Induction
INDUCTION
OBJECTIVES
1. Student should be able to understand the meaning of Principle of Mathematical Induction.
2. Students should be able to understand clearly each steps involved in different type of
induction.
3. Students should know how to use induction in daily lives.
What, Which, Where, When
1. Concept of induction (domino)
(Clear / Not Clear)
2. Difference between types of induction
Summation type
(Clear / Not Clear)
Division type
(Clear / Not Clear)
Comparison type
(Clear / Not Clear)
3. Identify basic step
(Clear / Not Clear)
4. Identify induction steps for
Summation type
(Clear / Not Clear)
Division type
(Clear / Not Clear)
Comparison type
(Clear / Not Clear)
1
Induction
MATHEMATICAL INDUCTION
-
One of the proof techniques.
-
It acts like a domino.
-
Let say the statement to be proved can be written as n  no P(n), where no is some
fixed integer.
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Suppose we wish to show that P(n) is true for all integers n  no.
-
Suppose also,
a) P(no) is true, and
b) If P(k) is true for some k  no, then P(k + 1) must also be true.
Then P(n) is true for all n  no.
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This result is called the Principle of Mathematical Induction.
Steps involved are:
a) Prove that P(no) is true.
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This is called as basic step.
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If this step is not true, then the next step is not relevant anymore.
b) Prove that P(k)  P(k + 1) is a tautology for all k  no.
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This is called as induction steps. .
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This step will prove that implication will always be true.
Types of induction.
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There are several types of induction:
a) Summation type
b) Division type
c) Comparison type
2
Induction
Summation type:
Basic Step
a) Prove that P(no) is true.
Induction Steps
b) Find P(k).
c) Find P(k + 1). Concentrate on the left side of the equation.
d) Identify P(k) in P(k + 1) left side.
e) Replace P(k) in P(K + 1) with P(k) right side from b).
f)
Get the right side of P(k+1).
Conclusion
g) Simplify.
Ex 1: Summation type.
Show by mathematical induction, for all n  1; 1 + 2 + 3 + … + n = n (n+1)
2
Basic step
a) Prove that P(no) is true.
no = 1, so
P(1) = 1 (1 + 1)
=1
(first element)
2
Therefore, it is true.
Induction steps
b) Find P(k).
Consider for any number k  1,
P(k) = 1 + 2 + 3 + … + k = k (k+1)
2
c) Find P(k + 1). Concentrate on the left side of the equation.
P(k+1) = 1 + 2 + 3 + …. + (k+1) = (k+1)[(k+1) + 1] = (k+1) (k+2)
2
2
left side
right side
d) Identify P(k) in P(k + 1) left side.
P(k+1) = 1 + 2 + 3 + …..+ k + (k+1)
P(k + 1) Left side
P(k) left side
3
Induction
e) Replace P(k) in P(K + 1) with P(k) right side from b).
P(k+1) = 1 + 2 + 3 + … + k + (k+1) = k (k+1) + (k+1)
2
P(k)
f)
P(k) right side
Work the left side of P(k+1) until the expression is similar to the right side of P(K + 1)
P(k+1) = 1 + 2 + 3 + …. + k + (k+1) = k (k+1) + 2 (k+1)
2
= k2 + k + 2k + 2
2
= k2 + 3k + 2
2
= (k+1) (k+2)
2
Conclusion
So, with Principle of Mathematical Induction, P(n) is true for all n  1.
Exercise 1:
1. Show that 2 + 4 + 6 + ….. + 2n = n(n+1) for all n  1.
2. Show that 1 + 3 + 5 + …. + (2n – 1) = n2; for all n  1.
3. Show that 1 + 5 + 9 + …. + (4n – 3) = n(2n – 1); n  1.
4. Show that 1 + 2 + 22 + 23 + … + 2n = 2n+1 – 1; n  0.
4
Induction
Division Type:
Basic Step
a) Prove that P(no) is true.
Induction Steps
b) Get P(k).
c) Get P(k + 1).
d) Manipulate P(k+1) so that it resemble closely to P(k).
e) Identify P(k) from b).
Conclusion
f)
Different parts.
Ex 2: Division type.
Show by mathematical induction, for n  1, 4n – 1 is divisible by 3.
Basic step
a) Prove that P(no) is true.
no = 1, so
P(1)
= 41 – 1
=3
Therefore, it is divisible by 3, so it is true.
Induction steps
b) Get P(k).
P(k)
= 4k – 1
c) Get P(k + 1)
P(k+1)
= 4k+1 – 1
= 4.4k - 1
d) Manipulate P(k+1) so that it resemble closely to P(k).
We know that 4 is 3 + 1, so
= 4.4k - 1
= (3 + 1).4k – 1
(same as if (a + b)c = ac + ab)
= 3.4k + 1.4k – 1
5
Induction
e) Identify P(k) from b).
= 3.4k + 4k –1
Conclusion
f)
From different parts
4k – 1
is divisible by 3 because P(k) is true.
3.4k
is also divisible by 3 because of whatever the value of k,
3.4k is always divisible by 3.
Therefore, 4n – 1 is divisible by 3 for all n  1.
Exercise 2:
1. Show that 22n – 1 is divisible by 3, for all integers n  1.
2. Show that 7n – 2n is divisible by 5, for all integers n  1.
3. Show that 6(7n) – 2(3n) is divisible by 4, for all n  1.
6
Induction
Comparison type.
Basic Step
a) Prove that P(no) is true.
Induction Steps
b) Find P(k).
c) Find P(k + 1).
d) Concentrate on the left side of P(k + 1) only. Make few comparison until we get the form of
left side of P(k).
e) Replace (with comparison) left side of P(k) with right side of P(k) from b).
f)
Make few comparison until we get the form of right side of P(k + 1).
Conclusion
g) Simplify and conclude.
Ex 3: Comparison type
Show by mathematical induction that for all n  2; 1 + 2n < 3n.
Basic step
a) Prove that P(no) is true.
no = 2, so
P(2):
1 + 22
<
32
5
<
9
Therefore, it is true.
Induction steps
b) Find P(k).
P(k):
1 + 2k
<
3k
1 + 2k+1
<
3k+1
c) Find P(k + 1).
P(k+1):
d) Concentrate on the left side of P(k + 1) only. Make few comparison until we get the form of
left side of P(k).
1 + 2k+1
=
1 + 2k.21
=
1 + 21.2k
<
2 + 2.2k =
=
2(1 + 2k)
(if a < b, then (a + x) < (b + x))
1 < 2 so
1 + 2.2k
7
(form of P(k) left side)
Induction
e) Replace (with comparison) left side of P(k) with right side of P(k) from b)
From b), 1 + 2k < 3k
(if a < b, then x.a < x.b),
Therefore,
2(1 + 2k)
f)
<
2(3k)
(right side of P(k))
Make few comparison until we get the form of right side of P(k + 1).
2(3k)
(if a < b, then a.x < b.x)
2 < 3 so,
2(3k)
<
3(3k)
=
3k+1
Conclusion
g) Therefore, proved that 1 + 2n < 3n for all value of n  2.
Exercise 3:
1. Show that, n < 2n for all positive integer n.
2. Shoe that n! > n2 for all integer n  4.
3. Show that 2n + 1  2n; for all n  3.
8