The simplest and most common proof of mathematical induction is to prove that when n belongs to all natural numbers, an expression becomes. The method comprises the following two steps:

The basis of recursion: prove that the expression holds when n = 1

The basis of recursion: prove that it is true when n = m, and it is also true when n = m+ 1 (The "if" in the recursive basis is defined as inductive hypothesis. Don't call the whole second step inductive hypothesis. )

There are two key points to remember in mathematical induction.

1。 It is proved that the conclusion holds when n is a certain value.

2。 Assuming that n=k holds, it is proved that the conclusion also holds when n=k+ 1

For example:

It is proved that the product of five consecutive natural numbers can be divisible by 120.

Answer:

1, when n= 1, 1*2*3*4*5= 120, which is divisible by 120, the original proposition holds.

2. Assuming that the original proposition holds when n=k, then when n=k+ 1,

(k+ 1)(k+2)(k+3)(k+4)(k+5)

=k(k+ 1)(k+2)(k+3)(k+4)

+5(k+ 1)(k+2)(k+3)(k+4)

Because k(k+ 1)(k+2)(k+3)(k+4) is a multiple of 120.

Just prove that 5(k+ 1)(k+2)(k+3)(k+4) is a multiple of 120.

I want to prove that (k+ 1)(k+2)(k+3)(k+4) is a multiple of 24.

Four numbers must have multiples of 4, multiples of 3 and another even number, so they must be divisible by 4*2*3=24.

That is, the original proposition holds when n=k+ 1

So the original proposition of integral 1, 2,0 holds for any natural number.