Can Any Real Number in (0,1) Exceed 1 with a Natural Number Exponent?

[Amer]

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]

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]

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]

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?

It should read:

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]

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.