Question about the structure of mathematical induction

Question

I'm studying proof by mathematical induction, and I have a question about the step where you demonstrate that a predicate $P(n) \Rightarrow P(n+1)$. Consider the following generalized example:

Prove or disprove: $\forall n \in \mathbb N \, \Big(P(n)\Big)$.

Step 1: I show $P(1) \equiv True$.

Step 2: I must demonstrate that $\forall n \in \mathbb N \, \Big(P(n) \Rightarrow P(n+1)\Big)$.

Here's where I get confused. My professor told me that in order to demonstrate this, I assume that for any arbitrary n, $P(n)$ is true. Then I show that $P(n+1)$ logically follows from assuming $P(n)$.

However, if I assume $P(n)$ is true, am I not also assuming $P(n+1)$ is true automatically? Since $(n+1) \in \mathbb{N}$, n+1 is a natural number, and I've already assumed that for any arbitrary natural number, P(n) is true.

If I instead start by assuming that for a specific n, P(n) is true, then the principle of mathematical induction does not prove the original statement because that specific n might by n = 6 in which case P(2), P(3), P(4), P(5) might not be true and I've only proved the proposition for n = 1 and n >= 6, not for all n.

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I know there is a flaw in my reasoning here somewhere, appreciate the help!

Answer 1

Actually, in step $2$ you show that if $P(n)$ is true for any fix $n$ then it is true for the natural number that follow $n$ which it is $n+1$. Here $n$ is an arbitrary fix natural number. So because you have proved that it is true for $n=1$ (in step $1$) then it is true for $1+1=2$.

Now, we know that $P(n)$ is true for $n=2$ so it is true for $2+1=3$ (Because we have proved step $2$).

Again, because it is true for $n=3$ so it is true for $3+1=4$

If we continue this, then $P(n)$ is true for all the natural numbers.

Answer 2

To begin with, here's how the induction step should go:

We want to show that $(\forall n \in \mathbb{N})[P(n)\implies P(n+1)]$.

To do this, we pick, arbitrarily, a member $n$ of $\mathbb{N}$.

We assume $P(n)$ holds. Then, on the basis of our argument (which depends on what $P(n)$ actually is), we show that $P(n+1)$ must also hold.

Thus, we have shown that regardless of what natural number $n$ actually is (since we chose it arbitrarily), $P(n) \implies P(n+1)$. Thus, $(\forall n \in \mathbb{N})[P(n)\implies P(n+1)]$.

Now, your confusion arises from

My professor told me that in order to demonstrate this, I assume that for any arbitrary n, P(n) is true.

By this, your professor didn't mean for you to assume $(\forall n \in \mathbb{N})P(n)$, which would be assuming the very theorem you want to prove; rather, it was them wanting you to pick any natural number, chosen arbitrarily, call it $n$, and assume $P(n)$ is true for this $n$. At which point, the above argument can be used to complete the proof.

If I instead start by assuming that for a specific n, P(n) is true, then the principle of mathematical induction does not prove the original statement because that specific n might by n = 6 in which case P(2), P(3), P(4), P(5) might not be true and I've only proved the proposition for n = 1 and n >= 6, not for all n.

Assuming $P(n)$ for a specific $n$ would mean picking a fixed number like $5$ and assuming $P(5)$. In which case showing $P(n+1)$ would just prove $P(5)\implies P(6)$ and nothing else. Instead, you are meant to assume $P(n)$ for a particular but unspecified $n$, and show $(\forall n \in \mathbb{N})[P(n)\implies P(n+1)]$. The idea is, since we know $P(n)\implies P(n+1)$ for any $n$, that means $P(1) \implies P(2)$ and $P(2) \implies P(3)$, and so on; as long as we know $P(1)$ is true, the chain of implications guarantees $P(n)$ for every other $n$.