Mathematical Induction Definition

It is a process that is used to trueness of a given statement for all positive integers. This method includes proving the first statement true firstly from the series of infinites statement and then it is proved that the any one of the other statements from that series is also true.

The Principle of Mathematical Induction includes following steps to prove that a statement is true for all natural numbers:

1)Basis step: the statement is proved true for the least natural number which is normally 0 or 1.
2)Inductive step: In this step it is assumed that the statement is true for some natural number p that lies between least possible value of n to any natural number n, and then it is proved that the statement is true for the next natural number (p+1) also.
The value of n for basis step is decided from 0 and 1 on the basis of given question or statement.

Proof of Mathematical Induction
Let statement S(n) is false for some values of n. Let no be the least value of n such that S(no) is false. no cannot be 0, because S(0) is true. Therefore no should be 1+n1. As n1
Let us take some Examples of Mathematical Induction to solve some problems:

Mathematical Induction Proofs Examples

Q.1) Show that the sum of first natural numbers from 0 to n is given by n(n+1)/2.
Sol.) The above statement can be written mathematically as:
0+1+2…..+n = n(n+1)/2

Let us use mathematical indction process now:
Basis step: in this step we will put least possible value of n in above equation, which is 0 for this case.
S(0) =>
 0 = 0(0+1)/2
In above equation LHS is 0 and RHS also solves to 0. Hence basis step hold true and thus statement is true for n=0.

Inductive step: let us assume that the equation is true for a number p that lies between 0 and n. Now we will show that it is true for p+1 also.
S(p+1)=>
0+1+2…..+p + (p+1)= ((p+1)((p+1)+1))/2
LHS can also be written as
(p(p+1))/2 + (p+1) , as we assumed that the statement is true for p natural number.
(p(p+1))/2 + (p+1) = (p(p+1)+ 2(p+1))/2
= (p^2+p+2p+2)/2
= ((p+1)(p+2))/2  
=((p+1)((p+1)+1))/2 = RHS.
Therefore S(p+1) is true. Hence from basis and inductive steps we come to the conclusion that S(n) is true for all natural numbers.