# Handsome calculus n00b seeks reassuring relationship with numbers.

1. Aug 2, 2006

### 3trQN

Hi peeps! I have some minor calculus problem, well confusion is the problem.

I was playing about with some numbers while doing some differential calc problems, when i started to explore a little further one expression.

$$y=f(x)=\sqrt[m]{\frac{1}{x^{n}}}=x^{-\frac{n}{m}}$$ ----- (1)

Of course i thought about rational numbers and primes, and that a rational number is any number which can be expressed as the quotient of two integers.

So assuming:
$$n \in Z^+$$
$$m \in Z^+$$

I then thought about what if the exponent was a prime, so
$$\frac{p}{m}$$ where $$p=prime$$

Then for:
$$1 < m < p$$
The exponent would allways be irrational.

Upon seeing the inequality expresison i wrote down i wondered if there was a link between it and the triangle inequality expression. Is this the case?

Last edited: Aug 2, 2006
2. Aug 2, 2006

### star.torturer

god title, but i think its puting people off

3. Aug 2, 2006

### 3trQN

story of my life that

4. Aug 2, 2006

### StatusX

The last equality isn't true. Do you want:

$$\sqrt[m]{\frac{1}{x^n}}=x^{-\frac{n}{m}}$$

or:

$$\sqrt[m]{\frac{1}{n}}=n^{-1/m}$$ ?

What does this mean? The exponent is p/m, which is rational when m is rational. Are you saying the function is irrational? This depends on which of the above two functions you're talking about.

I don't see what you mean. What kind of link are you thinking of?

Last edited: Aug 2, 2006
5. Aug 2, 2006

### 3trQN

Oops i forgot the x, sorry im latex illiterate. I meant the first correction, edited my post.

Ok ill re-phrase the post to clear things up, apologies.

6. Aug 2, 2006

### 3trQN

hmm nevermind, its best you all forget i ever posted this......i feel so stupid now

7. Aug 2, 2006

### daveb

It doesn't work. If x = 27, p = 7 and m = 3, then x ^(-p/m) = 27 ^(-7/3) = 1/3^7. None of these are irrational.

8. Aug 2, 2006

### 3trQN

Yes, ok dont rub it in :rofl: