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

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Discussion Overview

The discussion revolves around the question of whether any real number in the interval (0,1) can exceed 1 when raised to a natural number exponent. Participants explore the implications of this question and clarify the notation used.

Discussion Character

  • Conceptual clarification, Debate/contested

Main Points Raised

  • Some participants propose that for any real number \(r\) in the interval (0,1), there exists a natural number \(n\) such that \(r^n > 1\).
  • One participant suggests using rational numbers between \(r\) and \(\frac{r}{2}\) to support their argument, indicating that \(1 \leq p < rq\) for some \(p/q\) in \(\mathbb{N}\).
  • Another participant expresses confusion about the notation "r (0,1)" and questions the logic behind seeking a number greater than 1, suggesting that simply taking 2 would suffice.
  • A later reply clarifies that the notation should read \(r \in (0,1)\) and corrects the interpretation of the expression \(r^n\) as being misread as "m".
  • One participant humorously notes a personal issue with vision that contributed to the misunderstanding of the notation.

Areas of Agreement / Disagreement

The discussion contains multiple interpretations and clarifications regarding the notation and the underlying question. There is no consensus on the original question's intent or the validity of the claims made.

Contextual Notes

Participants express uncertainty about the notation and the mathematical implications of the question, leading to potential misunderstandings in the discussion.

Amer
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how to prove that for any real number in r (0,1) there exist a natural number n in N such that
rn > 1
 
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Amer said:
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.
 
Amer said:
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?
 
Evgeny.Makarov said:
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
 
CaptainBlack said:
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.
 

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