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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.

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    $\begingroup$ $m$ is by assumption the smallest positive integer for which $P$ doesn't hold. $\endgroup$ – Daniel Fischer Sep 25 '13 at 23:04
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    $\begingroup$ This is quite a formal proof. The real intuition comes from the fact that if $P(k) \to P(k+1)$ is true for $k \ge 1$, you can plug in $k = 1$ to get $P(1) \to P(2)$. And if you assume $P(1)$ is also true, the conclusion is $P(2)$. Apply the same principle again but with $k = 2$, you get $P(2) \to P(3)$. Since we have already shown $P(2)$, we get $P(3)$. This process can continue to any $P(k)$ with $k \ge 1$. $\endgroup$ – Tunococ Sep 25 '13 at 23:04
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    $\begingroup$ But of course, your questions are about this formal proof. The answer to your first question is that this proof is a proof by contradiction. The author starts by assuming that there exists some $m \ge 1$ such that $P(m)$ is not true, then tries to find a contradiction. $\endgroup$ – Tunococ Sep 25 '13 at 23:06
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    $\begingroup$ To answer your second question, there is an assumption that $m$ is the "smallest" number such that $P(m)$ is not true. That means if $P(m - 1)$ is also not true, $m$ would not be the smallest element in $S$. This is where the well-ordering principle is used: if $S$ is non-empty, $S$ has a smallest element. In your example, if $5$ and $4$ are both members of $S$, $m$, which is defined to be the smallest element of $S$, cannot be $5$. $\endgroup$ – Tunococ Sep 25 '13 at 23:09
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    $\begingroup$ possible duplicate of What makes induction a valid proof technique? $\endgroup$ – Najib Idrissi Sep 2 '15 at 8:05
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You don't have to show that one positive integer is false in order to show that the proposition is false.

Notice that the quoted proof says:

assume that there is at least one positive integer for which $P(n)$ is false.

That's something of a clue already. In mathematics, you can't prove something by first assuming that it's true. Instead, this is an example of a proof by contradiction. We start out by assuming that what we are trying to prove is false, and then derive a contradiction. Since we assume that the axioms of mathematics are sound, this means that our original assumption must be false, and hence that what we are trying to prove is true.

In this case, we want to prove that $P(n)$ is true for all $n$. The negation of that statement is that $P(n)$ is false for some $n$. We now work with this assumption and try to derive a contradiction.

We now let $m$ be the smallest number such that $P(m)$ is false. It's very important that $m$ is the smallest such example, which is why we need the well-ordering principle. In order to make the dependence on the well-ordering principle explicit, we say:

Let $S$ be the set of $n$ such that $P(n)$ is false. By assumption, $S$ is non-empty, so it must have a least element $m$ by the well-ordering principle.

But you don't have to think that way. Intuitively, if there is one counterexample among the natural numbers, there must be a smallest one. (Note that the same is not true if we restrict attention to the integers, which do not satisfy the well-ordering principle. For example, $\{-1, -2, -3, \dots\}$ hasn't got a least element.

Now consider $P(m-1)$. Since $m$ is the smallest number with $P(m)$ false, and $m-1$ is smaller than $m$, $P(m-1)$ must be true.

The reason you can't have both $P(4)$ and $P(5)$ false when $m$ is $5$ is that $m$ has to be the smallest counterexample, and it clearly isn't in this case, as $4$ is a smaller counterexample. In this case $m=4$ (or some smaller number).

$P(m-1)$ implies $P(m)$ by assumption. We are trying to show that $P(n)$ is true for all $n$ when we are given these pieces of information:

  • $P(1)$ is true.
  • If $P(k)$ is true then $P(k+1)$ is true.

We obviously need some information about $P$ to prove that $P(n)$ is true for all $n$, and this is the information we are given. Now, setting $k=m-1$, and noting that $P(m-1)$ is true, we must have that $P(k+1)=P((m-1)+1)=P(m)$ is true.

We have now arrived at a contradiction. This serves to disprove our initial assertion that $P(n)$ is not true for some $n$. It does not disprove the theorem that we are trying to prove - that $P(n)$ is true for all $n$. Indeed, since it disproves the negation of that theorem, it proves the theorem and we win.

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Mathematicians are super fond of proof by contradiction. Maybe this will make more sense:

Assume we have an inductive argument, and assume we've arbitrarily decided it doesn't work for some numbers. Let's say the smallest number it doesn't work for is 17. Then it works for 16. But since we know that if it holds for $n$ it holds for $n+1$, it holds for $16+1=17$. Contradiction, there's no smallest number the induction argument doesn't hold for, and so induction holds for every positive integer. Hooray for contradictions!

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  • $\begingroup$ But how can you possibly know that n+1 always works without trying every n+1? What is the statement was for all x the function $\frac{1}{(10000000000000000000-x)} $is defined. It will work for 0, 1 and 1+1 for quite a while. $\endgroup$ – Paul the Pirate Sep 25 '13 at 23:15
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    $\begingroup$ What you do is prove that IF it works for n, THEN it works n+1. THen you go "Well it works for 1. If n is 1, then n+1 is 2. So it works for 2. If n is 2, then n+1 is 3, so it works for 3..." Try taking the function you gave and prove that if it works for some arbitrary n, it works for n+1. You can't. $\endgroup$ – Hovercouch Sep 25 '13 at 23:18
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    $\begingroup$ @PaulthePirate - In this case, it is very clear that $n+1$ does not always work: in the case $n+1=10000000000000000000$, it does not work, so we can't prove this (false) statement by induction (or by any other means). I think you need to invest some time learning how induction works before diving into why it's implied by the well-ordering theorem. Go and find a nice book of examples to work through. I recommend chapter 1 of R.P.Burn's 'Numbers and Functions', which is devoted to induction. $\endgroup$ – John Gowers Sep 25 '13 at 23:18
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    $\begingroup$ @PaulthePirate, They trying to show that an inductive argument is a proof if you can show those two assumptions are true. They're not saying they're true in every single case. That's where the proving part of doing an inductive proof comes in. Take the example of $1 + 2 + 3 + 4 + ... n = n(n+1)/2$. It obviously works for 1. Now imagine it works for n. What happens if you replace all instances of n in both sides of the equation with n+1? $\endgroup$ – Hovercouch Sep 25 '13 at 23:23
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    $\begingroup$ @PaulthePirate that proves it for specific values of x+1, not x+1 in general. $\endgroup$ – Hovercouch Sep 26 '13 at 0:23
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"why do I have to show that one positive integer is false to show that it must be true?"

In a proof by contradiction, you assume something is false in order to prove it is true. You do that by showing that the assumption that it is false leads to a contradiction.

Here's an example: Suppose we want to prove $\log_2 3$ is an irrational number, i.e. it is not equal to $m/n$, where $m$ and $n$ are some integers.

So we assume $\log_2 3$ is rational, i.e. it is $m/n$ for some integers $m$ and $n$. Since $\log_2 3$ is positive, $m$ and $n$ are both positive or both negative. If they're both negative, we can multiply them both by $-1$ and have them both positive. Then we have $$ \begin{align} \log_2 3 & = \frac m n \\[12pt] 2^{m/n} & = 3 \\[12pt] 2^m & = 3^n \\[12pt] \text{The even number }2^m & \text{is equal to the odd number }3^n. \end{align} $$ But a number cannot be both even and odd, so the assumption has led to a contradiction. Since the assumption that $\log_2 3$ is rational leads to a contradiction, it must be that $\log_2 3$ is not rational.

"they claim $P(m-1)$ must be true because it is less than $m$. Why is this true?"

Because $m$ is the SMALLEST number for which $P(m)$ is not true. Hence any smaller number is one for which $P$ is true.

"Why does $P(m-1)$ imply $P(m)$?"

That depends on WHICH proof by induction you're doing. The point is that their whole argument applies in contexts in which one can prove that if $P(m-1)$ then $P(m)$. Situations in which you cannot prove that certainly exist, but those are situations in which you cannot write a proof by induction, and their argument is not intended to apply to such cases.

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  • $\begingroup$ I don't follow your algebra, how is $log_2 3 = m/n = 2^\frac{m}{n} = 3$ true? $\endgroup$ – Paul the Pirate Sep 25 '13 at 23:23
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    $\begingroup$ It is not true that $\log_2 3=m/n=2^{m/n}=3$, and I didn't say it is. What I said was that IF $\log_2 3=m/n$ THEN $2^{m/n}=3$. That comes from the definition of logarithms. If $\log_a b = c$ then $a^c=b$. $\endgroup$ – Michael Hardy Sep 25 '13 at 23:25
  • $\begingroup$ And you can prove that they aren't equal but just plugging in 0 and 1 for m and n? $\endgroup$ – Paul the Pirate Sep 25 '13 at 23:27
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    $\begingroup$ That proves only that they're not equal in that one case. One must prove that equality fails in ALL cases. $\endgroup$ – Michael Hardy Sep 25 '13 at 23:29
  • $\begingroup$ How can you do that? What is there is some number that makes them true? How can you prove that there isn't? $\endgroup$ – Paul the Pirate Sep 25 '13 at 23:35
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The $P(x)$ notation can become daunting sometimes. Let me show you a problem that is in fact parallel to the one you are trying to solve but it is not full of $``P"$. Suppose you are given that $\mathbb N = \{1,2,3, \cdots\}$ is the set of natural numbers. I claim that, if the set $\mathbb S$ has the following properties, then $S = \mathbb N$.

$$S \subseteq \mathbb N \tag{1}$$

$$1 \in S \tag{2}$$

$$n \in S \implies n+1 \in S \tag{3}$$

To tie this problem to yours, let $S = \{n \in \mathbb N | P(n) = TRUE\}$.

Let $M = \mathbb N\backslash S$ be the set of all natural numbers that are not in $S$. Then, if we can show that $M = \varnothing$, it will follow that $S = \mathbb N$. We start by assuming that $M \ne \varnothing$. We will show that this assumption leads to a contradiction. Hence $M = \varnothing$.

Because of condition $(1)$, $M \subseteq \mathbb N$. By hypothesis, $M$ is not an empty set. Hence, by WOP, there exists a natural number $s$ that is the smallest member of $M$.

By condition $(2)$, $s \ne 1$. People tend to gloss over this part, but it is a very important detail. Because $s \in \mathbb N$ and because $s \ne 1$, then $s-1 \in \mathbb N$. Since $s$ is the smallest member of $M$, then $s - 1 \not \in M$. But, since $s-1 \in \mathbb N$, by condition $(3)$, $s \in S$. But, because $s \in M$, then $s \not \in S$.

So because our assumption that $M$ was a non empty set lead to a contradiction, we conclude that $M = \varnothing$. It follows that $S = \mathbb N$.

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