I am trying to work through an example in my book and it just seems nonsensical
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 $P(k)\to P(k + 1)$ 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 $P(m− 1)\to P(m)$ 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 $n$."
I am getting confused at a lot of parts of this. So assuming that $P(1)$ is true and assuming that $P(k)$ implies $P(k+1)$ for all positive integers, why do I have to show that one positive integer is false to show that it must be true? I don't understand that.
Anyways now my set S of $P(n)$ false inputs is now not empty. So that element is now $m$, and $m$ must be positive so then they claim $P(m-1)$ must be true because it is less than $m$. Why is this true? For example why can't I have $5$ and $4$ both being false where $m$ is $5$? Why does $P(m -1)$ imply $P(m)$? There doesn't seem to be proven anywhere. They then claim that $P(m)$ must now be true even though they previously claimed that it was false, so how can that be? It seems like the contradiction works to disprove the original assessment that $P(n)$ must be true.