2

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:

  1. The integer $1$ belongs to $S$.
  2. 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:

  1. The integer $1$ belongs to $S$.
  2. 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.

Souparna
  • 910
  • 1
  • 5
  • 23
  • If we apply the first then on base of $k\in S$ we are asked to prove that $k+1\in S$. Here $k\in S$ is our hypothese. If we apply the second then on base of $1,2,\dots,k\in S$ we are asked to prove that $k+1\in S$. Here $1,2,\dots,k\in S$ is our hypothese and is evidently a stronger hypothese than only $k\in S$. There are situations in which this stronger hypothese is very helpful. – drhab Jul 08 '22 at 14:40
  • @drhab What exactly do you mean by 'stronger' and why is it needed only in some cases but not in others? Plz clarify as I'm quite new to number theory. – Souparna Jul 08 '22 at 14:43
  • By stronger I just mean that - in our effort to prove $k+1\in S$ - we have the disposal of more knowledge. The answer of jjagmath gives you a nice example. – drhab Jul 08 '22 at 14:46
  • Also have a look at the answers to this question. – drhab Jul 08 '22 at 14:55
  • 2
    The two forms are also known as weak induction vs. strong induction. See this previous Question about the difference between the two forms. With a clever bit of phrasing the strong form of induction can be formulated as an instance of weak induction, so in that sense they are equivalent. – hardmath Jul 08 '22 at 14:59
  • 1
    Accepting your first answer so quickly almost guarantees you will not receive further answers. But there is much more that can be said, so I highly recommend that you unaccept and wait at least a few days before accepting. Likely you'll receive more answers that shed further light on the matter, – Bill Dubuque Jul 08 '22 at 17:25

2 Answers2

4

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.

jjagmath
  • 18,214
  • Logically, both are equivalent: you apply the first principle to $P(k) \quad\mbox{for}\quad k=1,\ldots,n$ or the second one to $P(n)$, where $P(n)$ is any proposition with a free variable $n$. – wasn't me Jul 08 '22 at 14:52
  • 3
    @wasn'tme Eh? "mathematical" (or rather "mathēmatikós") originally meant "related to learning", so I hope you're not so pedantic as to insist that it refers to any particular field of study. The mere fact that word $X$ derives from word $Y$ with meaning $M$ does not cause $X$ to have any meaning even related to $M$. – Patrick Stevens Jul 08 '22 at 20:20
1

(Weak) Induction proofs run in two steps:

  1. Show $P(1)$ is true
  2. 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.

tryst with freedom
  • 11,538