that $2^n + 1$ is not divisible by $7$ for any integer $n$ with $1 \le n \le 5$.
4 marks
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Disprove the statement "if $a$ and $b$ are irrational, then $a + b$ is irrational".
4 marks
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Prove by contradiction that $\sqrt{2}$ is irrational.
5 marks
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Prove by induction that $\displaystyle \sum_{r=1}^{n} r = \dfrac{n(n + 1)}{2}$ for every positive integer $n$.
5 marks
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Prove by induction that the sum of the first $n$ odd numbers is $n^2$. That is, $\displaystyle \sum_{r=1}^{n} (2r - 1) = n^2$.
6 marks
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Prove by induction that $8$ divides $3^{2n} - 1$ for every positive integer $n$. Hence find the remainder when $3^{10} - 1$ is divided by $8$.
7 marks
Mark scheme
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Answer. The converse is false. A counterexample is $n = 2$.
- The converse is "if $n$ is even, then $n$ is divisible by $4$".
- $n = 2$ is even, but $2$ is not divisible by $4$.
- The converse is false.
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Answer. The product $n(n + 1)$ is even.
- Let the integers be $n$ and $n + 1$.
- One of two consecutive integers is even.
- A product with an even factor is even.
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Answer. None of $3$, $5$, $9$, $17$ and $33$ is divisible by $7$.
- For $n = 1, 2, 3, 4, 5$ the values are $3, 5, 9, 17, 33$.
- None of these is a multiple of $7$.
- Every allowed value of $n$ has been checked, so the statement is proved.
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Answer. A counterexample is $a = \sqrt{2}$ and $b = -\sqrt{2}$, since the sum is $0$.
- $\sqrt{2}$ and $-\sqrt{2}$ are both irrational.
- Their sum is $0$, which is rational.
- One counterexample is enough to disprove the statement.
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Answer. $\sqrt{2}$ is irrational.
- Assume $\sqrt{2} = \dfrac{p}{q}$ where $p$ and $q$ are integers with no common factor and $q \ne 0$.
- Then $p^2 = 2q^2$, so $p$ is even. Write $p = 2k$.
- Then $4k^2 = 2q^2$, so $q^2 = 2k^2$ and $q$ is even.
- $p$ and $q$ are both even, which contradicts the assumption that the fraction is in lowest terms.
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Answer. The formula holds for every positive integer $n$.
- When $n = 1$, both sides equal $1$.
- Assume the result is true for $n = k$: the sum is $\dfrac{k(k + 1)}{2}$.
- For $n = k + 1$, the sum is $\dfrac{k(k + 1)}{2} + (k + 1) = \dfrac{(k + 1)(k + 2)}{2}$.
- The result is true for $n = 1$ and passes from $k$ to $k + 1$, so it is true for every positive integer $n$.
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Answer. The sum of the first $n$ odd numbers is $n^2$.
- When $n = 1$, both sides equal $1$.
- Assume the sum of the first $k$ odd numbers is $k^2$.
- The next odd number is $2k + 1$, and $k^2 + 2k + 1 = (k + 1)^2$.
- By induction the result holds for every positive integer $n$.
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Answer. $8$ divides $3^{2n} - 1$, so the remainder when $3^{10} - 1$ is divided by $8$ is $0$.
- When $n = 1$, $3^2 - 1 = 8$, which is divisible by $8$.
- Assume $3^{2k} - 1 = 8m$ for some integer $m$.
- $3^{2(k+1)} - 1 = 9 \times 3^{2k} - 1 = 9(3^{2k} - 1) + 8 = 8(9m + 1)$.
- By induction, $8$ divides $3^{2n} - 1$ for every positive integer $n$.
- For $n = 5$, $3^{10} - 1$ is divisible by $8$, so the remainder is $0$.