Worked Proofs and Proof Critiques for Induction
CPSC 121, Sections 202 and BCS, 2008-2009W2
Worked Proof: Introductions for a Group of Size n
Definitions: An “introduction” is one person saying hello to one other person.
Theorem: For a group of n people to introduce themselves requires n(n-1) introductions.
Proof:
Base case: A group of 1 person requires no introductions. So, when n = 0: 0 = 1*0 = 1(1-1).
Induction hypothesis: Assume that a group of k people requires k(k-1) introductions (for an arbitrary
integer k ≥ 1).
Inductive step: Under this assumption, we need to prove that a group of k+1 people takes (k+1)((k+1)-1)
introductions.
Consider a group of k+1 people. Remove one person from the group. What remains is a group of k
people, who can introduce themselves by the induction hypothesis using k(k-1) introductions. To finish
with the group of k+1, the remaining person must greet and be greeted by each of the k people, for 2k
additional introductions.
So, k+1 people take:
k(k-1) + 2k
= k2 - k + 2k
= k2 + k
= (k+1)k
= (k+1)((k+1)-1).
This concludes our inductive proof. Therefore a group of size n takes n(n-1) introductions.
Worked Proof: Maximum Nodes Per Height for a Binary Tree
Definitions: Let a binary tree be a “root” node with 0 to 2 children and all of its children’s nodes. Let the
height of a binary tree be the longest path from the root node to a node with no children. Let n(h) be
the maximum number of nodes in a binary tree of height h.
Theorem: n(h) = 2h+1 - 1.
Proof:
Base case: A binary tree of height 0 is a single node. So, n(0) = 1 = 2 - 1 = 21 - 1 = 20 + 1 - 1.
Induction hypothesis: Assume that n(k) = 2k+1 - 1 (for an arbitrary integer k ≥ 0).
Inductive step: Under this assumption, we need to prove that n(k+1) = 2(k+1)+1 - 1.
Consider a binary tree of height k+1. There are three cases for this tree:
(1) No children, but then its height would be 0, which contradicts the assumption that k ≥ 0 (and so
k + 1 ≥ 1). So, this case cannot occur.
(2) One child, but then the tree could have more nodes by adding a second child; thus, this is not
the tree of height k+1 with the maximum number of nodes. So, this case is not relevant.
(3) Two children. Then, to have the maximum number of nodes, each child’s height must be k (else,
we could increase the number of nodes by adding a new (great-great-great...grand) child to the
bottom of the child’s tree and increasing its height). Further, for the whole tree to have the
maximum number of nodes, each child must have the maximum number of nodes for a tree of
height k (else, we could replace it with such a tree and again increase the number of nodes
without changing the whole tree’s height). Thus, the number of nodes in the whole tree is the
number in each child tree plus the root node 2*n(k) + 1. By the induction hypothesis, we know
n(k) = 2k+1 - 1. Then:
n(k+1) = 2*n(k) + 1
= 2*(2k+1 - 1) + 1
= 2*2k+1 - 2 + 1
= 21+k+1 - 1
= 2(k+1)+1 – 1
This concludes our inductive proof. Therefore the maximum number of nodes in a binary tree of height
h is 2h+1 - 1.
Proof Critique: All Horses Are the Same Colour
What’s wrong with this proof? Anything?
Theorem: All horses are the same colour.
Proof:
Base case: All horses in any group of one horse are obviously the same colour.
Induction hypothesis: Assume that all horses in any group of size k are the same colour (for an arbitrary
integer k ≥ 1).
Inductive step: Under this assumption, we need to prove that all horses in any group of horses of size
k+1 are the same colour.
Consider an arbitrary group of k+1 horses.
Remove any one horse from it. What remains is a group of k horses, which are all the same colour by
the induction hypothesis. Only the set-aside horse may be a different colour.
Now, return the horse to the group and remove a different horse. Again, the remaining horses are all
the same colour, but from the previous step we already know that this time the set-aside horse is also
the same colour. Therefore, all horses in any group of size k+1 are the same colour.
This concludes our inductive proof. Therefore all horses in any group of any size are the same colour,
i.e., all horses are the same colour.
Worked Proof: Binary Search
Definitions: A list a has a length len(a). The elements in the list in order are number 0, 1, 2, ..., len(a) - 1.
Theorem: We can correctly search a sorted list of names by binary search:
Algorithm to search a list a of length len(a) for the name n:
If len(a) is 0, the name was not found; otherwise, check the element len(a)/2 and:
(1) If element len(a)/2 is the one sought; report it, and we’re done.
(2) If element len(a)/2 is larger than the one sought, repeat the algorithm on the list of
elements 0, ..., len(a)/2 - 1.
(3) Otherwise (element len(a)/2 is smaller than the one sought), repeat the algorithm on the
list of elements len(a)/2 + 1, ..., len(a) - 1.
Proof:
Base case: A list of length 0 contains no names; so, the algorithm correctly reports failure.
Induction hypothesis: Assume that we can correctly search a sorted list of names of any integer length 0
≤ k < n, where n is an integer ≥ 1, by binary search.
Inductive step: Under this assumption, we need to prove that we can correctly search a sorted list of
names of length n by binary search.
Since n > 0, the algorithm will check on element n/2 . Since 0 < n, 0 < n/2 < n; so, 0 ≤ n/2 < n, and
the list does contain such an element.
There are then three cases:
(1) The algorithm finds the name sought, in which case it correctly reports it.
(2) Element n/2 is too large. Since the list is sorted, the name sought has an index smaller than
n/2 , and the algorithm proceeds on the list of those elements (0, ..., n/2 - 1) with a length
of n/2. We know 0 ≤ n/2 < n (i.e., the list’s length is in the range covered by our induction
hypothesis); so, the algorithm operates correctly.
(3) If element n/2 is too small, the algorithm proceeds on the right portion of the list, with n n/2 - 1 elements (all but the first n/2 elements and element n/2 ). We know 0 ≤ n/2 ≤
n - 1; so, 0 ≤ n - (n - 1) - 1 ≤ n - n/2 - 1, and n - n/2 - 1 ≤ n - 1 < n (i.e., the list’s length is in the
range covered by our induction hypothesis). Thus, the algorithm operates correctly.
This concludes our inductive proof. Therefore, we can correctly search a sorted list of names by binary
search.
Proof Critique: All Integers Greater than or Equal to Two Are Even
What’s wrong with this proof? Anything?
Theorem: All integers greater than or equal to two are even.
Proof:
Base case: There exists an integer k=1 such that 2*k = 2. Therefore, two is even.
Induction hypothesis: Assume that all integers k where 2 ≤ k < n are even (for an arbitrary integer n > 2).
Inductive step: Under this assumption, we need to prove that n is even.
Consider the number n.
n = (n-2)+2
By the induction hypothesis, n-2 is even. So, there is an integer k’ such that n-2 = 2*k’.
n = 2*k’ + 2 = 2*(k’ + 1)
k’ + 1 is also an integer.
Therefore n+1 is even.
This concludes our inductive proof. Therefore all integers greater than or equal to two are even.