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Visvesvaraya Technological University VTU
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- Mathematical Induction - ( 1 - 7 )
- Propositional Logic - ( 8 - 12 )
- Predicate Logic - ( 13 - 14 )
- Rules Of Inference - ( 15 - 19 )
- Operators And Postulates - ( 20 - 22 )
- Group Theory - ( 23 - 28 )
- Counting And Probability - ( 29 - 38 )
- Mathematical Logic - ( 39 - 45 )
- Mathematical Induction And Recurrence Relations - ( 46 - 80 )

Topic:

12. MATHEMATICAL INDUCTIONDiscrete Mathematics
Mathematical induction, is a technique for proving results or establishing statements
for natural numbers. This part illustrates the method through a variety of examples.
Definition
Mathematical Induction is a mathematical technique which is used to prove a
statement, a formula or a theorem is true for every natural number.
The technique involves two steps to prove a statement, as stated below:
Step 1(Base step): It proves that a statement is true for the initial value.
Step 2(Inductive step): It proves that if the statement is true for the n
number n), then it is also true for (n+1)
th
th
iteration (or
iteration ( or number n+1).
How to Do It
Step 1: Consider an initial value for which the statement is true. It is to be shown that
the statement is true for n=initial value.
Step 2: Assume the statement is true for any value of n=k. Then prove the statement is
true for n=k+1. We actually break n=k+1 into two parts, one part is n=k (which is
already proved) and try to prove the other part.
Problem 1
n
3 -1 is a multiple of 2 for n=1, 2, ...
Solution
1
Step 1: For n=1, 3 -1 = 3-1 = 2 which is a multiple of 2
n
k
Step 2: Let us assume 3 -1 is true for n=k, Hence, 3 -1 is true (It is an
assumption) We have to prove that 3
3
k+1
k
k+1
-1 is also a multiple of 2
k
k
– 1 = 3 × 3 – 1 = (2 × 3 ) + (3 –1)
k
k
The first part (2×3 ) is certain to be a multiple of 2 and the second part (3 -1) is also
true as our previous assumption.
Hence, 3
k+1
– 1 is a multiple of 2.
n
So, it is proved that 3 – 1 is a multiple of 2.
Problem 2
2
1 + 3 + 5 + ... + (2n-1) = n for n=1, 2, ...
48

Discrete Mathematics
Solution
2
Step 1: For n=1, 1 = 1 , Hence, step 1 is satisfied.
Step 2: Let us assume the statement is true for n=k.
2
Hence, 1 + 3 + 5 + ... + (2k-1) = k is true (It is an assumption)
2
We have to prove that 1 + 3 + 5 + ... + (2(k+1)-1) = (k+1) also
holds 1 + 3 + 5 + ... + (2(k+1) – 1)
= 1 + 3 + 5 + ... + (2k+2 – 1)
= 1 + 3 + 5 + ... + (2k + 1)
= 1 + 3 + 5 + ... + (2k – 1) + (2k + 1)
2
= k + (2k + 1)
= (k + 1)
2
2
So, 1 + 3 + 5 + ... + (2(k+1) – 1) = (k+1) hold which satisfies the step
2
2. Hence, 1 + 3 + 5 + ... + (2n – 1) = n is proved.
Problem 3
n
n n
Prove that (ab) = a b is true for every natural number n
Solution
1
1 1
Step 1: For n=1, (ab) = a b = ab, Hence, step 1 is satisfied.
k
k k
Step 2: Let us assume the statement is true for n=k, Hence, (ab) = a b is true (It is
an assumption).
We have to prove that (ab)
k
k+1
=a
k+1 k+1
b
also hold
k k
Given,
(ab) = a b
Or,
(ab) (ab)= (a b ) (ab) [Multiplying both side by ‘ab’]
Or,
(ab)
k
k+1
k k
k
k
= (aa ) ( bb )
k+1
(ab)
Or,
= (ak+1bk+1)
Hence, step 2 is proved.
n
n n
So, (ab) = a b is true for every natural number n.
Strong Induction
Strong Induction is another form of mathematical induction. Through this induction
technique, we can prove that a propositional function, P(n) is true for all positive
integers, n, using the following steps:
Step 1(Base step): It proves that the initial proposition P(1) true.
Step 2(Inductive step): It proves that the conditional statement
[ (1) ⋀ (2) ⋀ (3) ⋀ … … … … ⋀ ( )] → ( + 1) is true for positive integers k.
49

13. RECURRENCE RELATIONDiscrete Mathematics
In this chapter, we will discuss how recursive techniques can derive sequences and be
used for solving counting problems. The procedure for finding the terms of a sequence in
a recursive manner is called recurrence relation. We study the theory of linear
recurrence relations and their solutions. Finally, we introduce generating functions for
solving recurrence relations.
Definition
A recurrence relation is an equation that recursively defines a sequence where the next
term is a function of the previous terms (Expressing F n as some combination of Fi with
i<n).
Example: Fibonacci series: Fn = Fn-1 + Fn-2, Tower of Hanoi: Fn = 2Fn-1 + 1
Linear Recurrence Relations
A linear recurrence equation of degree k is a recurrence equation which is in the format
xn= A1 xn-1+ A2 xn-1+ A3 xn-1+... Ak xn-k (An is a constant and Ak≠0) on a sequence of
numbers as a first-degree polynomial.
These are some examples of linear recurrence equations:
Recurrence
relations
Fn = Fn-1 + Fn-2
Initial values
Solutions
a1=a2=1
Fibonacci number
Fn = Fn-1 + Fn-2
a1=1, a2=3
Lucas number
Fn = Fn-2 + Fn-3
a1=a2=a3=1
Padovan sequence
Fn = 2Fn-1 + Fn-2
a1=0, a2=1
Pell number
How to solve linear recurrence relation
Suppose, a two ordered linear recurrence relation is: F n = AFn-1 +BFn-2 where A and B are
real numbers.
The characteristic equation for the above recurrence relation is:
2
x − Ax − B = 0
Three cases may occur while finding the roots:
Case 1: If this equation factors as (x- x1)(x- x1) = 0 and it produces two distinct real
n
n
roots x1 and x2, then Fn = ax1 + bx2 is the solution. [Here, a and b are constants]
2
Case 2: If this equation factors as (x- x1) = 0 and it produces single real root x1, then
n
n
Fn = a x1 + bn x1 is the solution.
Case 3: If the equation produces two distinct real roots x1 and x2 in polar form x1 = r ∠ θ and x2 = r ∠(- θ), then Fn = rn (a cos(nθ)+ b sin(nθ)) is
the solution.
50

Discrete Mathematics
Problem 1
Solve the recurrence relation Fn = 5Fn-1 - 6Fn-2 where F0 = 1 and F1 = 4
Solution
The characteristic equation of the recurrence relation is:
2
x – 5x + 6=0,
So,
(x-3) (x-2) = 0
Hence, the roots are:
x1 = 3 and
x2= 2
The roots are real and distinct. So, this is in the form of case 1
Hence, the solution is:
n
Fn = ax1 + bx2
n
n
n
Here, Fn = a3 + b2 (As x1 = 3 and x2=
2) Therefore,
0
0
1
1
1=F0 = a3 + b2 = a+b
4=F1 = a3 + b2 = 3a+2b
Solving these two equations, we get a = 2 and b = -1
Hence, the final solution is:
n
n
n
Fn = 2.3 + (-1) . 2 = 2.3 - 2
n
Problem 2
Solve the recurrence relation Fn = 10Fn-1 - 25Fn-2 where F0 = 3 and F1 = 17
Solution
The characteristic equation of the recurrence relation is:
2
x –10x -25 =0,
So,
2
(x – 5) = 0
Hence, there is single real root x1 = 5
As there is single real valued root, this is in the form of case
2 Hence, the solution is:
n
Fn = ax1 + bnx1
0
n
0
3 = F0= a.5 + b.0.5 = a
1
1
17 = F1= a.5 + b.1.5 = 5a+5b
Solving these two equations, we get a = 3 and b = 2/5
51

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