Chapter 5 - Proof by induction

balanced

A string of parentheses is balanced if the number of left parentheses is equal to the number of right parentheses

Show that for every x in a recursively defined set P, x has a particular property

Base case: Show that every element in the basis satisfies the property. Inductive step: The recursive rules show how to construct larger elements in P using smaller elements from P. Assume the property holds for the smaller elements and prove that the property holds for the larger elements.

Show that for every properly nested set of parentheses (every x in P), x is balanced (left[x] = right[x])

Basis: Show that () is balanced Inductive step: Case 1: Assume u is balanced, Prove that (u) is balanced. Case 2: Assume u and v are balanced, Prove that uv is balanced

Theorem 5.3.1: Explicit formula for a sequence defined by a recurrence relation.

Define the sequence {gn} as: - g0 = 1.gn = 3· - gn-1 + 2n, for any n ≥ 1. Then for any n ≥ 0, gn = 5/2 ⋅ 3^n − n − 3/2

Theorem 5.3.2: Explicit formula for a sequence defined by a recurrence relation.

Define the sequence {hn} as: - h0 = 7 - hn = (hn-1 )^3, for any n ≥ 1 Then for any n ≥ 0, hn=7(3n)

inductive hypothesis

In the statement "S(k) implies S(k+1)" of the inductive step, the supposition that S(k) is true

Theorem 5.2.1: Number of vertices in a perfect binary tree.

Let T be a perfect binary tree. Then the number of vertices in T is 2k - 1 for some positive integer k.

Structural induction

a type of induction used to prove theorems about recursively defined sets that follows the structure of the recursive definition

inductive step

establishes that if the theorem is true for k, then the theorem also holds for k + 1

base case

establishes that the theorem is true for the first value in the sequence

principle of mathematical induction

if the base case (for n = 1) is true and inductive step is true, then the theorem holds for all positive integers Let S(n) be a statement parameterized by a positive integer n. Then S(n) is true for all positive integers n, if: 1. Base case: S(1) is true . 2. Inductive step: For all k ∈ Z+, S(k) implies S(k+1)