MATHEMATICAL INDUCTION.ppt

Mathematical Induction
Number theory is the branch of mathematics that studies the
properties of, and the relationships between, particular types of
numbers. (Rosen, 2011).
Of the sets of numbers studied in number theory, the most
important is the set of positive integers.
More specifically, the primes, those positive integers with no
positive proper factors other than 1.
Fundamental theorem of arithmetic, tells us that every positive
integer can be uniquely written as the product of primes in
nondecreasing order. This shows that the primes are the
multiplicative building blocks of the positive integers.

A Bit of History of Number Theory
 Interest in prime numbers goes back at least 2500 years, to
the studies of ancient Greek mathematicians.
 Perhaps the first question about primes that comes to mind
is whether there are infinitely many. In The Elements, the
ancient Greek mathematician Euclid provided a proof, that
there are infinitely many primes.
 Interest in primes was rekindled in the seventeenth and
eighteenth centuries, when mathematicians such as Pierre
de Fermat and Leonhard Euler proved many important
results and conjectured approaches for generating primes.
 The study of primes progressed substantially in the
nineteenth century; results included the infinitude of
primes in arithmetic progressions.

A Bit of History of Number Theory

Key Problems in Number Theory
 Distinguishing primes from composite integers.
 Factoring a positive integer into primes.
 The search for integer solutions of equations is
another important part of number theory.
i.e. Diophantine Equation, after the ancient Greek
mathematician Diophantus.

Elementary Number Theory
Elementary number theory, is not dependent on advanced
mathematics, such as the theory of complex variables, abstract
algebra, or algebraic geometry. - Analytic number theory
(which takes advantage of the theory of complex variables) and
- Algebraic number theory (which uses concepts
from abstract algebra to prove interesting results
about algebraic number fields).
Elementary number theory is largely about the integers,
denoted by the symbol Z.
The integers are an example of an algebraic structure called an
integral domain.
This means that Z satisfies the axioms for an algebraic
structure.

The Well-Ordering Property of the integers
states:
A least element exist in any non empty set of
positive integers.
Every nonempty subset of the positive integers
has a smallest element.
It is one of the more important axioms for Z; it can
be used to prove the Principle of mathematical
induction.

The First Principle of Mathematical Induction: If a set of
positive integers has the property that, if it contains the
integer ,
then it also contains , and if this set contains 1 then it
must be the set of all positive integers. More generally,
a property concerning the positive integers that is true
for , and that is true for the integer whenever it is true
for the integer,must be true for all positive integers.

There are two aspects to mathematics,
discovery and proof.
Both are of equal importance.
• We must discover something before
we can attempt to prove it.
• We can only be certain of its truth
once it has been proved.

What is induction?
A method of proof
It does not generate answers: it only can prove them
Three parts:
• Base case(s): show it is true
for one element
• Inductive hypothesis: assume
it is true for any given element
– Must be clearly labeled!!!
• Show that if it true for the next
highest element
• “Proof by induction”

In this section, we examine
the relationship between these two key
components of mathematics more
closely.

Conjecture
A conjecture is a mathematical
statement that has not yet been
rigorously proved. Conjectures
arise when one notices a pattern
that holds true for many cases. ...
Conjectures must be proved for the
mathematical observation to be
fully accepted. When a conjecture
is rigorously proved, it becomes a
theorem.

Conjecture and Proof
Let’s try a simple experiment.
• We add more and more of the odd numbers
as follows:
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
• What do you notice about the numbers
on the right side of these equations?

They are, in fact, all perfect squares.
These equations say:
• The sum of the first 1 odd number is 12.
• The sum of the first 2 odd numbers is 22.

This leads naturally to the following
question:
Is it true that, for every natural number n,
the sum of the first n odd numbers is n2?
• Could this remarkable property be true?

Conjecture
Since we know that the nth odd number
is 2n – 1, we can write this statement more
precisely as:
1 + 3 + 5 + . . . + (2n – 1) = n2
• It’s important to realize that this is still a conjecture.
• We cannot conclude by checking a finite number
of cases that a property is true for all numbers
(there are infinitely many).

Conjecture
Let’s see this more clearly.
Suppose someone tells us he has added up
the first trillion odd numbers and found that
they do not add up to 1 trillion squared.
• What would you tell this person?

Conjecture
It would be silly to say that you’re sure
it’s true because you’ve already checked
the first five cases.
• You could, however, take out paper
and pencil and start checking it yourself.
(This would probably take the rest of your life.)

Conjecture
The tragedy would be that, after
completing this task, you would still not
be sure of the truth of the conjecture!
• Do you see why?

Proof
Herein lies the power of
mathematical proof.
• A proof is a clear argument that
demonstrates the truth of a statement
beyond doubt.

Let’s consider a special kind of
proof called mathematical induction.
• Now, let’s see how it works.

Suppose we have a statement that says
something about all natural numbers n.
Let’s call this statement P.
• For example,
P: For every natural number n, the sum
of the first n odd numbers is n2.

Since this statement is about all
natural numbers, it contains infinitely
many statements.
• We will call them P(1), P(2), . . . .
P(1): The sum of the first 1 odd number is 12.
P(2): The sum of the first 2 odd numbers is 22.
P(3): The sum of the first 3 odd numbers is 32.
. .
. .
. .

How can we prove all of these
statements at once?
• Mathematical induction is a clever way
of doing so.

The crux of the idea is:
• Suppose we can prove that, whenever one
of these statements is true, then the one following
it in the list is also true.
• That is,
For every k, if P(k) is true, then P(k + 1) is true.
• This is called the induction step since it leads us
from the truth of one statement to the next.

T- TEST
(We need to show that P(n) is true. )
A-ASSUME
(Assume that statement is true for n=k.)
P-PROVE
(Prove that the statement is true for n=k+1)
E-EXPLAIN

Now, suppose that we can also prove
that P(1) is true.
• The induction step now leads us through
the following chain of statements.
• P(1) is true, so P(2) is true.
P(2) is true, so P(3) is true.
P(3) is true, so P(4) is true.
. .
. .
. .

So, we see that, if both the induction
step and P(1) are proved, then
statement P is proved for all n.
• The following is a summary of
this important method of proof.

Principle of Mathematical Induction
For each natural number n, let P(n)
be a statement depending on n.
• Suppose that these conditions are satisfied.
1. P(n) for n=1 is true.
2. For every natural number k, Assume that
statement is true for n=k.
3. Prove that the statement is true for n=k+1
• Then, P(n) is true for all natural numbers n.

Applying Principle of Mathematical Induction
To apply this principle, three are
two steps:
TEST- P(n) for n=1 is true.
ASSUME- Assume that statement is true for n=k.
PROVE- Prove that the statement is true for
n=k+1

Induction Hypothesis
Notice that, in step 2, we do not prove
that P(k) is true.
• We only show that, if P(k) is true,
then P(k + 1) is also true.
• The assumption that P(k) is true
is called the induction hypothesis.

We now use mathematical induction
to prove that the conjecture we made at
the beginning of this section is true.

E.g. 1—A Proof by Mathematical Induction
Prove that, for all natural numbers n,
1 + 3 + 5 + . . . + (2n – 1) = n2
LHS RHS
• Let P(n) denote the statement.

E.g. 1—Proof by Mathematical Induction
TEST
We need to show that P(n) is true.
n=1
• However, P(1) is simply the statement
that 1 = 12.
• This is, of course, true.
Step 1

Assume that statement is true for
n=k.
• We assume that
1 + 3 + 5 + . . . + (2k – 1) = k2
is true.
Step 2

E.g. 1—Proof by Mathematical Induction Step 2
Prove that the statement is true for
n=k+1
• That is,
1 + 3 + 5 + . . . + (2k – 1) + [2(k + 1) – 1]
= (k + 1)2
• Note that we get P(k + 1) by substituting k + 1
for each n in the statement P(n).

We start with the left side and use the
induction hypothesis to obtain the right side
of the equation:
1 + 3 + 5 + . . . + (2k – 1) + [2(k + 1) – 1]
= [1 + 3 + 5 + . . . + (2k – 1)] + [2(k + 1) – 1]
(Group the first k terms)
Step 2

= k2 + [2(k + 1) – 1] Induction Hypothesis
= k2 + [2k + 2 – 1] Distributive property
= k2 + 2k + 1
= (k + 1)2
Therefore,
k2 + [2(k + 1) – 1]= (k + 1)2
is true by PMI.
Step 2

Having proved steps 1 and 2,
we conclude by the Principle of
Mathematical Induction that P(n)
is true for all natural numbers n.

E.g. 2—A Proof by Mathematical Induction
Prove that, for every natural number n,
• Let P(n) be the statement.
• We want to show that P(n) is true for
all natural numbers n.
( 1)
1 2 3 ...
2
n n
n

    

TEST
We need to show that P(n) is true.
n=1
• However, P(1) says that:
• This statement is clearly true.
 
1 1 1
1
2


Step 1

Assume that statement is true for
n=k.
• Thus, our induction hypothesis is:
• Prove that the statement is true for n=k+1
• That is,
( 1)
1 2 3 ...
2
k k
k

    
Step 2

So, we start with the left side and use the
induction hypothesis to obtain the right side:
1 + 2 + 3 + . . . + k + (k + 1)
[1 + 2 + 3 + . . . + k] + (k + 1)
Step 2
( 1)
( 1)
2
k k
k

  

E.g. 2—Proof by Mathematical Induction Step 2
Factorizing, we have,

EXAMPLE 4
1+5+9+……..+(4n-3)=n(2n-1)
TEST: n=1
4(1)-3=1(2(1)-1)
1=1 TRUE

EXAMPLE 4
ASSUME that the statement is true for n=k
1+5+9+……..+(4k-3)=k(2k-1)
PROVE that the statement is true for k+1
1+5+9+……..+(4k+3)+(4(k+1)-3)=k+1(2(k+1)-1)
k(2k-1)
LHS: k(2k-1)+(4(k+1)-3)

1+5+9+……..+(4k+3)+(4(k+1)-3)=k+1(2(k+1)-1)
IS TRUE.

MATHEMATICAL INDUCTION.ppt

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MATHEMATICAL INDUCTION.ppt