Download Principle of Mathematical Induction

Section 1.6
1
Proof by Induction
Section 1.6
1.6: Proof by Mathematical Induction
Purpose of Section:
Section To introduce the Principle of Mathematical Induction, both
weak and the strong versions, and show how certain types of theorems can be
proven using this technique.
Introduction
The Principle of Mathematical Induction is a method of proof normally
used to prove that a proposition is true for all natural numbers 1,2,3,…
, although there are many variations of the basic method. The method is
particularly important in discrete mathematics, and one often sees theorems
proven by induction in areas like computer science. The technique is so
intuitive and familiar that it sometimes is used without reference to its use.
For example, suppose someone tells you they are going to color the natural
numbers 1,2,3, …
with some color and that the number 1 will be colored
blue, and that if a given number is colored blue, then the next number will also
be colored blue. Is there any doubt in your mind that all the numbers will be
colored blue? Of course not. This is the induction axiom. And the good thing
is you don’t have to proof it. It is an axiom1.
1
In 1889 Italian mathematician Guiseppe Peano (1858-1932) published a list of five
axioms which define the natural numbers.
Peano’s 5th axiom is called the induction
axiom, which states that “any property which belongs to 1 and also to the successor of
any number which has the property belongs to all numbers.”
Section 1.6
2
Proof by Induction
Margin Note: Don’t confuse mathematical induction with inductive reasoning
associated with the natural sciences. Inductive reasoning in the sciences is a
scientific method whereby one induces general principles from observations.
Mathematical induction is not the same thing: it is a deductive form of
reasoning used to establish the validity of a proposition for all natural
numbers.
Mathematical induction provides a convenient way to establish that a
statement is true for all natural numbers 1,2,… . The following statements are
prime candidates for proof by mathematical induction.
▶
▶
▶
For all natural numbers n , 1 + 3 + 5 + + (2n − 1) = n 2
If A is a set contains n elements, then the collection of
all subsets of A contain 2n elements.
1 3 5
2n − 1
1
⋅ ⋅ ⋅⋅
≤
⋅ for all natural numbers n .
2 4 6
2n
3n + 1
Here, then is how the method of mathematical induction works.
Mathematical Induction
The Principle of Mathematical Induction is a method of proof for
verifying that a proposition P ( n) is true for all natural numbers n . The
methodology for proving theorems by induction is as follows.
Methodology of Mathematical Induction
To verify that a proposition P ( n) holds for all natural numbers n , the
Principle of Mathematical Induction consists of successfully carrying out the
following two steps.
i Base Case:
Case: Prove that P (1) is true.
i Induction Step: Assume that P (n) is true for an arbitrary n , then
prove that P ( n + 1) is true.
Note: There are several; modifications of the basic induction proof stated
here. For example, there is no reason the base case starts with P (1) . If the
base case is replaced by the verification of P ( a ) , where " a " is any integer
(positive or negative), one would conclude P ( n) true for all n ≥ a . Also, if the
3
Section 1.6
Proof by Induction
induction step is replaced by the implication P ( n) ⇒ P ( n + 2) , one would
conclude P ( n) true for P (1), P (3),..., P (2n + 1),... Also, sometimes the base case
consists in verifying more than one proposition, maybe P (1), P (2) and P (3) are
required for the initial step (look at Example 7 in this section).
Margin Note:
Note: Induction works like dominos. You tip over the first domino; the
first domino tips over the second one; the second one tips over the third one;
and so on. You get the idea. Eventually, all dominoes are tipped over.
Theorem 1 (Famous Identity by Induction
Induction)
nduction)
If n is a positive integer, then 1 + 2 + + n =
n ( n + 1)
.
2
Proof:
We denote P ( n) as the statement to be proved:
P ( n) : 1 + 2 + + n =
Base Case:
Case:
n ( n + 1)
2
Clearly P (1) is true since2
P (1) says 1 =
1 ⋅ (2)
2
Induction Step:
Step: We assume P (n) true for an arbitrary n :
P ( n) : 1 + 2 + + n =
n ( n + 1)
2
(assume true)
Adding n + 1 to each side of this equation, we get:
1 + 2 + + n + (n + 1) =
n ( n + 1)
+ (n + 1)
2
n(n + 1) + 2(n + 1)
=
2
(n + 1)(n + 2)
=
2
which is the statement P (n + 1) . Hence, we have proven that P (n + 1) is true.
By induction the result is proven.
▌
2
The reader can verify that P(2) and P(3) are also true, but that isn’t relevant to proof by induction.
4
Section 1.6
Proof by Induction
Note:
Do we have to prove that the principle of mathematical induction
holds? The answer is no. We accept mathematical induction as a logical axiom
in much the same way as we accept the classical rules of Aristotelian logic.
Famous Story in Mathematics
The identity
1+ 2 + + n =
n ( n + 1)
2
can also be proven using the idea Gauss had when he was 9 years old and
impressed his teacher by summing 1 + 2 + + 100 = 5050.
Indeed: 1 + 2 + + 100 = (1 + 100 ) + ( 2 + 99 ) + + ( 50 + 51) = 50 ⋅101 = 5050.
Theorem 2 (Induction in Calculus)
we have
P ( n) :
Prove that for every natural numbers n ,
d n ( xe x )
dx n
= ( x + n) ex
Proof: We show P (n) is true for all natural numbers n by induction.
Base Step:
Step If n = 1 and from the product rule for differentiation, we can write
d ( xe x )
dx
=x
d x x
e + e = ( x + 1) e x .
dx
Induction Step: Assuming
P ( n) :
d n ( xe x )
true for an arbitrary n , we compute
dx
n
= ( x + n) ex
5
Section 1.6
P(n + 1) :
d n +1 ( xe x )
dx n +1
Proof by Induction
n
x
d  d ( xe ) 

= 
dx  dx n 


d
= ( x + n ) e x 
dx
= ( x + n) ex + ex
(induction assumption)
(product rule)
=  x + ( n + 1)  e x
which proves P (n + 1) . Hence the theorem is proved. ▌
Theorem 3 (Proving Closed Forms of Series by Induction)
integer n :
For any positive
x n +1 − 1
x ≡ 1+ x + x + + x =
∑
x −1
k =0
n
2
k
n
Proof:
Letting P (n) be the statement
x n +1 − 1
P ( n) : 1 + x + x + + x =
x −1
2
n
we verify
Base Case:
Case: In this problem the initial step starts at n = 0 due to the way P (n)
is defined. It is not necessary, but we evaluate both P (0) and P (1) .
P(0) : 1 =
Induction Step:
x 0 +1 − 1 x − 1
=
=1
x −1
x −1
P(1) : 1 + x =
x2 − 1
= x +1
x −1
Assuming P (n) is true for any natural number n , we have
1 + x + x2 + + xn =
x n +1 − 1
x −1
Adding x n +1 to each side of this equation, gives
6
Section 1.6
Proof by Induction
x n +1 − 1 n +1
+x
x −1
x n +1 − 1 + x n +1 ( x − 1)
=
x −1
n +1
x − 1 + x n + 2 − x n +1
=
x −1
n+2
x −1
=
x −1
1 + x + x 2 + + x n + x n +1 =
which says the P (n + 1) is true. By induction the theorem is proven.
▌
Margin Note: How do we know if the proof of a theorem is correct? It
would be nice if we could feed a theorem into a computer and let the machine
verify if the proof, but except for very simple cases, this is not feasible. Many
theorems are easy to follow, but many are extremely difficult. When Andrew
Wiles proved Fermat’s Last Theorem in 1993, he presented his results to a
group of experts. No one could verify on the spot if the proof was correct due
to its length (100 pages) and complexity. Only after the proof was analyzed by
a committee of six specialists over several months was the theorem validated.
Theorem 4 (Inequality by Induction) The inequality 2n > n 2 holds for n ≥ 5 .
Proof: Defining P (n) : 2n > n 2 we prove:
Base Case:
P(5) : 25 = 32 > 25 = 52 .
Induction Step:
P ( n ) ⇒ P ( n + 1) for n ≥ 5 . We must show
2
2n > n 2 ⇒ 2 n +1 = ( n + 1) , n ≥ 5 . We write
2n +1 = 2 ⋅ 2n
> 2n 2
2
=n +n
(induction hypothesis)
2
≥ n 2 + 5n
(we assume n ≥ 5)
= n 2 + 2n + 3n
> n 2 + 2n + 1
= (n + 1) 2
Hence P (n + 1) is true and so by induction P (n) true for all n ≥ 5 .
▌
7
Section 1.6
Proof by Induction
Theorem 5 (Tower of Benares Theorem)
The Tower of Benares3 puzzle (or tower of Hanoi) consists moving a collection of disks
from one peg onto another, where one is only allowed to move one disk at a time and no
larger disk can ever be above a smaller disk.
The number of steps required to move n disks from one peg to another peg
(either one) is 2 n − 1 .
Proof:
Proof Let
P ( n ) = the number of moves for n disks is 2n -1
We prove the induction steps:
P (1) is true
P ( n ) ⇒ P ( n + 1)
i P (1) is true since for a single disk it takes 21 − 1 = 1 step to move
one disk from one peg to another.
i
Assume P ( n ) is true. That is, it takes 2 n − 1 steps to transfer n
disks. We now show it requires 2n +1 − 1 steps to move n + 1 disks.
Let n + 1 disks be stacked on pet A . We move the top n disks to one of the
other pegs, say peg C (which takes 2 n − 1 steps by assumption), and then
move the largest disk, still sitting on peg A , to peg B (which takes 1 step).
3
According to legend, the Temple at Benares in ancient India marked the center of the world.
Within the temple priests moved golden disks from one diamond needle to another. God placed
64 gold disks on one needle at the time of creation. It was said that when the temple priests
completed their task the universe would come to an end. Since it takes 264 -1 moves to complete
the task and assuming the priest move one disk per second, it would take roughly 585 billion
years to move the disks from one needle to another..
Section 1.6
8
Proof by Induction
Finally, we move the n disks sitting on peg C to the top of the largest disk
resting on peg B (another 2 n − 1 steps) . Hence, we have the n + 1 disks sitting
on pole B in the proper arrangement. We are done. Adding up these steps,
we find
number of steps required to move n + 1 disks = ( 2 n − 1) + 1 + ( 2n − 1)
= 2 ⋅ 2n − 1
= 2n +1 − 1
Hence, we have proven P ( n + 1) is true so by induction the number of steps
required to move n disks from one pole to another is 2 n − 1 for any natural
number n ∈ .
▌
Is Mathematics Based on Logic?
In the late 1800s and early 1900s a few
mathematicians and logicians like Dedekind, Frege, Hilbert, Russell, and
Whitehead tried to construct arithmetic from formal logical principles and
axioms. The Italian logician Giuseppe Peano (1858-1932) formulated five
axioms from which one could deduce the natural numbers 1, 2, 3, …. . This
brings up the philosophical question of what exactly should be the “starting
point” of mathematics and arithmetic? To some mathematical intuitionists like
Kronecker and Poincare, felt the natural numbers are as intuitive and basic as
one can get and should act as the starting point, rather than being formulated on
less intuitive logical axioms. Logicians would disagree.
The type of induction discussed thus far is called weak induction.
induction We
now introduce the concept of strong induction, although technically the two
methods are equivalent.
Strong Induction
The Principle of Mathematical Induction stated thus far is sometimes
called weak induction in contrast to a variation of it called “strong” induction.
The two types of induction are actually equivalent but sometimes weak
induction doesn’t fit into the proof in a natural way whereas strong induction
does.
9
Section 1.6
Proof by Induction
Methodology of
of (Strong) Mathematical Induction
To verify a proposition P (n) holds for all natural numbers n , the Principle of
(Strong) Mathematical Induction consists of successfully carrying out the
following steps.
1. Base Case:
Case: Prove that P (1) is true.
2. Induction Step:
Step Show that for all n ∈ P(1) ∧ P(2) ∧ ∧ P(n) ⇒ P(n + 1).
A fundamental result in number theory is the Fundamental
Fundamental Theorem of
Arithmetic,
Arithmetic which can be proven by strong induction.
Note:
Note To prove that a proposition P ( n ) is true for all natural numbers does not
mean you have to use induction, but generally induction is the most effective
route. Also, if you use some other method in lieu of induction, you are might be
using some fact in your proof that does require induction.
Theorem 6 (Fundamental
(Fundamental Theorem of Arithmetic)
Arithmetic) Every natural number n ≥ 2
can be written as the product of prime numbers.
For example
2
2
350 = 2 × 5 × 7, 1911 = 3 × 7 × 13 .
Proof: Denoting
P(n) = n can be written as the product of prime numbers
we prove:
Base Case:
Case: P (2) holds since 2 is a prime number.
Induction Step: For an arbitrary n = 2,3,... we prove
P(2) ∧ P(3) ∧ ∧ P(n) ⇒ P(n + 1) .
Assume P (2), P (3), P (n) is true. That is, each natural number 2,3,… n can
be written as the product of primes.
We now consider two cases.
Section 1.6
10
Proof by Induction
Case 1: Suppose n + 1 is a prime number. In this case P (n + 1) is true and
thus the induction step is proven since any prime number n + 1 can be written
as a product of primes, namely n + 1 = n + 1 .
Case 2: Suppose n + 1 is not prime, which means we can factor n + 1 = qp ,
where clearly both of the factors p, q must be less than n + 1 and greater and
or equal to 2. (For example if n + 1 = 15 , we would could factor 15 = 5 ⋅ 3 where
both factors 3,5 are less than 15 and greater than or equal to 2). But since
p, q are less than n + 1 , the induction hypothesis (all P ( k ) are true for
k < n + 1 ) states they can both be written as the product of primes, say
p = p1 p2 pm
q = q1q2 qn
But if both p and q can be written as the product of primes so can
n + 1 = pq = ( p1 p1 pm )( q1q2 qn ) .
Hence P ( n + 1) is true and so by the principle of strong induction P ( n) is true
for all n ≥ 2 .
▌
History of Mathematical Induction
Although some elements of mathematical
induction have been hinted at since the time of Euclid, one of the oldest
argument using induction goes back to the Italian mathematician Francesco
Maurolico, who used induction in 1575 to prove that the sum of the first n odd
natural numbers is n 2 . The method was later discovered independently by the
Swiss mathematician John Bernoulli, and French mathematicians Blaise Pascal
(1623-1662) and Pierre de Fermat (1601-1665). Finally, in 1889 the Italian
logician Guiseppe Peano (1858-1932) laid out five axioms for deducing the
natural numbers, of which his fifth axiom was the Principle of Mathematical
Induction. Hence, from that point of view induction is a formal axiom of
arithmetic.
Theorem 7
(Solution of a Recurrence Relation)
Suppose a sequence
u0 , u1 ,..., un ,... is defined by the recursion relation with initial conditions:
un +1 = 3un − 2un −1
u0 = 2, u1 = 3
Find the sequence un , n = 1, 2,... that satisfies these equations.
computations we find u2 = 3u1 − 2u0 = 5, u3 = 7, u4 = 9,
reasonable guess would be un = 2 n + 1 .
Doing a few
u5 = 33 and thus a
To show P ( n) : un = 2n + 1 satisfies the
11
Section 1.6
Proof by Induction
recurrence relation for all n ≥ 0 , we use strong induction starting with initial
step n = 0. .
P (0) : u0 = 20 + 1 = 3 (check)
Base Case:
Case:
Induction Step: Assuming P (0), P (1),..., P ( n) true, we can write
un
= 2 n + 1, un −1 = 2n −1 + 1
and from the recurrence relation, we have
un +1 = 3un − 2un −1
= 3 ( 2n + 1) − 2 ( 2n −1 + 1)
= 3 ⋅ 2n − 2n + 1
= 2n +1 + 1
which
proves
P (n + 1).
Hence,
by
strong
induction
we
have
that
n
P (n) : un = 2 + 1 satisfies the recurrence relation for all n ≥ 0 .
This next example shows a variation of the base step from previous
examples. Each problem is different and you must adjust the induction proof
accordingly.
Theorem 8 (Modifying the Base Step)
You are given two rulers: one is 3
units long, the other 5. Show that you can measure any unit distance greater
than or equal to 8 using only these two rulers.
Proof: Let P ( n) be the proposition
P ( n ) = any integer length n ≥ 8 can be measured with rulers of length 3 and 5
It is useful to see the following pattern that develops.
12
Section 1.6
Proof by Induction
P(8) = 5 + 3
P(9) = 3 + 3 + 3
P(10) = 5 + 5
P(11) = P(8) + 3 = ( 5 + 3) + 3
P(12) = P(9) + 3 = ( 3 + 3 + 3) + 3
P(13) = P(10) + 3 = ( 5 + 5) + 3
...
...
...
This pattern will serve as an aid in deciding the base and induction steps
which is often the most difficult part in an induction proof.
Base Step: For the base step, we verify the first three propositions:
P(8) = 5 + 3, P(9) = 3 + 3 + 3, P (10 ) = 5 + 5
Induction Step: We now prove the induction step
P(8), P(9), ..., P(n) ⇒ P ( n + 1) , n ≥ 10
To prove this we make the simple observation that if P ( n − 2 ) is true (i.e. a
length of n − 2 can be measured with rulers of length 3 and 5), then P ( n + 1) is
[ For example
P ( 9 ) = 3 + 3 + 3 and P (12 ) = ( 3 + 3 + 3) + 3 . Hence, we know P (11) is true since
P ( 8 ) is true, and P (12 ) is true since P ( 9 ) is true, and so on. ] Hence if
also true since a length of n + 1 is 3 units longer than n − 2 .
P (8), P (9), ..., P (n)
n ≥ 10
is true so is P ( n + 1) and so by induction P ( n ) is true for all natural numbers
n.
There are several variations of the basic method of mathematical
induction. One such variation is double induction.
Double Induction
Sometimes we would like to prove a proposition P ( m, n ) involving two
natural numbers m by iterating the inductive process. This is done by
performing an induction on one of the variables, say m , and then an induction
13
Section 1.6
Proof by Induction
on n .
This strategy is called double induction and is carried out by the
following steps.
1. Prove P (1,1) is true
2. Prove P ( m,1) ⇒ P ( m + 1,1)
3. Prove P ( m, n ) ⇒ P ( m, n + 1) for all natural numbers m
You can think of double induction as proving P ( m, n ) at all points in the first
quadrant of the xy -plane with integer coordinates.
14
Section 1.6
Proof by Induction
Problems
1.
(Proof by Induction)
propositions.
Prove by weak or strong induction the following
n ( n + 1)( 2n + 1)
6
a)
12 + 22 + 32 + + n 2 =
b)
n3 − n is divisible by 3 for n ≥ 1 .
c)
∑ ( a + ib ) = 2 ( n + 1)( 2a + nb )
n
1
i =0
d)
2n < n ! (n ≥ 4)
f)
un = un −1 + 2, u1 = 5 has the solution un = 2n + 3 for n ≥ 2 .
g)
∑ ( 4k − 3) = n ( 2n − 1)
n
k =1
h)
i)
j)
5 divides 7 n − 2n for all positive integers.
2n > n
( cos θ + i sin θ )
n
= cos nθ + i sin nθ
k)
sin nx ≤ n sin x
l)
a < b ⇒ an < bn
m)
dn
n +1
ln x = ( −1) ( n − 1) ! x − n
n
dx
(De Moivre’s Theorem)
( a, b positive real numbers)
2. (Graphical Induction) Draw some lines in the plane as shown in Figure 1
and color adjoining regions either red or blue. Show that it is possible to draw
an arbitrary number of lines in such a way that two adjoining regions are not
15
Section 1.6
Proof by Induction
colored the same color. Hint: Define P (n) as the proposition “it is possible to
draw n lines in the desired manner for arbitrary n ∈ ”.
Figure 1
3. (Clever Mary) To prove the identity
n
∑k =
k =0
n ( n + 1)
2
Mary evaluated the left-hand side of the equation for n = 0,1, 2 getting
n
F(n)
0
0
1
1
2
3
.
She then fit a polynomial to these three points, getting
f ( n) =
n ( n + 1)
.
2
Mary turned this into her professor. Is her proof4 valid?
4
This problem is based on a problem in the book A = B by Marko Petkovsek, Doron Zeilberger and
Herbert Wilf. (This amazing book, incidentally, can be downloaded free on the internet.)
Section 1.6
4.
a)
16
Proof by Induction
(Proofs
Proofs without Words)
Words) They say a good picture is worth a thousand words,
but in mathematics it might be closer to a million. For the figures below,
describe why the figure provides a visual proof of the statement.
a 2 + b 2 = c3
b)
c)
1 + 3 + 5 + + (2n − 1) = n 2
1+ 2 + + n =
d)
n ( n + 1)
2
x 2 − y 2 = ( x − y )( x + y )
17
Section 1.6
e)
x 2 − y 2 = ( x − y )( x + y )
2
g)
Proof by Induction
a a

x + ax =  x +  −  
2 2

2
f)
2
h)
ab ≤
1
∫ (t
0
p/q
a+b
2
+ t q / p ) dt = 1