I’ve never really understood why math induction is supposed to work.

You have these 3 steps:

Prove true for base case (n=0 or 1 or whatever)

Assume true for n=k. Call this the

induction hypothesis.Prove true for n=k+1,

somewhere using the induction hypothesisin your proof.In my experience the proof is usually algebraic, and you just manipulate the problem until you get the induction hypothesis to appear. If you can do that and it works out, then you say the proof holds.

Here’s one I just worked out,

Show $\displaystyle\lim_{x\to\infty} \frac{(\ln x)^n}{x} = 0$

So you go:

Use L’Hospital’s rule.

$\displaystyle\lim_{x\to\infty} \frac{\ln x}{x} = 0$.

Since that’s

$\displaystyle\lim_{x\to\infty} \frac{1}{x} = 0$.

Assume true for $n=k$.

$\displaystyle\lim_{x\to\infty} \frac{(\ln x)^k}{x} = 0$.Prove true for $n=k+1$. You get

$\displaystyle\lim_{x\to\infty} \frac{(\ln x)^{k+1}}{x} = 0.$Use L’Hospital again:

$\displaystyle\lim_{x\to\infty} \frac{(k+1)(\ln x)^{k}}{x} = 0$.Then you see the induction hypothesis appear, and you can say this is equal to $0$.

What I’m not comfortable with is this idea that you can just

assumesomething to be true ($n=k$), then based on that assumption, form a proof for $n=k+1$ case.I don’t see how you can use something you’ve assumed to be true to prove something else to be true.

**Answer**

The inductive step is a proof of an implication: you are proving that **if** the property you want holds for $k$, **then** it holds for $k+1$.

It is a result of formal logic that if you can prove $P\rightarrow Q$ (that $P$ implies $Q$), then from $P$ you can prove $Q$; and *conversely*, that if from assuming that $P$ is true you can prove $Q$, then you can in fact prove $P\rightarrow Q$.

We do this pretty much every time we prove something. For example, suppose you want to prove that if $n$ is a natural number, then $n^2$ is a natural number. How do we start? “Let $n$ be a natural number.” Wait! Why are you allowed to just *assume* that you already have a natural number? Shouldn’t you have to start by *proving* it’s a natural number? The answer is **no**, we don’t have to, because we are not trying to prove an absolute, we are trying to prove a *conditional* statement: that **if** $n$ is a natural number, **then** something happens. So we may begin by assuming we are already in the case where the antecedent is true. (Intuitively, this is because *if* the antecedent is false, then the implication is necessarily true and there is nothing to be done; formally, it is because the Deduction Theorem, which is what I described above, tells you that if you manage to find a formal proof that ends with “$n^2$ is a natural number” by assuming that “$n$ is a natural number” is true, then you can use that proof to produce a formal proof that establishes the implication “if $n$ is a natural number then $n^2$ is a natural number”; we don’t have to go through the exercise of actually producing the latter proof, we know it’s “out there”).

We do that in Calculus: “if $\lim\limits_{x\to x_0}f(x) = a$ and $\lim\limits_{x\to x_0}g(x) = b$, then $\lim\limits_{x\to x_0}(f(x)+g(x)) = a+b$.” How do we prove this? We begin by *assuming* that the limit of $f(x)$ as $x\to x_0$ *is* $a$, and that the limit of $g(x)$ as $x\to x_0$ *is* $b$. We assume the premise/antecedent, and proceed to try to prove the consequent.

What this means in the case of induction is that, since the “Inductive Step” is actually a statement that says that an implication holds:

$$\mbox{“It” holds for $k$}\rightarrow \mbox{“it” holds for $k+1$},$$

then in order to prove this implication we can begin by assuming that the *antecedent* is already true, and then proceed to prove the consequent. Assuming that the antecedent is true is precisely the “Induction Hypothesis”.

When you are done with the inductive step, you have in fact not proven that it holds for any *particular* number, you have *only* shown that **if** it holds for a particular number $k$, **then** it must hold for the next number $k+1$. It is a *conditional* statement, not an absolute one.

It is only when you combine that conditional statement with the base, which *is* an absolute statement that says “it” holds for a specific number, that you can conclude that the original statement holds for all natural numbers (greater than or equal to the base).

Since you mention dominoes in your title, I assume you are familiar with the standard metaphor of induction like dominoes that are standing all in a row falling. The inductive step is like arguing that all the dominoes will fall if you topple the first one (without actually toppling it): first, you argue that each domino is sufficiently close to the *next* domino so that **if** one falls, **then** the next one falls. You are not tumbling every domino. And when you argue this, you argue along the lines of “suppose this one falls; since it’s length is …”, that is, you *assume* it falls in order to argue the next one will *then* fall. This is the same with the inductive step.

In a sense you are right that it feels like “cheating” to assume what you want; but the point is that you aren’t really assuming what you want. Again, the inductive step does not in fact establish that the result holds for *any* number, it only establishes a *conditional* statement. **If** the result happens to hold for some $k$, **then** it would *necessarily* have to also hold for $k+1$. But we are completely silent on whether it *actually* holds for $k$ or not. We are not saying anything about that at the inductive-step stage.

*Added:* Here’s an example to emphasize that the “inductive step” does not make any absolute statement, but only a conditional statement: Suppose you want to prove that for all natural numbers $n$, $n+1 = n$.

**Inductive step.** *Induction Hypothesis:* The statement holds for $k$; that is, I’m assuming that $k+1 = k$.

*To be proven:* The statement holds for $k+1$. Indeed: notice that since $k+1= k$, then adding one to both sides of the equation we have $(k+1)+1 = k+1$; this proves the statement holds for $k+1$. **QED**

This is a perfectly valid proof! It says that **if** $k+1=k$, **then** $(k+1)+1=k+1$. *This is true!* Of course, the antecedent is never true, but the *implication* is. The reason this is not a full proof by induction of a false statement is that there is no “base”; the inductive step only proves the conditional, nothing more.

By the way: Yes, most proofs by induction that one encounters early on involve algebraic manipulations, but not all proofs by induction are of that kind. Consider the following simplified game of Nim: there are a certain number of matchsticks, and players alternate taking $1$, $2$, or $3$ matchsticks every turn. The person who takes the last matchstick wins.

**Proposition.** In the simplified game above, the first player has a winning strategy if the number of matchsticks is not divisible by $4$, and the second player has a winning strategy if the number of matchsticks is divisible by 4.

The proof is by (strong) induction, and it involves no algebraic manipulations whatsoever.

**Attribution***Source : Link , Question Author : bobobobo , Answer Author : mathworker21*