A-Level Edexcel Maths Pure Solving Equations: 15. (i) Use proof by exhaustion

15. (i) Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4 \newline (n + 1)^{3} > 3^{n} \newline \newline (ii) Given that m^{3} + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even. - Edexcel - A-Level Maths Pure - Question 3 - 2021 - Paper 1

Question 3

$15.-(i)-Use-proof-by-exhaustion-to-show-that-for-$n-\in-\mathbb{N},-n-<-4-\newline-(n-+-1)^{3}->-3^{n}-\newline-\newline-(ii)-Given-that-m^{3}-+-5$-is-odd,-use-proof-by-contradiction-to-show,-using-algebra,-that-$m$-is-even.-Edexcel-A-Level Maths Pure-Question 3-2021-Paper 1.png$

15. (i) Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4 \newline (n + 1)^{3} > 3^{n} \newline \newline (ii) Given that m^{3} + 5$ is odd, use proof... show full transcript

Worked Solution & Example Answer:15. (i) Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4 \newline (n + 1)^{3} > 3^{n} \newline \newline (ii) Given that m^{3} + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even. - Edexcel - A-Level Maths Pure - Question 3 - 2021 - Paper 1

Step 1

Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4$

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Answer

To use proof by exhaustion, we will evaluate the inequality for each natural number $n < 4$ , specifically for $n = 1, 2, 3$ :

For $n = 1$ : [(n + 1)^{3} = (1 + 1)^{3} = 2^{3} = 8 \quad \text{and} \quad 3^{1} = 3 \quad \Rightarrow \quad 8 > 3 \text{ (True)}]
For $n = 2$ : [(n + 1)^{3} = (2 + 1)^{3} = 3^{3} = 27 \quad \text{and} \quad 3^{2} = 9 \quad \Rightarrow \quad 27 > 9 \text{ (True)}]
For $n = 3$ : [(n + 1)^{3} = (3 + 1)^{3} = 4^{3} = 64 \quad \text{and} \quad 3^{3} = 27 \quad \Rightarrow \quad 64 > 27 \text{ (True)}]

Since we have shown that the inequality holds for all values of $n = 1, 2, 3$ , we conclude that for $n \in \mathbb{N}, n < 4$ , it is true that $(n + 1)^{3} > 3^{n}$ .

Step 2

Given that $m^{3} + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even.

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Answer

To use proof by contradiction, we start by negating the statement we want to prove:

Assume that $m$ is odd. Then we can express $m$ as: $m = 2k + 1$ for some integer $k$ .

Now, let's compute $m^{3} + 5$ : $m^{3} = (2k + 1)^{3} = 8k^{3} + 12k^{2} + 6k + 1$ Thus, $m^{3} + 5 = (8k^{3} + 12k^{2} + 6k + 1) + 5 = 8k^{3} + 12k^{2} + 6k + 6$

Notice that $8k^{3}$ , $12k^{2}$ , and $6$ are all even, so: $m^{3} + 5 = ext{even} + ext{even} + ext{even} + ext{odd} \ = ext{odd}$ This contradicts our original premise that $m^{3} + 5$ is odd. Thus, our assumption that $m$ is odd must be false, implying that: $m \text{ must be even.}$

15. (i) Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4 \newline (n + 1)^{3} > 3^{n} \newline \newline (ii) Given that m^{3} + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even. - Edexcel - A-Level Maths Pure - Question 3 - 2021 - Paper 1

Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4$

Given that $m^{3} + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even.

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