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As the question states, I want to prove that $ \exists n \in \mathbb{N}$ s.t. $10^{-n} < x <10^x$ for $x>0, x \in \mathbb{R}$. The hint given is to prove $10^n >n$.

I proved $10^n > n$ via proof by induction. The base case is trivial. The inductive step is as follows:

Assume $10^n > n$ is true. We can rewrite $10^{n+1}$ as $10^n \cdot 10 \rightarrow 10^n(9+1) \rightarrow 10^n + 9 \cdot 10^n$

Then we can add $9 \cdot 10^{n}$ to our inductive hypothesis and get

$10^{n+1} > n + 9 \cdot 10^{n}$

And so $10^{k}>k$ for $ \forall K \in \mathbb{N}$.

My problem now is that I don't see how this helps me prove $10^{-n} < x <10^x$. Any help or hints would be much appreciated.

Update: Apparently if we prove $10^n>n$, we can use this fact coupled with the Archimedian property: $n\cdot x>y$ for $x,y\in R$ and if we let x=1 and y=x, we have $n>x$. With fact 1, we have $10^n > n > x$ so $10^n>x$.

I am however, stuck on proving $10^{-n} < x$. I feel like I am over thinking and that the proof for it would be trivial.

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Nikitau
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    $x<10^x$ is always true for real $x$, so this has nothing to do with $n$. Also, what happened to the $y$ in the title? This question needs to be carefully edited. – David Oct 06 '16 at 01:23
  • If that was the hint then perhaps the problem was for all $x\in(0,1)$, in which case $x\leq 10^x$ is easily shown to be true. – NickC Oct 06 '16 at 01:26
  • @NickC Unfortunately it's not that easy. It's for $ x \in \mathbb{R}$, which is leaving me completely baffled. – Nikitau Oct 06 '16 at 02:00
  • This question is closely related. http://math.stackexchange.com/questions/741600/prove-that-logx-x-for-x-0-x-in-mathbbn – NickC Oct 06 '16 at 02:47
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    With fact 1, we have 10^n > n > x so 10^n > x How is that relevant? In your question (and title) is the second inequality $x \lt 10^x$ or is it maybe $x \lt 10^n$? – dxiv Oct 07 '16 at 00:46

2 Answers2

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This doesn't really help all that much. You need to prove two things:

  1. $x<10^x$ for all real $x$.

  2. For all real $x$, there exists an integer $n$ such that $10^{-n}<x$.

(Also, I think you want to restrict this to positive real numbers $x$ - (2) is false for nonpositive $x$.)

What you did is prove that for all integers $x$, $x<10^x$. That's fine, but you need to prove it for all real numbers.

Carl Schildkraut
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  • Exactly. I'm not really sure why that was given as a hint. Your answer definitely helped narrow things down, though. Thanks! – Nikitau Oct 06 '16 at 03:01
  • Hi! I have an update. Apparently if we prove $10^n > n$, we can use this fact coupled with the Archimedian property: $n \cdot x > y$ for $x,y \in \mathbb{R}$, and if we let x=1 and y=x, we have $n > x$. With fact 1, we have $10^n > n > x$ so $10^n > x$.

    I think the reverse proof for $10^{-n}$ might be analogous. I'm trying to figure that out.

     – Nikitau Oct 06 '16 at 23:08
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Let $n:=\lfloor\log(1/x)/\log10\rfloor+1.$ Then of course $10^{-n}<x.$ Now, the other part of the problem. Let $\lfloor x\rfloor$ be the largest integer that is not greater than $x.$ Then, if $x\geqslant0,$ we have $\lfloor x\rfloor\geqslant0.$ Now you can easily prove by induction on $m\geqslant0$ that $m+1\leqslant10^m.$ Thus $\lfloor x\rfloor+1\leqslant10^{\lfloor x\rfloor}$ and since $x<\lfloor x\rfloor+1$ and also $10^{\lfloor x\rfloor}\leqslant10^x$ then $x<10^x.$

CIJ
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