Math 120 Notes
1.3 Proof by Induction
Deductive vs. Inductive Reasoning
Deduction – drawing conclusions by reasoning from general principles
(using logic to go from the general to the specific)
e.g. When a bird is born, it hatches from an egg.
An ostrich is a bird.
Therefore, when an ostrich is born, it hatches from an egg.
e.g. Given: diagram, line 1 is parallel to line 2
Prove: <1 + <2 + <3 = 180o
Statements
Reasons
l1
l2
given
1  4
TPLT (alternate)
2  5
TPLT (alternate)
completion of straight angle
4  5  3  180
replacement property
1  2  3  180
Induction – drawing a general conclusion from a number of individual facts
(going from the specific to the general)
e.g. If the only swans you ever saw were white, it would be reasonable to conclude
that all swans are white. One black swan would, however, disprove your conclusion.
Mathematical Induction
There is a kind of inductive reasoning that allows us to be sure of the truth of our conclusion
involving only Natural numbers. It is called mathematical induction. If the method is carried out properly,
it must lead to a true conclusion. The rules for this four-step approach are as follows:
Step 1: Check that the statement is true for n = 1.
Step 2: Assume that the statement is true for n = k.
Step 3: Prove that the statement is true for n = k + 1, if it is true for n = k, where k  N.
Step 4: Write a conclusion for all Natural numbers.
Proving Summation Formulas
1  2  3  ...  n 
1. The Sum of the First n Natural numbers:
n(n  1)
2
Step 1: Check that the statement is true for n = 1.
L.S. = 1
R.S. =
1(1  1) 1(2)

1
2
2
L.S. = R.S.  True for n =1.
Step 2: Assume that the statement is true for n = k.
Assume 1  2  3  ...  k 
k (k  1)
2
Step 3: Prove that the statement is true for n = k + 1.
L.S.  1  2  3  ...  k  (k  1)
k (k  1)

 (k  1)
2
k (k  1) 2(k  1)


2
2
2
k  3k  2

2
(k  1)[(k  1)  1]
2
(k  1)(k  2)

2
2
k  3k  2

2
R.S. 
Step 4: Write a conclusion for all Natural numbers.
1  2  3  ...  n 
n(n  1)
,n N
2
L.S. = R.S.  True for n = k + 1.
2. The Sum of the Squares of the First n Natural Numbers:
12  22  32  ...  n 2 
n(n  1)(2n  1)
6
Step 1: Check that the statement is true for n = 1.
L.S. = 12  1
R.S. =
1(1  1)[2(1)  1] 1(2)(3) 6

 1
6
6
6
L.S. = R.S.  True for n =1.
Step 2: Assume that the statement is true for n = k.
Assume 12  22  32  ...  k 2 
k (k  1)(2k  1)
6
Step 3: Prove that the statement is true for n = k + 1.
L.S.  12  22  32  ...  k 2  ( k  1) 2
k (k  1)(2k  1)

 (k  1) 2
6
k (k  1)(2k  1) 6(k  1) 2


6
6
2
2
k (2k  3k  1) 6(k  2k  1)


6
6
3
2
2
2k  3k  k  6k  12k  6

6
3
2
2k  9k  13k  6

6
(k  1)[(k  1)  1][2(k  1)  1]
6
(k  1)(k  2)(2k  3)

6
2
(k  3k  2)(2k  3)

6
3
2
2k  3k  6k 2  9k  4k  6

6
3
2
2k  9k  13k  6

6
R.S. 
L.S. = R.S.  True for n = k + 1.
Step 4: Write a conclusion for all Natural numbers.
12  22  32  ...  n 2 
n(n  1)(2n  1)
,n N
6
3. The Sum of the Cubes of the first n Natural Numbers:
 n(n  1) 
1  2  3  ...  n  
 2 
3
3
3
2
3
Step 1: Check that the statement is true for n = 1.
L.S. =
R.S. =
Step 2: Assume that the statement is true for n = k.
Assume
Step 3: Prove that the statement is true for n = k + 1.
Step 4: Write a conclusion for all Natural numbers.
The previous three proofs may be summarized using sigma notation.
n(n  1)
j

2
j 1
n
n(n  1)(2n  1)
j 

6
j 1
n
2
n

j 1
 n(n  1) 
j 
 2 
3
2