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November 28th, 2017, 10:51 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 n1 is m^n I dont know how to go on with the induction step any ideas? Last edited by sita; November 28th, 2017 at 10:57 AM. 
November 28th, 2017, 11:05 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
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 11:08 AM. 

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