- #1

- 1,357

- 0

Let P(x) be some statement concerning the integer x. If

- P(m) is true, and

- P(k) is true implies P(k + 1) is true, where m ≤ k < n,

then P(x) is true for all integers x = m, ..., n.

- Thread starter e(ho0n3
- Start date

- #1

- 1,357

- 0

Let P(x) be some statement concerning the integer x. If

- P(m) is true, and

- P(k) is true implies P(k + 1) is true, where m ≤ k < n,

then P(x) is true for all integers x = m, ..., n.

- #2

- 104

- 0

It's just induction.

- #3

- 1,357

- 0

- #4

- 1,425

- 1

- #5

uart

Science Advisor

- 2,776

- 9

In the context of mathematics it's just plain induction but if you're googling then obviously you'll do better to use the full name "mathematical induction" because outside of Math the term "induction" has lots of other applications.

- #6

- 1,357

- 0

I didn't just google "induction". I tried "finite induction types" and variations thereof.

- #7

uart

Science Advisor

- 2,776

- 9

- #8

- 104

- 0

Whilst we may use the terms 'induction' and 'strong induction' for two different things they are entirely equivalent notions (i.e. each one implies the other), and your notion of induction in post 1 falls between the two. It is a simple exercise to show that your description is equivalent to ordinary induction.

- #9

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

If P(1) is true and

whenever P(n) is true for all [itex]n\le k[/itex], then P(k+1) is true

Then P(n) is true for all P

It is "Strong" because the hypotheses are easier- you only have to show P(k+1) is true when P(n) is true for all [itex]n\le k[/itex] so you have "more information". It is easy to see that "strong induction" implies regular induction: if P(k+1) is true whenever P(k) is true then it is certainly the case that P(k+1) is true whenever P(n) is true for all n less than or equal to k.

The remarkable thing is that regular induction implies strong induction:

Suppose P(n) is a statement such that:

P(1) is true and

whenever P(n) is true for all n less than or equal to k, P(k+1) is true.

Let Q(n) be the statement "P(m) is true for all m less than or equal to n".

Q(1) is the statement "P(m) is true for all m less than or equal to 1". But the only natural number "less than or equal to 1" is 1 itself. Q(1) just says P(1) is true- and that's true.

Now suppose Q(k) is true. That means P(m) is true for all m less than or equal to k and so, by the induction hypothesis, P(k+1) is true. But since we already have that P(m) is true for all m less than or equal to k, we now know that P(m) is true for all k less than or equal to k+1: Q(k+1) is true. Therefore, by regular induction, Q(n) is true for all n. But Q(n) says P(m) is true for all m less than or equal to n. If Q(n) is true for all n, then P(n) is true for all n.

The statement given in the first post looks like "strong" induction restricted to n greater than or equal to m. As everyone else has said, it is equivalent (as long as n is larger than or equal to m) to regular induction.

- #10

- 246

- 1

there is also Transfinite Induction which is an extension to well-ordered sets

http://en.wikipedia.org/wiki/Transfinite_induction

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 8K

- Last Post

- Replies
- 3

- Views
- 1K

- Replies
- 2

- Views
- 441

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 912

- Last Post

- Replies
- 5

- Views
- 10K

- Last Post

- Replies
- 8

- Views
- 708

- Last Post

- Replies
- 1

- Views
- 525