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.

• do you know derivatives? – Exodd Sep 15 '18 at 12:49
• One needs additional information beyond the proof that $2^n>n$ to prove $2^{2.5} > 2.5$. Perhaps prove $2^x/x$ is increasing for $x>2$ and use other arguments for $x<2$. – GEdgar Sep 15 '18 at 12:49
• @Exodd, sure. Like we find derivatives of both fucntions and then? Compare them to each other? In other words, we return to the same problem if two derivatives are difficult to compare. Then to find derivatives again? What is the technique after all? – Turkhan Badalov Sep 15 '18 at 12:52
• related : For all reals $x$, prove $2^x > x$ – mathlove Sep 15 '18 at 12:52
• An arbitrary function $f$, even a strictly increasing smooth one, can satisfy $f(n)>n$ for all $n\in \Bbb N$ but not $f(x)>x$ for all $x$. More information about the function $f(x)=2^x$ is needed. – DanielWainfleet Sep 15 '18 at 13:13

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.

• To the proposer: For non-negative integer $a$ we have $2^a>a\implies 2^a\geq a+1.$.... So if $r>0$ and $a+1>r>a$ then $r\geq 2^r\implies a+1>r\geq 2^r>2^a\geq a+1,$ implying $a+1>a+1.$ The reason this doesn't work for an arbitrary strictly increasing function is that for the function $2^x$ the value of $2^a$ is an integer when $a$ is a non-negative integer, so the inequlaity $2^a>a$ can be strengthened to $2^a\geq a+1$. – DanielWainfleet Sep 15 '18 at 13:32
• That's a very nice way to prove it without using the statement "$2^k = k$ has no real solution for $k \ge 1$". I will edit my answer so that OP also reads your comment, thank you :) – ArsenBerk Sep 15 '18 at 13:38
• I was about to post an answer but your answer seemed to have a similar idea . BTW. The proposer is automatically notified of comments to answers . – DanielWainfleet Sep 15 '18 at 15:12

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.

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*}

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.

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 $x\ge 2$.