3

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.

Botond
  • 11,938
Turkhan Badalov
  • 1,083

5 Answers5

2

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.

ArsenBerk
  • 13,211
  • 1
    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
1

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.

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
1

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*}

primes.against.humanity
  • 2,221
0

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.

ajotatxe
  • 65,084
0

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 $x\ge 2$.

user
  • 154,566