Real numbers natural

Amer

Active member
how to prove that for any real number in r (0,1) there exist a natural number n in N such that
rn > 1

CaptainBlack

Well-known member
how to prove that for any real number in r (0,1) there exist a natural number n in N such that
rn > 1
Since between any two real numbers there is a rational, let $$p/q, \ p,q \in \mathbb{N}$$ be such that:

$\frac{r}{2}<\frac{p}{q}<r$

Then multiplying through by $$q$$ we get:

$1\le {p}<rq$

CB.

Evgeny.Makarov

Well-known member
MHB Math Scholar
how to prove that for any real number in r (0,1) there exist a natural number n in N such that
rn > 1
I don't understand the question. What is "r (0,1)"? You want a number n such that m > 1? If you want a number > 1, why not take 2?

CaptainBlack

Well-known member
I don't understand the question. What is "r (0,1)"? You want a number n such that m > 1? If you want a number > 1, why not take 2?
Prove that for any real number $$r \in (0,1)$$ there exist a natural number $$n \in N$$ such that $$r n > 1$$