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I honestly have no idea where to start with the following proof, and I was wondering if anyone could help me get started. I don't want the whole idea, I just need to know where to start with this proof.

The question: Let x be a real number. Suppose 1 < x. Given any M > 0; prove there is a natural number N such that n $\geq$ N implies M $\leq$ $x^n$.

My book has absolutely no examples (it's a custom text), and I honestly have just stared at this problem for a few hours now. If someone would be willing to just help me get going, I think I should be able to work out the rest from there. I appreciate any help anyone would be willing to offer, thank you.

edit: Here is the solution I came up with after the generous help I received below.

Let M > 0 and let x = 1 + t. Since x > 1, we have 1 + t > 1 if and only if t > 0. We then have $x^n$ = $(1 + t)^n$. Using the Bernoulli Inequality, we have $(1 + t)^n$ $\geq$ 1 + nt for all n $\geq$ 0. We then have $x^n$ $\geq$ 1 + nt > nt, thus $x^n$ > nt for all n $\geq$ 1. Let N be a natural number with N = $\frac{M}{t}$. Then n $\geq$ N = $\frac{M}{t}$ implies nt $\geq$ M. Since $x^n$ > nt, we have that n $\geq$ N implies $x^n$ > M.

I know my proof-writing skills are very bad and I'm sure there are a million ways I can improve what I've done, but I am very thankful for the help I received here from Andre Nicolas.

benty
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1 Answers1

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A start: Let $x=1+t$. Then $t\gt 0$ and $x^n=(1+t)^n$. Now use the Bernoulli Inequality $$(1+t)^n \ge 1+nt.$$ This can be proved by induction, or by using the Binomial Theorem.

André Nicolas
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  • Thanks for the help, I really, really appreciate it. I absolutely never would have known to use the Bernoulli Inequality, my book hasn't even mentioned it. I'm going to give this a go, thank you very much for the help. I can't up vote your post since I don't have 15 reputation unfortunately, but I really appreciate the help. Thank you. – benty Sep 23 '14 at 02:50
  • You are welcome. Your comment is much better than an upvote. You can use the Binomial Theorem, $(1+t)^n=1+nt+\cdots$, but that is technically less "elementary" than the Bernoulli Inequality, which has a one line induction proof. Another way to solve your problem is to use the logarithm. We have $x^n\gt M$ if $n\gt \frac{\ln M}{\ln x}$. Perfectly good, except that it uses far more machinery. – André Nicolas Sep 23 '14 at 02:56
  • I believe I was able to correctly provide a proof after your generous assistance, thank you very much. My text hasn't yet covered logarithms, if I reworked the proof for practice using them would I simply be able to apply the rules of logs I learned in pre-calc, or would I need to look up specific definitions for logarithms to have a correct proof? Regardless, thank you again for your help, I would never have been able to arrive at a solution without it. – benty Sep 23 '14 at 03:47
  • You are welcome. Whether you can use school properties of logarithms is course-dependent. In a typical "proof" or "analysis" course you cannot, unless they have already been proved reasonably rigorously in the course. – André Nicolas Sep 23 '14 at 03:55