MATH 365 HANDOUT 1 - DIVISIBILITY & INDUCTION
1. Divisibility
(1) Suppose that n is an integer such that 5 | (n + 2). Which of the following are divisible by 5?
(a) n2 − 4
(b) n2 + 8n + 7
(c) n4 − 1
(d) n2 − 2n
(2) Prove that the square of any integer of the form 5k + 1 for k ∈ Z is of the form 5k 0 + 1 for some
k 0 ∈ Z.
(3) Show that if ac | bc and c 6= 0, then a | b.
2. Induction
We show the general outline for a proof by induction via the following example.
Example. Show that for every positive integer n we have
1 + 2 + 3 + ··· + n =
n(n + 1)
.
2
Proof Outline:
a) The property that we want to prove for a general positive integer n is
P(n) :
1 + 2 + 3 + ··· + n =
n(n + 1)
2
b) In order to proceed by induction, we need to show that it holds for our “base case,” which in this
case is P(1), because 1 is the first positive integer. So we need to verify that
P(1) :
1=
1(1 + 1)
2
is a true statement.
c) Now we proceed to the “inductive step” or the “induction hypothesis,” which is assumption that
P(k) holds for some positive integer k. We show using this hypothesis that P(k + 1) holds. This
allows us to build up to any positive integer, because we know it holds for the first one, therefore
showing that P(n) is true for every positive integer!
Proof: We will show that
1 + 2 + 3 + ··· + n =
n(n + 1)
2
holds by induction on n.
The base case is when n = 1. Since 1(1 + 1)/2 = 2/2 does equal one, so it holds for n = 1.
1
2
MATH 365 HANDOUT 1 - DIVISIBILITY & INDUCTION
Now assume that for some positive integer k we have that
1 + 2 + 3 + ··· + k =
k(k + 1)
.
2
We will show that the claim is true for n = k + 1. We have
k(k + 1)
k(k + 1) 2(k + 1)
+k+1=
+
2
2
2
k(k + 1) + 2(k + 1)
(k + 1)(k + 2)
=
=
,
2
2
1 + 2 + 3 + · · · + k + (k + 1) =
where the finally equality holds by factoring out (k + 1).
Therefore the property holds for k + 1, so by induction, it holds for all positive integers n.
Exercises:
(1) Prove that for every positive integer n, we have
1 3 + 2 3 + 3 3 + · · · + n3 =
n2 (n + 1)2
.
4
(2) Prove that every integer is either even or odd.
(3) Prove that for any integer n ≥ 1, n2 is the sum of the first n odd integers. (For example, 32 = 1+3+5.)
(4) Show that 7n − 1 is divisible by 6 for all integers n ≥ 0.