MACM 101 Lecture 18: Lecture 18 Part 2_ Mathematical Induction

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Lecture 18 Part 2: Mathematical Induction
- Take a herd of k + 1 horses, and pick horses a and b there
Take two `subherds’ containing k horses each such that one contains a, but not b, and
the other contains b, but not a
The intersection contains some horse c. By inductive hypothesis a has the same color
as c and b has the same color as c
The Cardinality of the Power Set
- We have proved that, for any finite set A, it is true that | P(A) | = 2|A|
- Let Q(n) denote the statement `an n-element set has 2n subsets’
- Basis step: Q(0), and empty set has only one subset, empty
- Inductive step. We make the inductive hypothesis, a k-element set A has 2k subsets
- We have to prove Q(k + 1), that is if a set A contains k + 1 elements, then | P(A) | = 2k+1
Fix an element a ∈ A, and set B = A – {a}.
The set B contains k elements, hence | P(B) | = 2k
Every subset X of B corresponds to two subsets of A
Therefore, |P(A)| = 2x|P(B)| = (2)(2k) = 2k+1
Odd Pie Fights
An odd number of people stand in a yard at mutually distinct distances. At the same time each
person throws a pie at their nearest neighbor, hitting this person. Show that there is at least one
survivor, that is, at least one person who is not hit by a pie.
- Let P(n) denote the statement `there is a survivor in the odd pie fight with 2n + 1 people’
- Basis step: P(1), there are 3 people
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