**Mathematical induction is a way to find whether a given statement is true for all the natural Numbers or not.**Natural numbers are those ordinary counting numbers 1, 2, 3, 4 and so on. These numbers are also referred to as the positive integers or non-negative integers are only positive numbers are included.

The simplest form of Mathematical Induction identify that whether a statement which include a natural number n holds for all values of n is true or not.

**The Proof of Mathematical Induction contains two steps which are shown below:**

1. The basic case of induction mathematical is used to show weather the statement holds the value of n is equal to the lowest value of n. Generally, we take the value of n as 0 or 1.

2. The inductive case of mathematical induction is used to show weather the given statement holds the value of n then the full statement also holds when the value of n is substituted to n + 1.

This given method is applicable by proving the statement is true for the starting value. Then we have to prove the process which is used to go from one value to the other value which is valid. If both the given values are taken, then any value can be calculated by performing the process.

To prove Mathematical Induction Divisibility property let us go through the example below:

We are given that

**P(k): 7**…………………………………………………………………... (1)

^{k}- 2^{k}is divisible by 5…Then 7

^{k}- 2

^{k}= 5x [for some x ∈Z. (By definition of mathematical induction)......... (2)

Now we have to show that P

^{(k+1)}:

**7**………………………………………………………………………. (3)

^{k+1}- 2^{k+1}is divisible by 5**7**

= 7·7

= 5·7

= 5·7

= 5·7

= 5·(7

^{k+1}- 2^{k+1}= 7·7

^{k}- 2·2^{k}= 5·7

^{k}+ 2·7^{k}- 2·2^{k}= 5·7

^{k}+ 2·(7^{k}- 2^{k})= 5·7

^{k}+ 2·5a (by (2))= 5·(7

^{k}+ 2a) which is divisible by 5.Thus, P(n) is true by induction hypothesis.