1. SUBTOPIC 5 : MATHEMATICAL INDUCTION
Mathematical induction can be used to prove statements that assert that ∀ n ∈ , P (n) is
true. But, it is important to note that mathematical induction can be used only to prove results
obtained in some other way. It is not a tool for discovering formula or theorem.
Principle of Induction. To prove that P (n) is true for all positive integers n, where P (n)
is a propositional function. A proof by mathematical induction has two parts:
Basic step: We verify that P (1) is true.
Inductive step: We show that the conditional proposition ∀ k ∈ , P (k) ⇒ P (k + 1) is
true.
In the induction step, the assumption that P (n) holds is called the Induction Hypothesis
(IH). In more formal notation, this proof technique can be stated as
[P (0) ˄∀ k (P (k) ⇒ P (k + 1) ∀n P (n)
You can think of the proof by (mathematical) induction as a kind of recursive proof:
Instead of attacking the problem directly, we only explain how to get a proof for P(n + 1) out of a
proof for P(n).
How would you prove that the proof by induction indeed works??
Proof (by contradiction) Assume that for some values of n, P(n) is false. Let be the least such
n that P_n0_ is false. Cannot be 0, because P(0) is true. Thus, must be in the form =1
+ . Since < then by P ( is true. Therefore, by inductive hypothesis P ( + 1) must be
true. It follows then that P ( is true.
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2. The base case can start with any nonnegative number. If that number is then you would prove
that assertion P(n) holds for all n ≥ .
[P ( ) ˄∀ k ≥ (P (k) ⇒ P (k + 1) ∀n P (n)
The induction step not necessarily should start with n. You can change the step from n -1 to n,
where n > 0. Sometimes this yields slightly shorter expressions. However, you cannot make a
step from n - 1 to n + 1.
Example:
Prove for ≥ 1
1 x 1! + 2 x 2! + 3 x 3! + ... + n x n! = (n + 1)! - 1
This could be also written by using ∑ notation
If you take 15-355 (http://www.andrew.cmu.edu/course/15-355/) you will learn a more general
approach for deriving and proving combinatorial identities.
Proof
Base case: n _ 1
The left hand side is 1x1! The right hand side is 2! - 1. They are equal.
Inductive hypothesis: Suppose this holds
We need to prove
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3. Consider the left hand side
(n+1)! – 1 + (n+1) x (n+1) =
(n+1)! (1+n+1) -1 = (n+2)! -1
Suppose we have a sequence of propositions which we would like to prove:
P (0), P (1), P (2), P (3)… P (n)
We can picture each proposition as a domino:
P (k)
A sequence of propositions is visualized as a sequence of dominos.
P (0) P (1) P (2) P (k) P
….. (k+1)
When the domino falls (to right), the corresponding proposition is considered true:
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4. P (k)
true
Suppose that the dominos satisfy two constraints.
1. Well positioned: if any domino falls to right, the next domino to right must fall also.
P (k) P
true (k+1)
true
2. First domino has fallen to right.
P (0) true
Than can conclude that all the domino fall.
P (0) P (1) P (2) P (k) P
….. (k+1)
P (0) P (1) P (2) P (k) P
….. (k+1)
true
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5. P (1) P (2) P (k) P
….. (k+1)
P (0) true true
true
P (2) P (k) P
….. (k+1)
P (0) true P (1) true True
true P (k) P
….. (k+1)
P (0) true P (1) true P (2) true
P (k) P
….. (k+1)
P (0) true P (1) true P (2) true true true
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6. P (0) true P (1) true P (2) true true ….. P (k) true P (k+1) true
Principle of Mathematical Induction.
If:
1. Basis P(1) is true
2. Induction ∀ n P(k) P(k+1) is true
P (0) true P (1) true P (2) true true ….. P (k) true P (k+1) true
Then, ∀ n P (n) is true.
Use induction to prove that the sum of the first n odd integers is
Yeah!
Base case (n=1): the sum of the first 1 odd integer is
Assume p (k): the sum of the first k odd integers is =1
Where,
1 + 3 + … + (2k+1) =
Prove that
1 + 3 + … + + (2k-1) + (2k+1) =
1 + 3 + … + + (2k-1) + (2k+1) =
=
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7. Prove that: 1 x 1! + 2 x 2! + … + n x n! = (n+1)! -1, ∀ n
Base case (n=1): 1 x 1! = (1 x 1)! -1? 1 x 1! = 1
Assume P (k): 1 x 1! + 2 x 2! + … + k x k! = (k+1)! -1
2! – 1 = 1
Prove that: 1 x 1! + 2 x 2! + … + k x k! + (k+1) (k+1)! = (k+2)! -1
1 x 1! + 2 x 2! + … + k x k! + (k+1) (k+1)! = (k+1)! -1+ (k+1) (k+1)
= (1 + (k+1)) (k+1)! – 1
= (k+2) (k+1)! -1
= (k+2)! - 1
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8. EXERCISE:
1. For every n ≥ 4, n! >
2. Proof that,
We have seen that for any natural number n ,
1 + 2 + ... + n = n( n + 1 )/2
and
12 + 22 + ... + n2 = n( n + 1 )( 2n + 1 )/6
hold.
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9. ANSWER:
1. Proof:
In this problem .
Basis Step: If n = 4, then LHS = 4! = 24, and RHS = .
Hence LHS > RHS .
Induction: Assume that n! > for an arbitrary n ≥ 4 . -- Induction Hypothesis
To prove that this inequality holds for n+1, first try to express LHS for n+1 in terms of
LHS for n and try to use the induction hypothesis.
Note here (n + 1)! = (n + 1) n!
Thus using the induction hypothesis, we get (n + 1)! = (n + 1)n! > (n + 1) .
Since n ≥ 4 , (n+1) > 2.
Hence, (n + 1) > .
Hence, (n + 1)! > .
2. One might then wonder what 13 + 23 + ... + n3 would be equal to.
One way to find that out is to make a conjecture (i.e. educated guess) and prove the
conjecture.
Let us first take a guess.
By looking at the sums 1 + 2 + ... + n = n( n + 1 )/2 --- a function of n2
and 12 + 22 + ... + n2 = n( n + 1 )( 2n + 1 )/6 --- a function of n3 ,
one can guess that 13 + 23 + ... + n3 is equal to some function of n4, that is
13 + 23 + ... + n3 = an4 + bn3 + cn2 + dn + e ---- (1) for some constants a, b, c, d and e.
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10. Now to determine the value of a, b, c, d and e, we compare the value of the both sides of
the equation (1) for five values of n.
For n = 0, LHS = 0, and RHS = e. Hence we get e = 0.
Similarly for n = 1, 1 = a + b + c + d,
for n = 2, 9 = 16a + 8b + 4c + 2d,
for n = 3, 36 = 81a + 27b + 9c + 3d , and
for n = 4, 100 = 256a + 64b + 16c + 4d .
Solving this system of equations we obtain
a = c = 1/4, b = 1/2, and d = 0.
Hence our conjecture is
13 + 23 + ... + n3 = 1/4 n2 ( n2 + 2n + 1 ) = ( 1/2 n ( n + 1 ) )2 .
What we do next is to try to prove it by mathematical induction.
Proof by Mathematical Induction:
Basis Step: For n = 0, LHS = 03 = 0 , and
RHS = ( 1/2* 0 ( 0 + 1 ) )2 = 0 .
Hence LHS = RHS.
Induction: Assume that 13 + 23 + ... + n3 = ( 1/2 n ( n + 1 ) )2 for an arbitrary n. ---------
Induction Hypothesis
Then for n + 1, LHS = 13 + 23 + ... + n3 + ( n + 1 )3
= (13 + 23 + ... + n3) + ( n + 1 )3
= ( 1/2 n ( n + 1 ) )2 + ( n + 1 )3 by the induction hypothesis.
= ( 1/4 ( n + 1 )2( n2 + 4*( n + 1 ) )
= ( 1/2 ( n + 1 )( n + 2 ) )2 ,
which is equal to RHS.
End of Proof
Thus we now know for sure that
13 + 23 + ... + n3 = ( 1/2 n ( n + 1 ) )2 .
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