Mathematical Induction: For all n∈N and p∈R with p>=-1; 1+np<=(1+p)^n

Question

For all n∈N and p∈R with p≥-1; 1+np≤$(1+p)^n$; prove by mathematical induction.

So I know that you start with n=1, then assume P(k) is valid and then prove P(k+1). How do I prove that with the p≥-1. Do I need to prove for p+1 as well?

Answer 1

You SHOULDN'T work with $p+1$, since $p\in\mathbb{R}$ is not a natural number. Basically you would be proving for $p$ in a discrete subset of $\mathbb{R}$.

When using induction, you should choose one variable in $\mathbb{N}$ and fix the other variables, even if they belong to $\mathbb{N}$. In this particular case, you apply induction on $n\in\mathbb{N}$ (your only option). That implies you have consider $p\ge-1$ a "fixed variable" (if you forgive me the oxymoron). Let $p\ge-1$ be fixed, you start with $n=1$, i.e., $1+1\cdot p\le(1+p)^1$, which is evidently true. Then you suppose true for $n=k$, i.e., $1+kp\le(1+p)^k$. You just had to prove it to be true for $n=k+1$, i.e., $1+(k+1)p\le(1+p)^{k+1}$. (This last step is easy, so I leave it to you.)

Another alternative (but equivalent) version of induction is: Let $p\ge-1$ be fixed; you prove for $n=1$ (trivial, as above), then suppose the assertion to be true for all $n\le k$, and prove for $n=k+1$. (In this case, I don't think it is much useful than the previous version, but it is important that you know this alternative.)

I hope this helps!

Answer 2

$p$ has nothing to do with $n$. You do not prove it for different values of $p$. You assume only one thing about $p$; that $p \ge -1$. When you prove your different values of $n$ you prove them for $p \ge -1$. As you prove it for all $p\ge -1$ you don't need to prove it for any other $p$.

It it's true for $p \ge -1$ then it will be true for $r = p+1$ because $r \ge -1$ and you proved it for all $p \ge -1$. You don't need to prove it twice.

The only reason we do it for $n=1$ and then show $P(n=k) \implies P(n=k+1)$ is because we don't know how to prove it for all $n$ at once.