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Prove W.O.P.M.I., i.e. prove W.O.P.M.I. and prove M.I. ⇒ W.O.P¹8

Prove W.O.P.M.I., i.e. prove W.O.P.M.I. and prove M.I. ⇒ W.O.P¹8

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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The Well-Ordering Principle and (the theorem of) Mathematical In- duction. The Well-Ordering Principle (W.O.P.) for the (set of) natural numbers states: every nonempty subset of the (set of) natural numbers has a least element 17 (The theorem of) Mathematical Induction (M.I.) states: If a subset S of the (set of) natural numbers contains 0 and, for all n in the (set of) natural numbers, n € S ⇒ n+ 1 € S, then S is equal to (the set of) the natural numbers. Prove W.O.P. M.I., i.e. prove W.O.P. ⇒ M.I. and prove M.I. W.O.P¹8 Here is a(n unfortunately somewhat oblique) strategy toward the goal: ● First, prove W.O.P. ⇒ M.I. by proving its contrapositive: NOT M.I. NOT W.O.P. Toward a proof of NOT M.I. ⇒ NOT W.O.P., suppose NOT M.I. and consequently, obtain a subset S of the (set of) natural numbers that con- tains 0 and such that for all s € S, s € S ⇒ s + 1 € S, yet S does not equal the (set of) natural numbers. Use this as ammunition to construct a (nonempty) subset T of the natural numbers that does not have a least element. To construct such a T is to prove the negation of W.O.P. So, you'll have proved NOT M.I. ⇒ NOT W.O.P., and by contraposition, W.O.P. M.I., as desired. • Next, prove strong mathematical induction (S.M.I.) implies W.O.P. 19
Transcribed Image Text:The Well-Ordering Principle and (the theorem of) Mathematical In- duction. The Well-Ordering Principle (W.O.P.) for the (set of) natural numbers states: every nonempty subset of the (set of) natural numbers has a least element 17 (The theorem of) Mathematical Induction (M.I.) states: If a subset S of the (set of) natural numbers contains 0 and, for all n in the (set of) natural numbers, n € S ⇒ n+ 1 € S, then S is equal to (the set of) the natural numbers. Prove W.O.P. M.I., i.e. prove W.O.P. ⇒ M.I. and prove M.I. W.O.P¹8 Here is a(n unfortunately somewhat oblique) strategy toward the goal: ● First, prove W.O.P. ⇒ M.I. by proving its contrapositive: NOT M.I. NOT W.O.P. Toward a proof of NOT M.I. ⇒ NOT W.O.P., suppose NOT M.I. and consequently, obtain a subset S of the (set of) natural numbers that con- tains 0 and such that for all s € S, s € S ⇒ s + 1 € S, yet S does not equal the (set of) natural numbers. Use this as ammunition to construct a (nonempty) subset T of the natural numbers that does not have a least element. To construct such a T is to prove the negation of W.O.P. So, you'll have proved NOT M.I. ⇒ NOT W.O.P., and by contraposition, W.O.P. M.I., as desired. • Next, prove strong mathematical induction (S.M.I.) implies W.O.P. 19
• At this stage, you'll have proved both S.M.I. ⇒ W.O.P. and W.O.P. ⇒ M.I., and thus S.M.I. ⇒ M.I. It suffices now to prove M.I. ⇒ S.M.I. To see this, a proof of M.I.W.O.P. is constructed via M.I. ⇒S.M.I. W.O.P., and so M.I.W.O.P., and W.O.P. = M.I. So, prove M.I. S.M.I. to finish off the equivalence of M.I. and W.O.P. 16 We'll take the (set of) natural numbers to be the set {0, 1, ...} in this discussion, although it doesn't matter whether 0 is included or not, as long as its inclusion/exclusion is consistent. In fact, for any subset of the natural numbers, order them from least to smallest (assuming the validity of the W.O.P.!), and rename them 0, 1, ... to get an equivalent story. 17 where, for a (nonempty) subset S of the (set of) natural numbers, s is a least element iff for all t € S, s ≤ t, and where ≤ denotes the usual ≤ for natural numbers, e.g. 2 ≤ 3. 18 A request: aim for proofs not by contradiction, but by contraposition, as I've suggested all proofs by contradiction are! You will make me sad if you write a "proof by contradiction”. 19S.M.I. states: If a subset S of the (set of) natural numbers contains 0 and, for all n in the (set of) natural numbers, 0, 1, ..., n - 1, and n are in S, then n+1 € S, then S equals the (set of) natural numbers. So, instead of only getting n € S as a hypothesis, as in M.I., we get all of 0, 1,..., n - 1, and n in S as a hypothesis in S.M.I. (hence, strong mathematical induction: we get a stronger hypothesis to work with).
Transcribed Image Text:• At this stage, you'll have proved both S.M.I. ⇒ W.O.P. and W.O.P. ⇒ M.I., and thus S.M.I. ⇒ M.I. It suffices now to prove M.I. ⇒ S.M.I. To see this, a proof of M.I.W.O.P. is constructed via M.I. ⇒S.M.I. W.O.P., and so M.I.W.O.P., and W.O.P. = M.I. So, prove M.I. S.M.I. to finish off the equivalence of M.I. and W.O.P. 16 We'll take the (set of) natural numbers to be the set {0, 1, ...} in this discussion, although it doesn't matter whether 0 is included or not, as long as its inclusion/exclusion is consistent. In fact, for any subset of the natural numbers, order them from least to smallest (assuming the validity of the W.O.P.!), and rename them 0, 1, ... to get an equivalent story. 17 where, for a (nonempty) subset S of the (set of) natural numbers, s is a least element iff for all t € S, s ≤ t, and where ≤ denotes the usual ≤ for natural numbers, e.g. 2 ≤ 3. 18 A request: aim for proofs not by contradiction, but by contraposition, as I've suggested all proofs by contradiction are! You will make me sad if you write a "proof by contradiction”. 19S.M.I. states: If a subset S of the (set of) natural numbers contains 0 and, for all n in the (set of) natural numbers, 0, 1, ..., n - 1, and n are in S, then n+1 € S, then S equals the (set of) natural numbers. So, instead of only getting n € S as a hypothesis, as in M.I., we get all of 0, 1,..., n - 1, and n in S as a hypothesis in S.M.I. (hence, strong mathematical induction: we get a stronger hypothesis to work with).
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