Are there two meanings to induction?

Question

I've seen mathematical induction in two forms.

1. First form: It seems that if $P(0)$ holds and $\displaystyle \overbrace{\frac{k(k+1)}{2}+(k+1)}^{adding}=\overbrace{\frac{[k+1]([k+1]+1)}{2}}^{\text{Switching k for [k+1]}}$, then we can suppose that it holds for arbitrary $n\in\mathbb{N}$. In this case, it seems that the result holds because of the arithmetical laws allowed for that expressions. In here, it seems that the induction forces the exibition of the truth by showing the behavior of the expression under some given laws.

2. Second form: Proving that $m+0=m=0+m$. To prove that, I had only a few laws:

   • $S:\mathbb{N}\to\mathbb{N}$ is an injection; $0\in\mathbb{N}\setminus S(\mathbb{N})$ and the principle of finite induction.
   • $m+0:=m \quad\quad\quad m+S(n):=S(m+n)$

   The proof given in this book is:

   PROOF: By induction, $0+0=0$ follows from the definition, and $[[$if $0+n=n$, then $0+S(n)=S(0+n)=S(n)$$]]$.

   In this case, it seems that the sentence enclosed in double brackets is not something that follows from the given laws, it seems more like something that we want to be true and if it's true, then the conclusion holds. In this case, it seems that induction is something that forces something to be true instead of showing it is a consequence of the laws given earlier. Is that correct? I don't see how that could follow from the laws given in $2.$

Rephrasing it a little shorter:

1. Induction is used to force some behavior under some given laws and then show that under that laws, it actually holds.

2. Induction is used to force something that we actually want to be true, instead of showing it as a consequence of previous laws.

Answer 1

Looks like there's a mistake. It should be $$0+S(n)=S(0+n)=S(n).$$ The first equality follows from $m+S(n):=S(m+n),$ and the latter follows from the inductive hypothesis $0+n=n.$

It's the same kind of reasoning in both cases.

Answer 2

Mathematical induction is a theorem that demonstrates sufficient conditions for a property to hold on all elements of a particular infinite set. As a theorem, it can only prove what is a consequence of given laws. There is no potential for forcing anything.

Answer 3

Proof by (weak) induction takes an enumerated premise $P(n)$, and if we can show both that it holds for a basis, and that holding for an arbitrary step implies it holds for the successive step, then it is proven to hold for all steps subsequent to the basis.

That is: If $\underbrace{\;P(0)\;}_\text{basis}$ and $\underbrace{\forall n\in \Bbb N: [P(n)\to P(n+1)]}_\text{iterative step}$, then $\underbrace{\;\forall n\in\Bbb N: P(n)\;}_\text{conculsion}$ .

There's no 'forcing'. The iterative step involves demonstrating how the truth of each successive premise could be established if the prior were true, and the basis establishes a foundation for building the tower.

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So in your first case, the premise is $P(k):= \left[\sum_{j=1}^k j = \frac{k(k+1)}2\right]$.

We can see that that $P(0)$ is true because $\left[0 = \frac{0(0+1)}{2}\right]$

Similiarly, for an arbitrary $k$, $P(k+1)$ is $\left[\sum_{j=1}^{k+1} j = \frac{(k+1)(k+2)}{2}\right]$ which you can show is equivalent to $\left[k+1+\sum_{j=1}^k j = \frac{k(k+1)}{2}+(k+1)\right]$. So if $P(k)$ is true, then $P(k+1)$ must be.

We have shown $P(0)$ is true. Then by the iterative step $P(1)$ must also be true. This means $P(2)$ is true, and in turn that means $P(3)$ is true, and so on, and so forth, ad infinitum, et cetera. Thus $P(k)$ is shown to be true for any arbitrary natural number.

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For your second case, your premise is: $P(m) \equiv \left[0+m = m\right]$, and you are given these laws to use:

• $S:\Bbb N→\Bbb N$ is an injection; $0∈\Bbb N∖S(\Bbb N)$
• the principle of finite induction.
• $m+0:= m$
• $m+S(n):=S(m+n)$

The basis is: $P(0) \equiv \left[0+0=0\right]$, so we can see $P(0)$ is true.

Now, here's the tricky bit. Rather that the successor to $m$ being $m+1$ we're using $S(m)$. So we show for an arbitrary $m$ that $P(m)\to P(S(m))$, and that $P(0)$, then we prove $\forall m\in \Bbb m: P(m)$

Answer 4

Mathematical induction is a quick way of showing that many things are a consequence of some previously given laws.

In the case where the book wanted to prove that adding zero to a natural number is commutative, you started with

$$m+0:=m$$ $$m+S(n):=S(m+n)$$

which are given to be true for all $m,n \in \mathbb N.$

Now, since these statements are true for all $m,n \in \mathbb N,$ they must be true when $m = 0 \in \mathbb N,$ so

$$0+0=0$$ $$0+S(n)=S(0+n)$$

and these are still true for every $n \in \mathbb N.$ In particular,

$$0 + S(0) = S(0 + 0).$$

But since $0+0=0$ (as already shown), $S(0 + 0) = S(0)$, so

$$0 + S(0) = S(0).$$

Similarly,

$$0 + S(S(0)) = S(0 + S(0)) = S(S(0)),$$ $$0 + S(S(S(0))) = S(0 + S(S(0))) = S(S(S(0))),$$ $$0 + S(S(S(S(0)))) = S(0 + S(S(S(0)))) = S(S(S(S(0)))),$$

and so forth, at each step using the result of the previous step to simplify the expression $S(0 + S(\cdots S(0)\cdots)).$

It should be fairly evident that if we were to write the natural number $1717$ (for example) in this notation, we could use this method to prove that $0 + 1717 = 1717$. It would be quite tedious to do, however.

The inductive step in the book's proof is simply a summary of every step of the derivation above after $0+0=0$, using $n$ to stand for $0$, $S(0)$, $S(S(0))$, or whatever number we take at the start of each step.

This is a shallow and incomplete justification of mathematical induction, intended more as a guide to understanding than as a proof. Some interesting further explorations are in the answers to these questions:

• Mathematical induction question: why can we "assume $P(k)$ holds"?
• Why is mathematical induction a valid proof technique?