This document contains solutions to 5 math problems proved using mathematical induction and other techniques: 1) It proves via induction that the sum of the given fractions equals n/(n+1) for all n in the natural numbers. 2) It proves via induction that 5n-1 is divisible by 4 for all n in the natural numbers. 3) It proves that if sets A and B have sizes n and m respectively, with m>n, then no function from A to B can be onto. 4) It proves that the function f(a,b) = (b,a) mapping A x B to B x A is both 1-1 and onto, and therefore the

1. Math 189, Section 2, Exam 3 Solutions
Tyler Murphy
April 4, 2014
1
Use the Principle of Mathematical Induction to prove that:
1
1 ∗ 2
+
1
2 ∗ 3
+
1
3 ∗ 4
+ · · · +
1
n(n + 1)
=
n
n + 1
for all n ∈ N
Proof. (By induction)
Base Case: n = 1.
1
1(1 + 1)
?
=
1
1 + 1
.
1
1(1 + 1)
=
1
1(2)
=
1
2
=
1
1 + 1
.
So the base case is true.
I.H. Assume that
1
1 ∗ 2
+
1
2 ∗ 3
+
1
3 ∗ 4
+· · ·+
1
k(k + 1)
=
k
k + 1
is true for k ∈ N, k ≤ n.
WTS:
1
1 ∗ 2
+
1
2 ∗ 3
+
1
3 ∗ 4
+ · · · +
1
k(k + 1)
+
1
(k + 1)(k + 1 + 1)
=
k + 1
(k + 1) + 1
1
1 ∗ 2
+
1
2 ∗ 3
+
1
3 ∗ 4
+ · · · +
1
k(k + 1)
+
1
(k + 1)(k + 1 + 1)
By assumption,
1
1 ∗ 2
+
1
2 ∗ 3
+
1
3 ∗ 4
+ · · · +
1
k(k + 1)
=
k
k + 1
.
So we have:
1

2. k
k + 1
+
1
(k + 1)(k + 1 + 1)
.
=
k
k + 1
+
1
(k + 1)(k + 2)
.
=
k(k + 2)
(k + 1)(k + 2)
+
1
(k + 1)(k + 2)
.
=
k2 + 2k + 1
(k + 1)(k + 2)
.
=
(k + 1)(k + 1)
(k + 1)(k + 2)
.
=
k + 1
k + 2
.
=
k + 1
k + 1 + 1
.
This what we wanted to show.
2
Use the Principle of Mathematical Induction to prove that:
5n − 1 is divisible by 4 for all n ∈ N.
Proof. By Induction.
Base Case: n = 1.
51 − 1 = 4, which is clearly divisible by 4.
I.H. Assume 5k − 1 is divisible by 4 for k ≤ n, k ∈ N.
WTS. 5k+1 − 1 is divisible by 4.
5k+1
− 1 = 5 ∗ 5k
− 1.
= 5 ∗ 5k
− 5 + 5 − 1.
= 5(5k
− 1) + 4.
By assumption, 5k − 1 is divisible by 4, so 5 times that quantity is also. Clearly, 4 is
divisible by 4. So 5(5k − 1) + 4 is divisible by 4.
So, 5k+1 − 1 is divisible by 4.
2

3. 3
Suppose A and B are finite sets with A containing n elements and B containing m
elements.
Prove that if m > n, then no function A → B can be onto.
Proof. For any a ∈ A, if f : A → B is a function, then there is only one b ∈ B such that
f(a) = b.
Since A has only n elements, then there are n elements in B that are used by f. Since
B has m elements and m > n, there are m − n elements in B that are not hit by the
function. So f cannot be onto and still be a function.
4
Let A and B be sets.
Define f : A x B → B x A by the rule f(a, b) = (b, a).
(a) Prove that f is 1-1.
(b) Prove that f is onto.
(c) Prove that | A x B |=| B x A |.
(a)
Prove that f is 1-1.
Proof. Suppose that f(a, b) = f(m, n).
f(a, b) = (b, a) and f(m, n) = (n, m).
So (b, a) = (n, m).
So b = n and a = m.
So (a, b) = (m, n).
So f is 1-1.
(b)
Prove that f is onto.
Proof. Let (t, s) ∈ B x A.
Then t ∈ B and s ∈ A.
So (s, t) ∈ A x B.
3

4. So f(s, t) = (t, s).
So (t, s) ∈ ran(f).
So f is onto.
(c)
Prove that | A x B |=| B x A |.
Proof. Since f : AxB → BxA is 1-1 and onto, then we have a bijection from AxB to BxA.
By theorem, if a bijection exists between two sets, then the two sets have the same
cardinality.
So | A x B |=| B x A |.
5
Let S be the set of all real numbers in the interval (0,1) whose decimal expansions
contain only 0’s and 1’s. That is,
S = {0.ai1ai2ai3ai4ai5 · · · ∈ (0, 1) | aij = 0 or 1, i, j ∈ N}
Prove that S is uncountable
Proof. By contradiction
Suppose S is countable.
Then S is listable. Generate this list.
Let si = 0.ai1ai2ai3ai4ai5 · · · ∈ (0, 1) | aij = 0 or 1, i, j ∈ N.
We now find a number x ∈ S that is not in this list.
Define x = 0.yi1yi2yi3yi4yi5 . . . where if aij = 1 ∈ si, then yij = 0. If aij = 0, then
yij = 1.
So x differs from each si in the list in the ijth position.
So x is not in the list.
So S is uncountable.
4