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수학적 귀납법

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Sundstrom, Mathematical Reasoning · 4.1절 · 14번영어 원문

Use mathematical induction to prove that the sum of the cubes of any three consecutive natural numbers is a multiple of 9.

정답·풀이 보기 (영어 원문)

Three consecutive natural numbers may be represent by nn, n+1n + 1, and n+2n + 2, where nn is a natural number. For the inductive step, think before you try to do a lot of algebra. You should be able to complete a proof of the inductive step by expanding the cube of only one expression.

한 번 더 — 비슷한 문제Sundstrom, Mathematical Reasoning · 4.2절영어 원문

Consider the following question:

For which natural numbers

n

do there exist nonnegative integers

x

and

y

such that

n = 3x + 5y

?

To help answer this question, we will let Z∗={x∈Z∣x≥0}\mathbb{Z}^* = \left\{ {x \in \mathbb{Z} \mid x \geq 0} \right\}, and let P(n)P \left( n \right) be There exist x,y∈Z∗x,y \in \mathbb{Z}^* such that n=3x+5yn = 3x + 5y. Notice that P(1)P(1) is false since if both xx and yy are zero, then 3x+5y=03x + 5y = 0 and if either x>0x > 0 or y>0y > 0, then 3x+5y≥33x + 5y \geq 3. Also notice that P(6)P(6) is true since 6=3⋅2+5⋅06 = 3 \cdot 2 + 5 \cdot 0 and P(8)P(8) is true since 8=3⋅1+5⋅18 = 3 \cdot 1 + 5 \cdot 1.

Explain why P(9)P(9), P(10)P(10), P(11)P(11), and P(12)P(12) are true.

다 풀었으면 정답 확인

Construct the following table and use it to answer the first two questions. The table shows that P(3)P\left( 3 \right), P(5)P\left( 5 \right), and P(6)P\left( 6 \right) are true. We can also see that P(2)P(2), P(4)P(4), and P(7)P(7) are false. It also appears that if n∈Nn \in \mathbb{N} and n≥8n \geq 8, then P(n)P\left( n \right) is true.

| xx | 0 | 1 | 2 | 3 | 4 | 0 | 1 | 2 | 0 | 1 | 1 | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | yy | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 2 | 3 | 3 | | 3x+5y3x + 5y | 0 | 3 | 6 | 9 | 12 | 5 | 8 | 11 | 10 | 13 | 18 |

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