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 November 28th, 2017, 11:51 AM #1 Member   Joined: Oct 2017 From: Rumba Posts: 39 Thanks: 0 Proof by induction Would i be ok to use base step as n = 1 for a function A -> B for one in A A (x) mapping to B (a1,a2,a3.... am) No of functions possible for n-1 is m^n I dont know how to go on with the induction step any ideas? Last edited by sita; November 28th, 2017 at 11:57 AM. November 28th, 2017, 12:05 PM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 To prove by weak mathematical induction the proposition $P(n) \text { is true for any } n \in \mathbb N^+$, first prove $P(1) \text { is true.}$ This is frequently very easy to do. Moreover it justifies the following statement: $\exists \text { at least one number } k \in \mathbb N^+ \text { such that } P(k) \text { is true.}$ This is frequently called the assumption step although it is clearly a true statement. Notice however, that you may NOT assume that k = 1. You may assume that $k \ge 1.$ Now prove in the second step that $P(k + 1) \text { is true.}$ What is the intuition? P(1) is true by the first step so P(1 + 1) = P(2) is true by the second step, so P(2 + 1) = P(3) is true by the second step, so P(3 + 1) = P(4) is true by the second step, and so on forever. Last edited by JeffM1; November 28th, 2017 at 12:08 PM. Tags induction, proof Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Jimi Algebra 2 March 10th, 2014 07:41 AM walter r Calculus 3 April 26th, 2013 05:25 AM 3,14oner Abstract Algebra 4 April 11th, 2012 01:37 AM Airmax Applied Math 9 May 8th, 2009 01:02 PM MaD_GirL Number Theory 5 November 14th, 2007 07:34 PM

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