An Introduction to Proof by Mathematical Induction
Principle of Mathematical Induction:
Let Sn be a statement about the positive integer n. Suppose that
1. S1 is true
2. Sk+1 is true whenever Sk is true.
Then Sn is true for all positive integers n.
To make the idea clearer, let's look at a row of dominoes, lined up and ready to be pushed over.
We know from experience that if we push over one domino, the rest of the dominoes will fall over.
How can we prove this?
A. We know from experience that if we push over one domino, it should fall over.
B. We also know that if a domino is falling and has been placed correctly, it will knock over its neighbor.
Intuitively, it should be very clear that the fall should cascade all the way up to the last domino. That
is, if the next to the last domino falls, the last domino also falls.
But now we need to think about increasing the rigor of this argument by doing a real proof.
A Formal Proof:
Let's look at the behavior of the dominoes. Assume that there's some domino k, which doesn't fall over.
Since k is the first such domino, the domino right before k must have fallen over. But we know from B
that a falling domino always knocks its neighbor over. So domino k will fall over and we have a
contradiction.
What we have shown above is that because
1. We can knock over the first domino,
2. and a falling domino knocks over its neighbor.
then all the dominoes will fall over.
Now, if we think of each domino as an instance of a proposition. If a given instance is true, the
corresponding domino will fall over. Given a sequence of instances (row of dominoes):
If we can prove:
1. The proposition is true in the first instance.
2. And if a given instance is true, the next one in the sequence will also be true.
Then the proposition will be true in all instances.
This is called a Proof By Induction.
The general strategy is this:
1.
2.
3.
4.
State Sn
Show S1 is true – this is called the Initial Step. (Really just the first – need not be S1.)
Assume Sk is true (state Sk) and show Sk+1 is true (state Sk+1) – this is called the Induction Step.
You’re done! State the conclusion: Since S k  S k 1 , Sn is true.
Example 1
n
Prove that
i  1 2  3  n 
i 1
Proof: Let S n : 1  2  3    n 
?
Initial Step: S1 : 1 
nn  1
.
2
nn  1
(Note: This was not S n  1 2... , but S n : 1  2... )
2
11  1
yes
2
Induction Step:
k k  1
,
2
k  1k  2
show S k 1 : 1  2  3    k  k  1 
2
k k  1
Sk is true, so 1  2  3    k 
. If we add k  1 to both sides, we get:
2
k k  1
1  2  3    k  k  1 
 k  1
2
k k  1 2k  1 k k  1  2k  1 k 2  3k  2 k  1k  2





2
2
2
2
2
So S k  S k 1
Assume S k : 1  2  3    k 
 S n is true by the Principle of Math Induction
Example 2:
Prove that if n is a positive integer, then 7 n  1 is divisible by 6.
Proof: Let S n : 7 n  1 is divisible by 6.
Initial Step: S1 : 71  1  6 divisible by 6? yes
Induction Step:
Assume S k : 7 k  1 is divisible by 6, show S k 1 : 7 k 1  1 is divisible by 6.
If 7 k  1 is divisible by 6, 7 k  1  6i , where i is some integer.
7 k  1  6i

7 k  6i  1
 7 k 1  76i  1  7  6i  7  67i   6  1  67i  1  1
 7 k 1  1  67i  1
but 7i  1 is an integer, so 7 k 1  1  6 j , where j is an integer, so 7 k 1 is divisible by 6.
So S k  S k 1
 S n is true by the Principle of Math Induction
Example 3:
Prove that for n  5 , 2 n  n 2 .
Proof: Let S n : for n  5 , 2 n  n 2 .
?
Initial Step: S 5 : 2 5  5 2 . yes 32  25 .
Induction Step:
2
Assume S k : for k  5 , 2 k  k 2 , show S k 1 : 2 k 1  k  1 .
We know Sk is true, so 2 k  k 2 . Multiplying both sides by 2,
2  2k  2  k 2
 2 k 1  k 2  k 2  k 2  2k  1
 2 k 1  k  1
(Since k 2  2k  1 for k  5)
2
So S k  S k 1
 S n is true by the Principle of Math Induction