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For a natural number n, we define n! = 1·2·3· ... · (n − 1) · n. For example: 2! = 1 2,3! = 1 2 3 = 6,5! = 1·2·3·4·5 = 120. . . (a) Compute 6! and 7!. (In each case, simplify and express your final answer as a single integer.) (b) Prove that if n, m E N, and n > m, then n! > m!. (c) Let x E R. Prove that for all n E N, if x ≥n, then x" >n!.

For a natural number n, we define n! = 1·2·3· ... · (n − 1) · n. For example: 2! = 1 2,3! = 1 2 3 = 6,5! = 1·2·3·4·5 = 120. . . (a) Compute 6! and 7!. (In each case, simplify and express your final answer as a single integer.) (b) Prove that if n, m E N, and n > m, then n! > m!. (c) Let x E R. Prove that for all n E N, if x ≥n, then x" >n!.

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter2: Working With Real Numbers
Section2.7: Problem Solving: Consecutive Integers
Problem 14OE
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For a natural number n, we define n! = 1 · 2 · 3 · ... · (n − 1) · n. For example: 2! = 1·2, 3! = 1·2·3=6,5!= 1.2.3.4.5 = 120. (a) Compute 6! and 7!. (In each case, simplify and express your final answer as a single integer.) (b) Prove that if n, m E N, and n > m, then n! > m!. (c) Let x E R. Prove that for all n E N, if x ≥n, then x" ≥n!.
Transcribed Image Text:For a natural number n, we define n! = 1 · 2 · 3 · ... · (n − 1) · n. For example: 2! = 1·2, 3! = 1·2·3=6,5!= 1.2.3.4.5 = 120. (a) Compute 6! and 7!. (In each case, simplify and express your final answer as a single integer.) (b) Prove that if n, m E N, and n > m, then n! > m!. (c) Let x E R. Prove that for all n E N, if x ≥n, then x" ≥n!.
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