Explanation Given information: A is totally order set. Concept used: Mathematical induction Proof: Let A be a set that is totally ordered with respect to a relation ≤ . And let the property P ( n ) be the sentence “Every subset of A with n elements has both a least element and a greatest element” Show that the property is true for n = 1 . If A = ϕ then the property is true by default. So assume that A has at least one element and suppose S = { a 1 } is a subset of A with one element Because ≤ is reflexive a 1 ≤ a 1 . By the definition of least element and greatest element a 1 ¡s both the least and greatest element of S . Thus, the property is true for n = 1 . Show that for all integers k ≥ 1 . If the property is true for n = k then it is true for n = k + 1 . Let k be any integer with k ≥ 1 and suppose that any subset of A with k elements has both a least element and a greatest element. We must show that any subset of A with k + 1 elements has both a least element and a greatest element. If A has fewer than k + 1 elements, then the statement is true by default

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Chapter 8.5, Problem 39ES

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For any integer n≥1 every subset of A with n elements has both a least element and a greatest element, using mathematical induction.

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Chapter 8.5, Problem 38ES

Chapter 8.5, Problem 40ES

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To Prove: For any integer n ≥ 1 every subset of A with n elements has both a least element and a greatest element, using mathematical induction.

Solution Summary: The author proves that A is totally order set using mathematical induction. If nge 1 has a least element and greatest element, then the statement is true.

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To Prove:

For any integer n≥1 every subset of A with n elements has both a least element and a greatest element, using mathematical induction.

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Chapter 8 Solutions

To Prove: For any integer n ≥ 1 every subset of A with n elements has both a least element and a greatest element, using mathematical induction.

Solution Summary: The author proves that A is totally order set using mathematical induction. If nge 1 has a least element and greatest element, then the statement is true.

Videos

To Prove: For any integer n≥1 every subset of A with n elements has both a least element and a greatest element, using mathematical induction.

Want to see the full answer?

Chapter 8 Solutions

To Prove:

For any integer n≥1 every subset of A with n elements has both a least element and a greatest element, using mathematical induction.