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Title: 4Induction1

1
Induction

Let P(n) be a predicate (propositional
function) regarding integers nn0 for some n0
(typically n00 or 1).
Normal Induction
If you prove two things
1) Basis step P(n0) is true
and 2) Inductive step For all kn0 If P(k)
true then P(k1) true
then it follows that P(n) is true for all nn0.
Example Show that P(n) 12...nn(n1)/2 is
true for all n1.
Basis (for n01) 11(11)/2 is true.
Inductive step Assuming P(k) that is, assuming
12...kk(k1) prove
P(k1) that is, prove 12...k(k1)(k1)(k2)/
2
Ref http//en.wikipedia.org/wiki/Mathematical_ind
uction

The inductive step is a proof of an
implication P(k) gt P(k1)
2
Induction Examples

For n0, n2lt4n. P(n)n2lt4n
Basis step P(0)0lt401 is true
Inductive step Assuming k2lt4k, prove
(k1)2lt4k1.
The sum of the first n odd positive integers is
n2
Basis step 112 is true
Inductive step Assuming 135...(2k-1)k2,
prove that 135...(2k-1)(2k1)
(k1)2
A set with n elements has 2n subsets
(P(S)2s)
Basis step P(?)?120
Inductive step Assuming that every set with
(exactly) k elements has (exactly) 2k subsets,
prove that any set with (exactly) k1 elements
has (exactly) 2k1 subsets.

3
Other Numeric Examples of 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 (2n1) of people, at
least 1 will not be hit.
Basis step With 3 people (n1), the 2 closest to
each other will hit each other and the 3rd person
will not be hit,
Inductive step Assume the statement holds for
2k1 people.
Consider a collection, S, of 2(k1)12k3
people.
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 2k1 pies thrown at the 2k1 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
Tiling with Triominos

Every 2n x 2n checkerboard with one square
removed can be tiled by right triominos.
Basis step n1, 2x2 checkerboard
Inductive step

6
Induction Gone Wrong

Quasi-Theorem If a,bgt0 and max(a,b)n, then ab.
Basis stepIf a,bgt0 and max(a,b)1, then a1b.
Inductive step Assume the result true for k.
If max(a,b)k1, then max(a-1,b-1)k.
() By the assumption underlying the inductive
step, it follows that a-1b-1. Therefore ab.
Whats wrong here?
Note that P(k1) for all a,bgt0 if max(a,b)k1
then ab
Note that P(k) for all c,dgt0 if max(c,d)k
then cd
Its true that if max(a,b)k1 then
max(a-1,b-1)k,
but its not true that if a,bgt0 then a-1gt0 and
b-1gt0,
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 k1. Values a,b can be in 1,2, and
so c,d are in 0,1. The assumption P(1) applies
only if cd1, and therefore only when ab2, in
which case obviously ab, but that does not mean
that ab in all cases.

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, lets look at any set S of k1
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 k1 horses, with all the
horses in it of the same color as was
established in the subsets S1,S2.
Whats wrong here??!
Answer This works only if the two sets S1 and
S2 intersect, i.e. if kgt1.