The numbers in the red box will have annotations below 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 (setof) natural numbers, n € S ⇒ n+ 1 € S, then S is equal to (the set of) thenatural numbers.Prove W.O.P.M.I., i.e.prove W.O.P. ⇒ M.I. and prove M.I. W.O.P¹8Here 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. andconsequently, 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 notequal the (set of) natural numbers. Use this as ammunition to constructa (nonempty) subset T of the natural numbers that does not have a leastelement. 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. Tosee 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, althoughit 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 (assumingthe 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 elementiff 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 suggestedall 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 (setof) natural numbers. So, instead of only getting n € S as a hypothesis, as in M.I., we get allof 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).

Answered: Image /qna-images/answer/bd6d68df-7ffe-401f-9665-532be4057a40.jpg

Answered: Prove W.O.P.M.I., i.e. prove W.O.P.M.I.…

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

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

Similar questions

please add a lot of expl
Please answer no. 5 and DO NOT use EXCEL
Calculate the volume of the code
Find the number of people who read exactly one magazine
Write as a decimal
Suppose that If"(x)| ≤ 27 and f(4)(x)| ≤ 27 for all x in [0, π]. How large do we have to choose n so that the approximations Tn Mnr and Sn to the integral [² nT= nM= ns= f(x) dx are accurate to within 0.00001?
Please explain thoroughly on paper thanks
Please help me solve on a ti 84 plus calculator
do number 15