I am going to confine this answer to what is often called "The Principle of Mathematical Induction" (PMI). There are other flavors (you will likely see "Complete Induction" soon, for example). In this case, we are trying to show that some property $P(n)$ holds for each member of $\mathbb{N} = \{1,2,3,...\}$.
First, for the specific:
And does every induction proof follow these steps? Or is there any other way to approach to Mathematical Induction.
For PMI, yes, in the sense that there is always a special case -- you show explicitly that $P(1)$ is in fact true. This is the base case.
You then make the inductive hypothesis, where you assume that $n$ is a number for which $P(n)$ holds. [If you are worried about why you can do that, notice that you have at least one such number -- the base case! (*)]
Then, you do the inductive step: You now use the fact that $P(n)$ is true to show that $P(n+1)$ is also true.
All PMI proofs have this form. But, not all are like the "algebra" trick/formula version in your example and likely that make up the examples you may have seen so far. For example, if you know about the idea of a tree many of their properties can be proven inductively, where we induct on the number of nodes. These proofs have the same basic form:
- show that the tree with a single node has the property
- assume all trees with $n$ nodes have the property
- use this to show that all trees with $n+1$ nodes have the property -- typically, unlike the algebraic examples, you accomplish this by (carefully!) selecting a node of the larger tree to remove, noting that the resulting smaller tree has $n$ nodes and so has the property which you (hopefully) prove is preserved when you add the node back.
But, yes, most initial induction proofs do look like your sum of odds example.
Now, why does it actually work? You say:
$n$ belongs to integers or any other number group:
You actually have to be careful. You need a well-ordered set to induct -- that is, one where each non-empty subset has a least element (so, without doing something fancy, you cannot directly induct on the integers, for example, let alone the rationals or reals).
As an analogy: You are trying to show that you can reach every rung on a ladder, where the ladder has a rung for each member of $\mathbb{N}$. To do this:
- show that you can step onto the first rung directly
- suppose you are on some rung
- use that to show how to take a single step up
This is exactly the three parts of a PMI proof.
Here's a good way to try to understand why this works. A proof (of a true statement) is a roadmap to show why a purported counterexample fails to be a counterexample. Suppose that you believe that the sum of the first $k$ odd whole numbers is not $k^2$, at least not all the time. Then, since $\mathbb{N}$ is well-ordered, there would need to be a first number $k$ for which the statement fails (a smallest counterexample). But:
- You show that it works for $k = 1$, so that is not the counterexample.
- Suppose that $k$ is the smallest counterexample. Then the sum of the first $k-1$ odds is $(k-1)^2$ and your proof $P(n) \to P(n+1)$ shows how I can move from the sum of the first $k-1$ odds to the sum of the first $k$. This invalidates the purported counterexample.
But, my experience teaching induction is that often it feels like magic (or maybe cheating!) to students. They see it as we assumed $P(n)$ to prove $P(n)$. We did not. We assumed $P(n)$ and proved $P(n+1)$ and that is a big difference!
(*) Picky detail from way above: I can show the fact that $P(n) \to P(n+1)$ without regard to whether or not there is an $n$ for which $P(n)$ is true. The logic of Inductive Hypothesis -> Inductive Step does not care. But, if you don't have a base case, then you do not have a full induction proof.
