I have a set of problems I am trying to complete. They are all proofs by induction. I have made attempts at most of them and always get stuck doing the inductive step.
Can someone explain in layman's terms what the inductive step does and how we are supposed to know the proper equation manipulation? Any tips on how to complete the Inductive Step portion of proof by induction would be very helpful!
Think of induction as dominoes being knocked over. The inductive step shows that if the statement (whatever it is) is true for $N$, it is true for $N+1$. But then applying the hypothesis to $N+1$, the statement is true for $N+2$, and so forth. So what the inductive statement is really doing is showing that if a statement is true for some $N$, it is true for all the ones bigger than $N$. So then it comes to the base case. Suppose you show the statement is true for $N= 5$, then combining this with the inductive hypothesis, you have shown the statement is true for $N= 5,6,7,\ldots$. As for the manipulation, you will assume that the statement is true for $N= k$, then you should manipulate the necessary equations or statements to show that the statement is true for $N= k+1$. But notice that in (most) of the cases, you are starting with the information you have for $N=k$ then pushing towards $N= k+1$, not the other way around. A common error is to assume the statement for $N=k$, but then write down the statement for $N= k+1$, hence tacitly assuming the case for $N= k+1$ (how else could you write down the statement for $N= k+1$ unless one is assuming what they have written is correct). For a lot of introductory induction problems, you can write the statement for $N=k+1$ and work towards $N=k$. Then reversing your steps will show the argument you should be making. This is because many introductory problems involve streams of $=$'s, which are reversible. But again, you start with what you know, $N=k$ (by assumption) and show the statement for $N=k+1$ holds.
You can think of this as knocking over dominos. The inductive hypothesis shows that if you knock over one of the dominos in the line, all the ones after it will eventually be pushed over. The base case is the one you push over first. So the two parts together to show that the domino you begin with and all subsequent dominos are knocked over.
If you wanted to read more about induction generally and see lots of examples, all very well done, I would strongly recommend Keith Conrad's Examples of Proof by Induction. If you wanted more specific advice, I would add the problem you are working on to the question.
The inductive step, when you are proving a statement $P(n)$ for all $n \in \Bbb N$ (or something similar like "all integers greater than $4$" or whatever), is showing that, if $P(k)$ is true, then $P(k+1)$ is also necessarily true.
This establishes a domino effect of sorts, when combined with the validation of the base case. You know, for instance, $P(0)$ holds, and that $P(k)$ implies $P(k+1)$. Since $P(0)$ is known to be true, the $P(k) \implies P(k+1)$ property shows (for $k=0$) that $P(1)$ holds. And then it shows $P(2)$ holds. And $P(3)$. And $P(4)$. And so on and so forth, never-ending. The truth of any value ends up implying the one "next up" in the chain. It's really quite a powerful domino effect.
So, as for your other question, when do you recognize how to use $P(k)$ in proving $P(k+1)$ holds? (After all, you assume $P(k)$ in proving $P(k+1)$ holds, so you should hopefully use it at some point!) Sadly, there's no general technique. When looking at the $P(k+1)$ statement, your goal should be to somehow manipulate the expression into a form that somehow resembles $P(k)$. For instance, a common introductory induction proof might consider
$$P(n) : \sum_{i=1}^n i = \frac{n(n+1)}{2}$$
If you assume $P(k)$, how might you prove $P(k+1)$? Notice that
$$P(k+1) : \sum_{i=1}^{k+1} i = \frac{(k+1)(k+2)}{2}$$
and you are assuming
$$P(k) : \sum_{i=1}^{k} i = \frac{k(k+1)}{2}$$
What you are looking to do is, in the $P(k+1)$ statement, to turn the sum into something referenced in $P(k)$. And notice that you can pull out the $(k+1)^{th}$ term -- that is,
$$\sum_{i=1}^{k+1} i = (k+1) + \sum_{i=1}^{k} i$$
And look, we see a reference to $P(k)$, the sum from $1$ to $k$! Since we assume $P(k)$ is true, we can substitute that summation with $k(k+1)/2$, and then work with the $k+1$ summand as well and try to validate $P(k+1)$.
Not all induction proofs are this easy, but that's the idea: manipulating $P(k+1)$ to reference $P(k)$, and validate its truth. How easily the manipulation comes to mind can only be honed with practice, however, because a wide variety of fields in math use induction to some degree.
