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Decide which of the following are valid proofs of the following statement: Let n be any integer. Then n³ n is divisible by 3.

Decide which of the following are valid proofs of the following statement: Let n be any integer. Then n³ n is divisible by 3.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Decide which of the following are valid proofs of the following statement: Let n be any integer. Then n³-n is divisible by 3. a) Assume n³ — n is divisible by 3. Let n³ n = 3b for some integer b. Let n = k + 1. Then n³ − n = (k + 1)³ – (k + 1) = k³ + 3k² + 3k + 1 − (k+ 1) = k³ + 3k² + 3k + 1 − k − 1 = (k³ − k) + 3k² + 3k = 3b + 3k² + 3k because k³ - k divisible by 3. = 3(b + k² + k) which is clearly divisible by 3. b) Note that n³ n = n(n² - 1) = n(n+1)(n − 1) which is the product of 3 consecutive integers, one of which must be a multiple of 3. Thus the product is a multiple of 3. c) Note that n³ - n = = n(n² — 1) and 3 is prime, so 3 | (n³ − n) if and only if either 3 | n or 3 | (n² — 1) (i.e. 3 divides one of its factors.) We have 3 cases depending on the remainder when n is divisible by 3 (also known as n mod 3). Let us address each case separately. case n = 0 case n = 1 mod 3. Then 3 n so 3 | n(n² — 1). mod 3. Then n = 3k + 1 for some integer k. Note that (3k + 1)² − 1 = 9k² + 6k+1 − 1 = 9k² + 6k = 3(3k² + 2k) . Thus n³-n is divisible by 3 since one of its two factors (n² - 1) is divisible by 3. case n = 2 mod 3. Then n = 3k + 2 for some integer k. Note that n² − 1 = (3k + 2)² − 1 = 9k² + 12k + 4 − 1 = 3(3k² + 4k+ 1), which is divisible by 3 (and thus so is n³ n since one of its factors is divisible by 3.) -
Transcribed Image Text:Decide which of the following are valid proofs of the following statement: Let n be any integer. Then n³-n is divisible by 3. a) Assume n³ — n is divisible by 3. Let n³ n = 3b for some integer b. Let n = k + 1. Then n³ − n = (k + 1)³ – (k + 1) = k³ + 3k² + 3k + 1 − (k+ 1) = k³ + 3k² + 3k + 1 − k − 1 = (k³ − k) + 3k² + 3k = 3b + 3k² + 3k because k³ - k divisible by 3. = 3(b + k² + k) which is clearly divisible by 3. b) Note that n³ n = n(n² - 1) = n(n+1)(n − 1) which is the product of 3 consecutive integers, one of which must be a multiple of 3. Thus the product is a multiple of 3. c) Note that n³ - n = = n(n² — 1) and 3 is prime, so 3 | (n³ − n) if and only if either 3 | n or 3 | (n² — 1) (i.e. 3 divides one of its factors.) We have 3 cases depending on the remainder when n is divisible by 3 (also known as n mod 3). Let us address each case separately. case n = 0 case n = 1 mod 3. Then 3 n so 3 | n(n² — 1). mod 3. Then n = 3k + 1 for some integer k. Note that (3k + 1)² − 1 = 9k² + 6k+1 − 1 = 9k² + 6k = 3(3k² + 2k) . Thus n³-n is divisible by 3 since one of its two factors (n² - 1) is divisible by 3. case n = 2 mod 3. Then n = 3k + 2 for some integer k. Note that n² − 1 = (3k + 2)² − 1 = 9k² + 12k + 4 − 1 = 3(3k² + 4k+ 1), which is divisible by 3 (and thus so is n³ n since one of its factors is divisible by 3.) -
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