Why mathematical induction is valid?

Question

I'm teaching out of Rosen's discrete math book, and for mathematical induction he says:

"Why is mathematical induction a valid proof technique? The reason comes from the well-ordering property, listed in Appendix 1, as an axiom for the set of positive integers, which states that every nonempty subset of the set of positive integers has a least element. So, suppose we know that P(1) is true and that the proposition P(k) → P(k + 1) is true for all positive integers k. To show that P(n) must be true for all positive integers n, assume that there is at least one positive integer for which P(n) is false. Then the set S of positive integers for which P(n) is false is nonempty. Thus, by the well-ordering property, S has a least element, which will be denoted by m. We know that m cannot be 1, because P(1) is true. Because m is positive and greater than 1, m − 1 is a positive integer. Furthermore, because m − 1 is less than m, it is not in S, so P(m − 1) must be true. Because the conditional statement P(m − 1) → P(m) is also true, it must be the case that P(m) is true. This contradicts the choice of m. Hence, P(n) must be true for every positive integer n."

Everything makes sense except for this sentence:

"Furthermore, because m − 1 is less than m, it is not in S..."

That doesn't appear to be part of the well-ordering property, or anything else I could find in the book. Where did that come from and how can it be explained? I see no logical reason m-1 cannot be in S.

As he says $m$ is the least element of $S$, since $m-1$ is smaller than $m$ it can't be in $S$ otherwise $m$ wouldn't be least. — Demophilus, Jun 08 '17 at 20:41
Ah I see, dumb oversight. m has to be the least. I think this could be explained in a clearer way though, the writing is thick and convoluted. — wordsforthewise, Jun 08 '17 at 20:44
I don't necessarily agree. It also begs the question of course, how would you write it in a more clear fashion? — Demophilus, Jun 08 '17 at 20:48

score 1 · Answer 1 · answered Jun 08 '17 at 23:37

1

Part of what you quoted says this:

$S$ has a least element, which will be denoted by $m$

If $m$ is the smallest (or "least") element in $S,$ then everything smaller than $m$ is not in $S$.

answered Jun 08 '17 at 23:37

Michael Hardy

1

wordsforthewise · Answer 2 · 2017-06-08T21:48:45.327

The answer as Demophilus said is m is the smallest element of S, so m-1 is not in S because m-1<m. I think this can be written more clearly as:

We are trying to prove P(k) is True for all positive integers, n.
Assume P(k) → P(k + 1) is True for all positive integers, n.
Assume P(1) is True.
Assume P(k) is False for some positive integer, m.
Assume m is the smallest element of the set S, which is related to the well-ordering property.
The set S contains all positive integers where P(k) is False.
We know that m cannot be 1, because we said P(1) is True.
m must be greater than 1, because 1 is the smallest positive integer.
If m>1, then m - 1 is not in the set S, because m is the smallest element of S.
Remember we assumed P(k) → P(k + 1) is True for all positive integers.
P(m) → P(m + 1) is True can also be written as "if P(m) is True, then P(m + 1) is True"
P(m) → P(m + 1) can be re-written P(m - 1) → P(m), so "if P(m - 1) is True, then P(m) is True"
m - 1 is not in S, meaning P(m - 1) is True.
P(m) is False.
But because P(m - 1) is True, P(m - 1) → P(m) says "if P(m - 1) is True, then P(m) is True"
Therefore we have a contradiction (P(m) is True and P(m) is False), so m does not exist
Therefore the set S of positive integers where P(k) is False is empty, so P(k) is True for all positive integers. We have proved our proposal via induction.

I think the bullet points also make it easier to go back to previous steps and re-read them, rather than having to scan through a dense paragraph. I'm sure there are clearer ways to write this than I did.

Why mathematical induction is valid?

2 Answers2