What is the need for the Second Principle of Finite Induction? I was studying Elementary Number Theory by David Burton and came across the first and second principle of Finite Induction.
The First Principle of Finite Induction was stated as -

Let $S$ be a set of positive integers with the following properties:

*

*The integer $1$ belongs to $S$.

*Whenever the integer $k$ is in $S$, the next integer $k + 1$ must also be in $S$.

Then $S$ is the set of all positive integers.

The Second Principle of Finite Induction was stated as -

Let $S$ be a set of positive integers with the following properties:

*

*The integer $1$ belongs to $S$.

*If $k$ is a positive integer such that $1 , 2, \dots , k$ belong to $S$, then $k + 1$ must also be in $S$.

Then $S$ is the set of all positive integers.

Going by the definition of First Principle, if $1$ belongs to $S$ then $1+1=2$, $2+1=3$ and all other positive integers should also belong to $S$. This is the requirement for Second Principle. Thus, they look quite similar to me. So, why is the Second Principle needed after all?
Edit:-
I gathered from the answer and comments that the First Principle is weak induction and the Second Principle is strong induction. But, I am still confused why this distinction is made if they are equivalent.
 A: The first principle is often useless in number theory when proving something about the multiplicative structure of the integers.
If you want to prove something about the prime factors of $n$, it's not very useful to know something about the prime factors of $n-1$ (the induction hypothesis in the first principle) because the prime factorizations of $n-1$ and $n$ are not related.
However, if $n$ factors as $n = a b$ with $a,b<n$, then knowing about the prime factors of $a$ and $b$ (by the induction hypothesis in the second principle) will give you knowledge about the prime factors of $n$.
For several proofs of theorems of elementary Number Theory, using the second principle is the obvious choice.
A: (Weak) Induction proofs run in two steps:

*

*Show $P(1)$ is true

*Assume $P(n)$ is  true and prove $P(n+1)$ from it

It may sometimes turn out that as we are converting $P(n)$ into the statement of $P(n+1)$, we may require a step in the equality which can be done if $P(n-1)$ is true. However, in normal induction, this is useless(you could make it useful... with some effort). So, here, it is more natural to use strong induction.
Why did I say useful with some effort? Well the thing is actually you can prove strong from weak and weak from strong, so the two inductions are equivalent. So, by adding some steps you can do whatever you did with strong induction just from weak induction.
