   Mathematical Induction Proof
  I cannot get my head around this at all.Suppose that v is of type Vector of Integers and we know the following:-1. v is sorted.2. No two items in v are the same.3. v.at(1) is 12.For an integer n, where 1 <= n <= v.size(), let p(n) be the following proposition: p(n): v.at(n) => 11 + n.A. Explain why p(1) is true?B. suppose that p(k - 1) is true (where 1 < k <= v.size()). Explain why p(k) must then be true.C. Complete a proof that p(n) is true for all integers n with 1 <= n < v.size(). For A. I have,In plain English the proposition means,The n at position n in v is greater than or equal to 11 + n.Let n = 1.P(1): v.at(1) ≥ 11 + 1 = 12.(B)From statement 3 which indicates v.at(1) is 12 and statement 2 which indicates no two items in v are the same, the above statement proves p(1) is true.I'm struggling with the Inductive hypothesis (question b).   
  Something tells me this isn't strictly game related...As for the proof, try to write out p(k): p(k) = 11 + k = 11 + k - 1 + 1 = ...;  Ok, so here's how you go along with these type of stuff.I) You typically need all of the suppositions at some point. Thats why they're there.II) You need them in mathematical lingo :)So: <=> is equivalency n,m,k are all from the set of natural numbers [1..infinity)Supposition 1)The vector is sorted.sorted(v): for each n,m in [1;size(v)] : v.at(n) ≤ v.at(m) <=> n ≤ mor any other representation you like.Supposition 2)unique(v): for each n,m in [1;size(v)] : v.at(n) ≠ v.at(m) <=> n ≠ mor any other representation you likeSupposition 3)v.at(1) = 11Induction hypothesis:there exists a k such that the following is trueInd_1: v.at(k) >= 11 + kWhat we need to show now is that Ind_2: v.at(k+1) >= 11 + k + 1because of Supposition 1) we know v.at(k+1) ≥ v.at(k), since k + 1 ≥ kbecause of Supposition 2) we even know v.at(k+1) > v.at(k), since k + 1 ≠ kusing this in (Ind_1)  (I): v.at(k+1) > 11 + kAt this point we needSupposition 0:v stores Integers :)(as I already said you typically need each and every supposition)(Ind_2) and Supposition 0:v.at(k+1) >= 11 + k + 1which is what we were trying to prove.To complete the prove we need to show that such a k we assumed to exist in (Ind_1) is not only wishful thinking,so Ind_0: v.at(1) ≥ 11 + 1Supposition 3 makes this12 ≥ 12 which is obviously true.Ind_0 and Ind_2 complete the induction for any k.Without using integers you could easily have had v.at(2) == 11.1, wich of course is not ≥ 12 :)Also, I have hidden 10 flaws in this text. If you find them all, you might win a can opener.   ;  This is off topic. The original post looks to be out of a textbook, like a homework/schoolwork question. So I am closing the thread.   