- #1

- 9

- 0

## Main Question or Discussion Point

Just a quick question for you guys, I've been unable to find the answer to this. Can all real numbers be written as n^p, where p is a real number?

- Thread starter kulix
- Start date

- #1

- 9

- 0

Just a quick question for you guys, I've been unable to find the answer to this. Can all real numbers be written as n^p, where p is a real number?

- #2

- 22,097

- 3,281

What is n??

Anyway, negative numbers can't be written in that form.

Anyway, negative numbers can't be written in that form.

- #3

- 9

- 0

Or

if n < n * sqrt(n^2-1) < n^2, does that mean there exists p such that n * sqrt (n^2 -1) = n^p ?

- #4

- 22,097

- 3,281

- #5

- 9

- 0

The better question;

if n < n * sqrt(n^2-1) < n^2, does that mean there exists p such that n * sqrt (n^2 -1) = n^p ?

- #6

- 22,097

- 3,281

Take [itex]n=\sqrt[3]{x}[/itex] and p=3.

- #7

- 9

- 0

let S be a series from 2 to infinity of 1 / (n*sqrt(n^2 -1)).

Can you write S as 1/n^p?

- #8

- 22,097

- 3,281

This would forceif n < n * sqrt(n^2-1) < n^2, does that mean there exists p such that n * sqrt (n^2 -1) = n^p ?

[tex]p=\log_n{n\sqrt{n^2-1}}[/tex]

This is ok if the logarithm exists. For that, we need n>0, n≠1 and [itex]n\sqrt{n^2-1}>0[/itex].

- #9

- 695

- 2

P.S.: Oops, two posts got in the middle while I was writing this. I was intending to continue after post #6.

- #10

- 9

- 0

Perfect, thank you!

- #11

- 22,097

- 3,281

This question makes no sense. In the series S, your variable n is the dummy variable and thus ranges over all positive integers. In your last sentence, n has become a fixed number.

let S be a series from 2 to infinity of 1 / (n*sqrt(n^2 -1)).

Can you write S as 1/n^p?

- #12

- 9

- 0

My bad again, what I meant to say is can you write it as a series of 1/n^p.

- #13

- 22,097

- 3,281

[tex]\sum C\frac{1}{n^2}[/tex]

where C is a constant.

- #14

- 9

- 0

- #15

- 22,097

- 3,281

So if you follow Dodo's and my hints, then you won't end up with

[tex]\sum \frac{1}{n^p}[/tex]

but rather with something in the exponent which is also dependent of n. This makes the situation harder.

- #16

- 9

- 0

- #17

Office_Shredder

Staff Emeritus

Science Advisor

Gold Member

- 3,750

- 99

- Last Post

- Replies
- 3

- Views
- 348

- Last Post

- Replies
- 18

- Views
- 4K

- Last Post

- Replies
- 11

- Views
- 7K

- Replies
- 7

- Views
- 8K

- Last Post

- Replies
- 12

- Views
- 948

- Last Post

- Replies
- 29

- Views
- 3K

- Replies
- 5

- Views
- 5K

- Last Post

- Replies
- 6

- Views
- 476

- Replies
- 2

- Views
- 10K

- Replies
- 5

- Views
- 2K