(A) the first mathematical induction:

Generally speaking, to prove that a proposition p(n) is related to the natural number n, there are the following steps:

(1) proves that the proposition holds when n takes the first value n0. The value of n0 is generally 0 or 1, but there are some special cases.

(2) Assuming that the proposition holds when n=k(k≥n0, k is a natural number), it is proved that the proposition also holds when n=k+ 1.

Synthesizing (1)(2), the proposition p(n) holds for all natural numbers n(≥n0).

Prove:

Let m be a set of natural numbers that make proposition P correct, so:

The number (1) 1 belongs to m because the proposition p is correct for 1.

(2) Assume that the number n belongs to m, that is, the proposition of the number n is correct. At this time, the proposition p can also be proved to be correct for the direct successor number n', that is, n' also belongs to m.

So the set M has the properties of the above inductive axioms (1) and (2), so the set M should contain all natural numbers. In other words, the proposition p is correct for any natural number n.

(2) the second mathematical induction:

For a proposition p(n) related to natural numbers,

(1) Verify that p(n) holds when n=n0;

(2) suppose n0 ≤ n.

Synthesizing (1)(2), the proposition p(n) holds for all natural numbers n(≥n0).

In order to prove this, we need the minimum number theorem: any nonempty set A of natural numbers must have a minimum number.

Prove:

By reduction to absurdity, if proposition P is not true for all natural numbers, then the set m of natural numbers that makes proposition P not true is not an empty set. According to the minimum number principle in the preliminary theorem, there must be a minimum number L in M, so the proposition P is not valid for L. Since 1 can make the proposition valid, then l≠ 1 is L, that is, the proposition P is valid for all natural numbers less than L. Therefore, according to the proposition of this theorem, it can be proved that the proposition P is also valid for all natural numbers. This contradiction shows that the proposition P is valid for all natural numbers.

In the first chapter of advanced algebra textbooks, there is generally a detailed proof process.