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I need to show using Bernoulli's inequality and the Archimedean property that

For all $x,y \in \mathbb{R}$ with $x \gt 1$ there exists an $n \in \mathbb{N}_0$ such that $x^n \ge y$.

My attempt:

Since $x \gt 1$ we can define $x := 1 + a$ for some $a \in \mathbb{R}_+$. Then for all $n \in \mathbb{N}$ by Bernoulli's inequality $x^n = (1+a)^n \ge 1+na \ge na$.

By the Archimedean property for every $b \in \mathbb{R}$ there exists $n \in \mathbb{N}$ so that $n \gt b$. Let's choose $b := \frac{y}{a}$ then there is an $n \in \mathbb{N}$ so that $na \gt y$.

Using this $n$ we get $x^n = (1+a)^n \ge 1+na \gt na \gt y$ which means $x^n \ge y$.


Is the proof okay? Is there a better way which is maybe shorter or simpler? Thanks!

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    $\begingroup$ The proof is okay and I would consider it simple enough. Maybe you can get rid of $1+na\gt na$ bit by choosing $n$ so that $1+na\gt y$, i.e. $na\gt y-1$, so you may start with $b=\frac{y-1}{a}$ ... but I am nitpicking here - it is good, really. $\endgroup$ – user491874 Dec 14 '17 at 20:53
  • $\begingroup$ This is the simplest proof I know of, and your presentation was fine. Of course once you become more knowledgeable about calculus you see that $x^n\rightarrow\infty$ as $n\rightarrow\infty$, so knowing what those terms mean proves it immediately. This isn't a proof per se, because it's not immediately clear why $x^n\rightarrow\infty$, and the best justification I can give is that the derivative is larger than the constant $\log x>0$. Anyway, rambling aside, I think your proof is fine as long as you don't need to prove Bernoulli. $\endgroup$ – Bob Jones Dec 14 '17 at 20:56
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If $y \leq 0$ then it is obvious.

If $0<y\leq 1$ then $n=1$ works.

If $y>1$ and such $n$ does not exist then $x^n<y$ for every $n \in \mathbb N$ but because $x^n>x^m$ if $n>m$ and $x>1$ this would mean that $x^n$ has an upper bound ($y$ is one of them), which is not true since $x^n$ grows without bounds for $x>1$ and $n \to + \infty$, so we arrived at the contradiction, which means that such $n$ always exists.

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