Why does proving inductively $n < 2^n$ for $n \geq 1$ imply it is true for real values of $n$? If we prove by induction that $2^n > n$ for $n \geq 1$ where $n \in N^+$, how can one know this inequality holds for real values of n like $2^{2.5} > 2.5$?
Maybe a bit silly question but I can't find answer by myself. I think I need to show that the function $2^n$ is larger than $n$ analytically rather by induction but I don't know how and if induction is sufficient, then why? Thanks in advance.
 A: It is probably not the most appropriate way but since we know that both $f(x) = x$ and $g(x) = 2^x$ is continuous for $x \ge 1$, we can say if there is a real number $r$ such that $r > 2^r$, then this number should be between two adjacent integers, say $a$ and $a+1$. We know that $2^a > a$ and $2^{a+1} > a+1$ by induction. So if there exists such $r$, then there should be some real value, say $k$, where $2^k = k$ and $a < k < a+1$. But for $k \ge 1$, this equality doesn't have a real solution.
EDIT: DanielWainfleet proposed a better way in comments, I strongly recommend you to read that proof as well.
A: It doesn't work this way. 
You have to prove it for real number as well. 
Possibly by considering the derivative, find it's minimum point
$$f(x) = 2^x - x$$
$$f'(x) = 2^x \ln 2 - 1$$
and note that $f$ is convex.
A: We have $2^n > n \Leftrightarrow 2^n \geqslant n+1$. If $n < x < n+1$ for some integer $n$, then
\begin{align*}
2^x > 2^n \geqslant n+1 > x.
\end{align*}
A: It is not immediate. A formula for integers may not hold for reals. For example:
$$\sin n\pi=0$$
But in this case you can "extend" the formula using the fact that the function $f(x)=2^x-x$ is increasing in $[1,\infty)$.
Indeed, if $x\ge 1$ is real, take $n=\lfloor x\rfloor$. Then
$$2^x-x>2^n-n>0$$
The key point here is how to prove that $f$ is increasing. If you know derivatives, it is easy.
A: We have that
$$n < 2^n \iff 2^n-n>0$$
and since for any $x>1$ we have $x\in(n,n+1)$
$$2^{n}-(n+1)\le 2^x-x$$
then it suffices to show by induction that also $2^{n}-n-1>0$ holds for $n\ge 2$.
Therefore from $n+1<2^n$ proved by induction we can conclude that $x<2^x$ for any $x\ge 2$.
