proof by induction - explanation on it

Question

Proof by induction. It's pretty useful, and the purpose of it makes a lot of sense. However one thing has always bothered me concerning it. So when you apply induction, one has a base case where you choose a case and apply it to show it works. But then you assume it works for $k$ in the inductive case. Here is my question. Why is it that I can assume it works for $k$? Certainly the base case holds true but whose to say that some other case doesn't work. Maybe I am over thinking induction but that has always bothered me.

Answer 1

It works in the first case.

If it works in the first case, then it works in the second case.

If it works in the second case, then it works in the third case.

If it works in the third case, then it works in the fourth case.

$\ldots$ and so on.

That's what mathematical induction does. But all of the statements above beginning with "if" are proved in one argument, rather than one by one.

Answer 2

Suppose you are making a mathematical argument. You assume $A$, and then from this you show $B$. Your question is, why are you allowed to assume $A$?

The answer is, you may not conclude from this argument that $B$ is true. Because, as you observed, $A$ might not be true, since it was only an assumption.

What you can conclude is that if $A$ is true, then $B$ is true.

For example: Assume that all farmers are liars, and that Joe is a farmer. Then Joe is a liar. Can you conclude from this argument that Joe is a liar? No, because maybe not all farmers are liars, or maybe Joe is not a farmer. But you can conclude that if all farmers were liars, and Joe was a farmer, then Joe would be a liar.

You're allowed to assume anything you want, as long as you remember at the end that the conclusion is only true if the assumption holds. You can assume that the moon is made of green cheese, as long as your conclusions all have the form "If the moon were made of green cheese, then..."

In mathematical induction you assume that $P(n)$ is true, and using this you show that $P(n+1)$ is also true. Can you actually conclude from this that $P(n+1)$ is true? No, all you can conclude is that if $P(n)$ is true, then $P(n+1)$ is also true.

But mathematical induction does something wonderful here: You know (from the previous paragraph) that if $P(1)$ is true, then $P(2)$ is also true. And you do know that $P(1)$ is true, because you showed it separately as the base case. So you can conclude that $P(2)$ is true, because you do know that $P(1)$ is true.

And because you now know that $P(2)$ is true, and you know that if $P(2)$ is true, then $P(3)$ is true, you can conclude that $P(3)$ is true.

And so on.

Answer 3

Assuming this and then showing the assumption means that must follow, proves that "if this then that". The important part is the demonstration.

This is the second step in the induction proof:

$1. \quad P(1) \\2. \quad P(k) \implies P(k+1) \\\therefore \quad \forall k\in \mathbb Z^+ : P(k)$

You assume that the predicate holds for a general iteration in order to demonstrate that if it does so then it also holds for the next iteration.

Answer 4

A typical proof goes like this. Say you want to prove an assertion $A(n)$ is true. You prove that $A(0)$ is true, and then that if $A(k)$ is true, then $A(k + 1)$ must be true as well. That is, if it happens to be true for some number, then it must be true for the next.

If you manage to prove this, then since it's true for $0$, it must be true for the next number $1$. And since it's true for $1$, it must be true for the next number $2$. And so on.

Your question is why, when we do the induction step of proving $A(k) \Longrightarrow A(k+1)$, we are entitled to assume $A(k)$. The answer is that this is only a temporary, hypothetical assumption. Say that, while we are temporarily assuming $A(k)$ is true, we are able to prove $A(k+1)$. When we return to our main line of reasoning afterwards, we are not permitted to say that we have proved $A(k)$, or that we've proved $A(k+1)$. We can say only that we've proved $A(k) \Longrightarrow A(k+1)$.

Here is another example of the way this kind of temporary hypothesis is used. Say you want to prove that if a number $n$ is divisible by $6$, then it is divisible by $3$. The way to prove this is that we temporarily assume that you have some number $n$ that is divisible by $6$. From, there, we say that $n$ must be of the form $n = 6k$ for some integer $k$. Therefore $n = 3(2k)$, so we conclude (temporarily) that $n$ is divisible by $3$. Returning to our main line of reasoning, this proves the assertion $6 | n \Longrightarrow 3|n$. However, we abandon any idea that we know that $6|n$ or that $3|n$. These were temporary hypotheses. We know only that $6 | n \Longrightarrow 3|n$.

Now, the question you've asked is like asking, in this situation, why we were permitted to assume that $n$ was divisible by $6$. We certainly hadn't (and haven't) proved that $n$ is divisible by $6$. But we assumed it temporarily in order to prove a logical implication.

Answer 5

I've found that a much easier (yet equivalent for $\mathbb N$ under usual logic) way to view induction is the foundational approach.

Instead of the

1. Prove A(base)
2. Assume A(k)
3. Prove A(k+1)

You only get one step and a proof by contradiction. Assume $\neg\forall x A(x)$ then $\exists x \neg A(x)$. Take least such $x$ and prove contradiction.

The difference is small but it extends induction to all well-ordered sets and moreover allows you to use all $y<x$ in your proof of $A(x)$. At the same time it avoids having to assume $k$ to prove $k+1$.