1. UCI ICS/Math 6D, Fall 2007 4-Induction-1 Induction Let P(n) be a predicate (“propositional function”) regarding integers n≥n 0 for some n 0 (typically n 0 =0 or 1). “Normal” Induction “Normal” Induction: If you prove two things: 1) Basis step: P(n 0 ) is true and 2) Inductive step: For all k≥n 0 : If P(k) true then P(k+1) true then it follows that P(n) is true for all n≥n 0. Example: Show that P(n) = “1+2+...+n=n(n+1)/2” is true for all n≥1. 1)Basis (for n 0 =1): “1=1·(1+1)/2” is true. 2)Inductive step: Assuming P(k) [that is, assuming 1+2+...+k=k(k+1) ] prove P(k+1) [that is, prove 1+2+...+k+(k+1)=(k+1)(k+2)/2 ] Ref: http://en.wikipedia.org/wiki/Mathematical_inductionhttp://en.wikipedia.org/wiki/Mathematical_induction The inductive step is a proof of an implication P(k) => P(k+1)

2. UCI ICS/Math 6D, Fall 2007 4-Induction-2 Induction Examples For n≥0, n 2 <4 n. P(n)=“n 2 <4 n ” Basis step: P(0)=“0<4 0 =1” is true Inductive step: Assuming k 2 <4 k, prove (k+1) 2 <4 k+1. The sum of the first n odd positive integers is n 2 Basis step: “1=1 2 ” is true Inductive step: Assuming 1+3+5+...+(2k-1)=k 2, prove that 1+3+5+...+(2k-1)+(2k+1)= (k+1) 2 A set with n elements has 2 n subsets (|P(S)|=2 |s| ) Basis step: |P(  )|=|{  }|=1=2 0 Inductive step: Assuming that every set with (exactly) k elements has (exactly) 2 k subsets, prove that any set with (exactly) k+1 elements has (exactly) 2 k+1 subsets.

3. UCI ICS/Math 6D, Fall 2007 4-Induction-3 Other Numeric Examples of Induction For all (integer) n≥0, n<2 n For all (integer) n≥4, 2 n < n! For all (integer) n≥0, 3 evenly divides n 3 -n For all (integer) n≥0, 1+2+4+…+2 n =2 n+1 -1 Let H n =1+1/2+1/3+1/4+…+1/n For all (integer) n≥1, H 2 n >1+n/2

4. UCI ICS/Math 6D, Fall 2007 4-Induction-4 Pie Fighting Rules: n people at mutually distinct distances. Each person hits his/her nearest neighbor with a pie. Thrm: With an odd number (2n+1) of people, at least 1 will not be hit. Basis step: With 3 people (n=1), the 2 closest to each other will hit each other and the 3 rd person will not be hit, Inductive step: Assume the statement holds for 2k+1 people. Consider a collection, S, of 2(k+1)+1=2k+3 people.  S be the 2 people closest to each other. Let a,b  S be the 2 people closest to each other. a and b will hit each other. If anyone in S-{a,b} throws at a or b, there will be less than 2k+1 pies thrown at the 2k+1 people in S-{a,b}. Someone not hit. Otherwise, people in S-{a,b} throw only at each other and so, by our assumption, someone in S-{a,b} is not hit.

5. UCI ICS/Math 6D, Fall 2007 4-Induction-5 Tiling with Triominos Every 2 n x 2 n checkerboard with one square removed can be tiled by right triominos. Basis step: n=1, 2x2 checkerboard Inductive step:

6. UCI ICS/Math 6D, Fall 2007 4-Induction-6 Induction Gone Wrong Quasi-Theorem: If a,b>0 and max(a,b)=n, then a=b. Basis step:If a,b>0 and max(a,b)=1, then a=1=b. Inductive step: Assume the result true for k. If max(a,b)=k+1, then max(a-1,b-1)=k. (*) By the assumption underlying the inductive step, it follows that a-1=b-1. Therefore a=b. What’s wrong here? Note that P(k+1) = “for all a,b>0 if max(a,b)=k+1 then a=b” Note that P(k) = “for all c,d>0 if max(c,d)=k then c=d” It’s true that if max(a,b)=k+1 then max(a-1,b-1)=k, but it’s not true that if a,b>0 then a-1>0 and b-1>0, and therefore the assumption that P(k) is true does not apply to (a-1,b-1). For example, you can easily see that the inductive step is not correct when you attempt to do it for k=1. Values a,b can be in {1,2}, and so c,d are in {0,1}. The assumption P(1) applies only if c=d=1, and therefore only when a=b=2, in which case obviously a=b, but that does not mean that a=b in all cases.

7. UCI ICS/Math 6D, Fall 2007 4-Induction-7 All Horses Have The Same Color Let P(n) be the proposition that all horses in a set of n horses have the same color. P(1) is clearly true. If P(k) is true, let’s look at any set S of k+1 horses Leaving out the first horse, we get subset S1 of k horses, all of which have the same color according to P(k). Leaving out the last horse, we get subset S2 of k horses, all of which have the same color according to P(k). Putting these 2 sets together we get our original set of k+1 horses, with all the horses in it of the same color as was established in the subsets S1,S2. What’s wrong here??! Answer: This works only if the two sets S1 and S2 intersect, i.e. if k>1.