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Real numbers natural

Amer

Active member
Mar 1, 2012
275
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
Jan 26, 2012
890
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
Jan 30, 2012
2,492
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
Jan 26, 2012
890
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

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
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.