SE 500 Mathematics for Software Engineering
Fall 2023 HW #9: Mathematical Induction
Sample Solutions

1. Prove by (weak) mathematical induction that  (+i | 1≤i≤n : 2i + 3) = n² + 4n  for all n ≥ 0.

Solution: The proof is by mathematical induction on n. Taking P to be the given equation, to prove the base case, P.0 (i.e., P[n:=0]), we transform its left-hand side to its right-hand side. Similarly, in completing the induction step, we transform the left-hand side of P[n:=n+1] into its right-hand side.

Base Case: n=0 (+i | 1≤i≤n : 2i + 3)[n:=0] = < textual substitution > (+i | 1≤i≤0 : 2i + 3) = < range is empty; (8.13); 0 is identity of addition > 0 = < arithmetic > 02 + 4·0 = < textual substitution > (n2 + 4n)[n:=0]

Inductive Case: Let n≥0 be arbitrary, and assume an an induction hypothesis (IH) that (+i | 1≤i≤n : 2i + 3) = n2 + 4n (+i | 1≤i≤n : 2i + 3)[n:=n+1] = < textual substitution > (+i | 1≤i≤n+1 : 2i + 3) = < (8.23) > (+i | 1≤i≤n : 2i + 3) + (2i+3)[i:=n+1] = < IH, textual substitution, arithmetic > (n2 + 4n) + (2n + 5) = < algebra > (n2 + 2n + 1) + (4n + 4) = < algebra > (n+1)2 + 4(n+1) = < textual substitution > (n2 + 4n)[n:=n+1]

Perhaps a more likely way to have carried out the proof would have been to transform P[n:=0] to true (for the base case) and similarly for P[n:=n+1] (for the induction step). Showing this for the induction step:

Inductive Case: Let n≥0 be arbitrary, and assume P an an induction hypothesis (IH) (i.e., (+i | 1≤i≤n : 2i + 3) = n2 + 4n) P[n:=n+1] = < defn. of P > ((+i | 1≤i≤n : 2i + 3) = n2 + 4n)[n:=n+1] = < textual substitution > (+i | 1≤i≤n+1 : 2i + 3) = (n+1)2 + 4(n+1) = < (8.23), textual substitution > (+i | 1≤i≤n : 2i + 3) + (2(n+1)+3) = (n+1)2 + 4(n+1) = < IH, textual substitution, arithmetic > (n2 + 4n) + (2n + 5) = (n+1)2 + 4(n+1) = < algebra, arithmetic > n2 + 6n + 5 = n2 + 6n + 5 = < = is reflexive > true

2. Prove by (weak) mathematical induction that  (+i | 1≤i≤n : 1/2ⁱ) = 1 - 1/2ⁿ  for all n ≥ 0.

Solution: Omitted, as it is very similar to a solution to Problem 1..

3. Prove by (weak) mathematical induction that  4ⁿ - 1 is divisible by 3, for all n ≥ 0.

Hint: If you divide 4ⁿ⁺¹ - 1 by 4ⁿ - 1, you get a quotient of 4 and a remainder of 3. That is, 4ⁿ⁺¹ - 1 = 4·(4ⁿ - 1) + 3

Solution: The proof is by mathematical induction on n.

Base Case: n=0 (3 | 4n-1)[n:=0] = < textual substitution > 3 | 40-1 = < arithmetic > 3 | 0 = < zero is divisible by every integer except itself > true

Inductive Case: Let n≥0 be arbitrary, and assume an an induction hypothesis (IH) that 4n - 1 is divisible by 3. Let m the quotient when 3 is divided into 4n - 1. That is, let 3m = 4n - 1. (3 | 4n-1)[n:=n+1] = < textual substitution > 3 | 4n+1-1 = < algebra > 3 | 4(4nn - 1) + 3 = < IH > 3 | 4(3m) + 3 = < algebra > 3 | 3(4m + 1) = < Divisibility 1 > true

4. Define f : ℕ → ℕ as follows:

f.0	=	1
f.1	=	3
f.n	=	4·f(n−1) − 3·f(n−2)    (for n≥2)

Prove by (strong) mathematical induction that f.n = 3ⁿ  for all n ≥ 0.

Solution: Omitted.

5. Assume as true the claim that every acyclic graph that contains at least one edge also includes a vertex of degree one. Let G be an connected acyclic graph (i.e., a (free) tree). Prove by mathematical induction that v(G) = e(G) + 1 (i.e., the number of vertices in G is one more than the number of edges in G).

Procedure: Show that the statement is true in the case that v(G) = 1. That is the base case, or basis. Then take as an induction hypothesis (IH) that the statement is true for all trees having exactly n vertices, where n≥1 is arbitrary. Let G be a tree having n+1 vertices. Use the IH in arguing that the statement is true for G. (That is the induction step.)

Solution: By mathematical induction on the number of vertices. For a graph G, let v_G and e_G, respectively, be the number of vertices and edges in G.

Base Case: vG = 1. Hence, vG = eG + 1 = < eG = 0, as any graph having only one vertex has zero edges. > 1 = 0 + 1 = < arithmetic > 1 = 1 = < reflexivity of = > true

Inductive Step: Let n≥1 be arbitrary and assume, as an IH, that every tree having exactly n vertices has exactly n-1 edges (so that vG = eG + 1 for all such trees). Let G be a tree having exactly n+1 vertices. Construct H from G by removing a vertex of degree one and its incident edge. (The problem said that you could assume that such a vertex must exist.) We can make these observations: H is a tree. (Why? Because the removal of a vertex of degree one from a graph cannot disconnect the graph.) vG = vH + 1. In particular, vH = n eG = eH + 1 vG = eG + 1 = < observations above > vH + 1 = eH + 1 + 1 = < vH = n and H is a tree, so by the IH we have vH = eH + 1> eH + 1 + 1 = eH + 1 + 1 = < = is reflexive > true

6.

* / \ / \ / \ * * / \ | / \ | * * * / \ / \ * * | / \ | / \ * * * | | | | * *

A rooted binary tree has one of the three forms shown below, where each of T_C, T_L, and T_R is itself a rooted binary tree. An example appears to the right.

Root is a leaf T0	Root has one child T1	Root has two children T2
*	* | | | | * / \ / \ / TC \ +-------+	* / \ / \ / \ / \ * * / \ / \ / \ / \ / TL \ / TR \ +-------+ +-------+

The height of a rooted tree is equal to the length of a longest path among all paths from the root to the leaves. Recursively, the height of a rooted tree is zero if the root is its only node and, otherwise, the height is one plus the height of its "tallest" proper subtree (which would be a subtree rooted at one of the children of the root).

Referring to the diagram above, we have

• height(T₀) = 0
• height(T₁) = 1 + height(T_C)
• height(T₂) = 1 + (height(T_L) max height(T_R))

Prove by (weak) mathematical induction that, for all n≥0, every rooted binary tree having no less than 2ⁿ nodes has height no less than n.

To clarify, what is to be proved is that, for all n≥0 and all rooted binary trees T,

where v(T) is the number of nodes/vertices in T and h(T) is the height of T.

Hint: Regarding the induction step, suppose that T is a rooted binary tree having at least 2ⁿ⁺¹ nodes. If T is of the form of T₁, how many nodes must be in T_C? If T is of the form of T₂, how many nodes must be in larger among T_L and T_R?

Solution: We offer two solutions, the first of which includes some Gries & Schneider flavor but otherwise is similar to what you would expect to find in most math books; the second solution is more strictly in the style of Gries & Schneider.

Induction step. Assume, as an IH, that for an arbitrary n≥0, every rooted binary tree having at least 2ⁿ nodes has height at least n.

Let T be a rooted binary tree with at least 2ⁿ⁺¹ nodes. To be demonstrated is that h(T) ≥ n+1.

Case 1: The root of T has only one child. Call the subtree rooted there T_C. Then T_C has at least 2ⁿ⁺¹ - 1 nodes, which is at least 2ⁿ. Which means that the IH can be applied to T_C. Also, the height of T_c is one less than that of T.

h(T) = < observation above > h(TC) + 1 ≥ < IH applied to TC says h(TC) ≥ n; number theory a≥b ⇒ a+c ≥ b+c > n + 1

Case 2: The root of T has two children. Call the subtrees rooted at those children T_L and T_R. Here are some observations:

• The height of T is one more than the larger among the heights of T_L and T_R.
• The nodes in T include (only) its root and the nodes in T_L and T_R, which is to say that, combined, T_L and T_R have at least 2ⁿ⁺¹ - 1 nodes. Without loss of generality, suppose that T_L has at least as many nodes as T_R. Then T_L has at least (⌈2ⁿ⁺¹-1)/2⌉ nodes, which is at least 2ⁿ. Which means that the IH applies to T_L, allowing us to conclude that h(T_L) ≥ n

h(T) ≥ < observation above > h(TL) + 1 ≥ < IH applied to TL says h(TL) ≥ n; number theory a≥b ⇒ a+c ≥ b+c > n + 1

Base Case: n=0.

P[n:=0] = < defn of P; text. sub> > v(T) ≥ 20 ⟹ h(T) ≥ 0 = < Every tree has height at least zero > v(T) ≥ 20 ⟹ true = < (3.72) > true

Induction Step: Assume P as an IH. Show P[n:=n+1].

Case 1: The root of T has a single child. Call the tree rooted at that child TC. LHS(P[n:=n+1]) = < defn of P; text. sub. > v(T) ≥ 2n+1 = < v(T) = v(TC) + 1 > v(TC) + 1 ≥ 2n+1 = < algebra > v(TC) ≥ 2n+1 - 1 ⟹ < 2n+1 - 1 ≥ 2n; ≥ is transitive > v(TC) ≥ 2n ⟹ < IH > h(TC) ≥ n = < h(T) - 1 = h(TC) > h(T) - 1 ≥ n = < algebra > h(T) ≥ n+1 = < defn of P; text. sub. > RHS(P[n:=n+1)

Case 2: The root of T has two children. Call the trees rooted at those children TL and TR LHS(P[n:=n+1]) = < defn of P; text. sub. > v(T) ≥ 2n+1 = < v(T) = v(TL) + v(TR) + 1 > v(TL) + v(TR) + 1 ≥ 2n+1 = < algebra > v(TL) + v(TR) ≥ 2n+1 - 1 ⟹ < number theory: a + b ≥ c ⟹ 2·(a max b) ≥ c > 2·(v(TL) max v(TR)) ≥ 2n+1 - 1 = < algebra; all quantities here are integers > v(TL) max v(TR) ≥ ⌈(2n+1 - 1) / 2⌉ = < algebra > v(TL) max v(TR) ≥ ⌈(2n - 1/2)⌉ = < number theory: 0≤α<1 ⟹ ⌈a - α⌉ = a > v(TL) max v(TR) ≥ 2n = < number theory: a max b ≥ c ⟹ a≥c ∨ b≥c > v(TL) ≥ 2n ∨ v(TR) ≥ 2n ⟹ < IH twice; (4.2) twice > h(TL) ≥ n ∨ h(TR) ≥ n = < number theory: a≥c ∨ b≥c ≡ a max b ≥ c > h(TL) max h(TR) ≥ n = < h(T) - 1 = h(TL) max h(TR) > h(T) - 1 ≥ n = < algebra > h(T) ≥ n+1 = < defn of P; text. sub > RHS(P[n:=n+1])