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If $x,y$ are rationals and both positive numbers, how can I show:

  1. if $x>1$ there is a positive interger $n$ such that $x^n>y$,
  2. if $0 < x < 1$ there is a positive interger $n$ such that $x^n<y$.

I'd particularly like a proof that uses the binomial theorem.

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depending on the relationship between $x$ and $y$ i.e if $x\le y$ or $x\ge y$. – user31280 Oct 25 '12 at 11:35
up vote 2 down vote accepted
  1. Write $x=(1+t)$, where $t>0$. Then $$x^n=(1+t)^n=\sum_k \binom nkt^k = 1+nt+\binom n2t^2+\dots \ge 1+nt$$ (so far this is called the Bernoulli inequation). In order that this be $>y$, try to find a corresponding $n\in\Bbb N$.

  2. The easiest if we reduce it to 1. by considering $1/x$ which is $>1$ in this case, and $1/y$ in place of ($y$ in 1).

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