Mathematics 220
Homework 7
Due March 3/4
1. Question 6.2. Prove that any non-empty subset of a well-ordered set of real numbers
is well-ordered.
Solution:
Proof. Let S be a non-empty subset of B. Since S ⊆ B, it follows that S ⊆ A. Since
A is well-ordered, every subset of A has a least element. Thus S has a least element.
Hence B is also well-ordered.
2. Prove that for all n ∈ N
1
1
1
n
+
+···+
=
.
1·2 2·3
n(n + 1)
n+1
Solution:
Proof. We proceed by induction. Let P (n) be the statement
• When n = 1 we have
1
2
=
1
1+1
1
1·2
1
+ · · · + n(n+1)
=
n
.
n+1
so the statement is true.
• Assume P (k) is true, so
1
k
1
+···+
=
1·2
k(k + 1)
k+1
thus
1
1
1
k
1
+···+
+
=
+
1·2
k(k + 1) (k + 1)(k + 2)
k+1
(k + 1)(k + 2)
2
k + 2k + 1
(k + 1)2
=
=
(k + 1)(k + 2)
(k + 1)(k + 2)
k+1
=
k+2
Thus P (k + 1) is true.
By the principle of mathematical induction, the statement is true for all n ∈ N.
3. Let a, b ∈ Z and m. Using mathematical induction, prove that if a ≡ b mod m then
an ≡ bn mod m for all n ∈ N.
Page 1 of 5
Mathematics 220
Homework 7
Due March 3/4
Solution: We proceed by induction. The base case is n = 1: then it is given that
a ≡ b mod m and there is nothing to prove. Induction step: assume we know that
if a ≡ b mod m, then an ≡ bn mod m. We need to prove that an+1 ≡ bn+1 mod m.
We have proved a theorem stating that congruences can be multiplied: if a ≡ b
mod m and c ≡ d mod m, then ac ≡ bd mod m. Use this theorem with c = an and
d = bn , and we obtain that an+1 ≡ bn+1 mod m. This completes the proof of the
induction step, and then by the Principle of Mathematical Induction, the statement
is proved.
4. Let A be a finite set, and let P(A) be the set of all its subsets. Prove the fact that
|P(A)| = 2|A| by mathematical induction.
Solution: We use induction on the number of elements of A (let |A| = n).
Base case: n = 0, that is, A = ∅. Then P (A) = {∅}, which consists of 1 element,
and 1 = 20 . Induction step: assume that for any set A of n elements, |P (A) | = 2n .
We need to prove that for a set of n + 1 elements, |P (A) | = 2n+1 .
Let A be a set of n + 1 elements. Let x ∈ A. Consider the set B = A − {x}. Now the
subsets of A can be broken into two classes: the subsets containing x and the subsets
not containing x. That is, P (A) is a disjoint union of two sets P1 and P2 , where
P1 = {S ⊆ A : S ∋ x}, and {S ⊆ A : x ∈
/ S}. (Note that A ∈ P1 and ∅ ∈ P2 ). By
definition, the set P2 is in fact the power set of B, since a subset of A not containing
x is precisely the same thing as a subset of A − {x}. Then by induction assumption,
|P2 | = 2n . It remains to observe that each element S ∈ P1 can be represented as
S = {x} ∪ S1 , where S1 = S − {x} ⊆ B. Thus, the number of elements in P1 is the
same as the number of subsets of B as well. We get:
|P (A) | = |P1 | + |P2 | = 2|P2 | = 2 · 2n = 2n+1 ,
which completes the proof of the induction step. By the Principle of mathematical
induction, |P (A) | = 2|A| for any finite set A.
5. Use induction to prove DeMorgan’s law for n sets (n ≥ 1): let A1 , . . . , An be sets; prove
that
A1 ∪ · · · ∪ An = A1 ∩ A2 ∩ · · · ∩ An .
Solution: The result for n = 1 simply states that A1 = A1 which is trivially true.
Assume now the results holds for n ∈ N and consider n + 1. Apply DeMorgan’s law
(for the usual n = 2 case) with A = A1 ∩ A2 ∩ · · · ∩ An and B = An+1 to conclude
that
A1 ∩ · · · ∩ An ∩ An+1 = A ∩ An+1 = A ∪ An+1 .
Page 2 of 5
Mathematics 220
Homework 7
Due March 3/4
Use the induction hypothesis to see that the righthand side of the above equals
(A1 ∪ A2 ∪ · · · ∪ An ) ∪ An+1 = A1 ∪ A2 ∪ · · · ∪ An+1 .
This proves the result for n+1. By the principle of mathematical induction the result
holds for all natural numbers n.
6. Prove that for all natural numbers n ≥ 10, 2n > n3 .
Solution: We prove this by induction. Base case: n = 10 (since we are trying to
prove the statement for n ≥ 10). Then 210 = 1024, n3 = 1000, and the statement
holds.
Induction step. Assume that k ≥ 10, and we know that 2k > k 3 . We need to prove
that 2k+1 > (k + 1)3 . We have: (k + 1)3 = k 3 + 3k 2 + 3k + 1. Note that for any k > 1,
3k < 3k 2 , and so for any k > 1, we have 3k 2 + 3k + 1 < 7k 2 . Since k > 10, we have
that 7k 2 < k 3 . Then we get:
k 3 + 3k 2 + 3k + 1 < k 3 + 7k 2 < k 3 + k 3 = 2k 3 .
Now, by induction assumption, k 3 < 2k . Then, by the above inequality, (k + 1)3 <
2k 3 < 2 · 2k = 2k+1 . Thus, the induction step is proved, and by the Principle of
mathematical induction, the statement holds for all n ≥ 10.
7. Let F1 , F2 , . . . , Fn . . . , be the sequence of Fibonacci numbers: by definition, F1 = F2 = 1,
and the sequence is defined recursively by the formula Fn = Fn−1 + Fn−2 for n ≥ 3.
(Thus, we get the sequence 1, 1, 2, 3, 5, 8, 13, 21, . . . . ) Prove the formula
√ !n !
√ !n
1− 5
1
1+ 5
.
−
Fn = √
2
2
5
Two remarks: (1) Note that it’s not even completely obvious why the expression on the
right is a rational number! But you will prove that
√ it is, in fact, an integer, namely, the
n-th Fibonacci number. (2) The number ϕ = 1+2 5 is called the golden ratio.
√
√
Solution: Let ϕ = 1+2 5 and let ψ = 1−2 5 . The key fact is that ϕ and ψ are solutions
of x2 −x−1 = 0. Note also that ϕ + ψ = 1. (check this using the quadratic formula!).
Now proceed by induction. If n = 1 then
1 √
1
√ (ϕ − ψ) = √ 5 = 1 = F1 .
5
5
If n = 2, then we have
1
1
√ (ϕ2 − ψ 2 ) = √ (ϕ − ψ)(ϕ + ψ) = 1 = F2 .
5
5
Page 3 of 5
Mathematics 220
Homework 7
Due March 3/4
Assume the result holds for Fk for k ≤ n and consider Fn+1 . We have already checked
the formula for F2 and so we may assume n ≥ 2. Then
1
Fn+1 = Fn + Fn−1 = √ (ϕn − ψ n + ϕn−1 − ψ n−1 )
5
1
= √ (ϕn−1 (1 + ϕ) − ψ n−1 (ψ + 1))
5
1
= √ (ϕn−1 ϕ2 − ψ n−1 ψ 2 ),
5
where in the last line we again use the quadratic equation satisfied by ϕ and ψ. The
above gives us Fn+1 = √15 (ϕn+1 −ψ n+1 ), and so the formula holds for n+ 1. By strong
induction the result holds for all natural numbers n.
8. A graph is called complete if any two vertices are connected by an edge. Prove that the
complete graph with n vertices contains precisely n(n − 1)/2 edges.
Solution: As we discussed in class, the number of edges in a graph is half the sum
of degrees of its vertices. In a complete graph, every vertex is connected with every
other vertex by an edge, so the degree of each vertex is n − 1. Then the sum of the
degrees is n(n − 1), and the number of edges is n(n − 1)/2.
Note that you could also use induction on n, but it would be longer.
9. A graph is called a tree if it is connected and contains no cycles (equivalently, there is
precisely one path from any vertex v1 to any other vertex v2 ). Prove that:
(a) Any tree must contain at least one vertex of degree 1 (that is, a vertex with only
one edge leading to it; sometimes such a vertex in a tree called “a leaf”). Hint:
consider the longest possible path in this graph, and think about its ends.
Solution: As the hint suggests, consider the longest possible path in this graph
(that is, the path with the greatest number of vertices in it; note that it can
never visit the same vertex twice because the graph is a tree). Since the graph
has no cycles, this path has to have two ends (cannot begin and end at the same
vertex). Let v1 and v2 be the ends of this path. We claim that both v1 and v2
are leaves. We prove this by contradiction: suppose v2 was not a leaf. Our path
arrives at v2 at the end via some edge. If v2 has degree greater than 1, then
there is another edge starting at v2 , and so we can continue walking on that
edge, and this would allow us to extend our path – a contradiction, we assumed
our path was the longest possible. The same argument works for v1 .
(b) Prove that any tree with n vertices contains precisely n − 1 edges.
Page 4 of 5
Mathematics 220
Homework 7
Due March 3/4
Solution: We prove this by induction on the number of vertices in a tree.
Base case: n = 1, then there are no edges, and the statement is true. If n = 2,
since the graph is required to be connected, there has to be an edge between the
two vertices, so again we get that the statement holds.
Induction step. Assume that any tree with n vertices has n − 1 edges. Consider
a tree with n + 1 vertices. We need to prove that it has n edges. It would be
very convenient to be able to erase a vertex; but we have to be careful – the
graph that is left has to be a tree; of course it would not have cycles, but it
would need to be connected. So we cannot just erase an arbitrary vertex to
reduce the question to the induction assumption. Luckily, in Part (a) we proved
that our tree with n + 1 vertices has to have a leaf (in fact, at least two, but we
only need one). Let us remove one leaf and its edge. Then since the edge was
leading only to this leaf, the graph remains connected, and so it is a tree. Then
the remaining graph has n vertices and n − 1 edges, and adding our vertex and
its edge back, we see that the original graph had n edges, and tis completes the
proof of the induction step.
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