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 $$r n > 1$$
What you have taken to be an "m" is in fact "r n" but with no space so that in the default font it looks like m

CB

Evgeny.Makarov

Well-known member
MHB Math Scholar
What you have taken to be an "m" is in fact "r n" but with no space so that in the default font it looks like m
Wow, talk about keming. It is true, I recently changed contact lenses and my vision went down a bit.