- Thread starter
- #1

#### conscipost

##### Member

- Jan 26, 2012

- 39

Any feedback appreciated,

- Thread starter conscipost
- Start date

- Thread starter
- #1

- Jan 26, 2012

- 39

Any feedback appreciated,

- Admin
- #2

- Mar 5, 2012

- 9,416

I'm afraid not, since $(-2)^x=-(2^x)$ does not hold for $x=2/3$.

Any feedback appreciated,

That is because $(-2)^{2/3} = 2^{2/3} \ne -(2^{2/3})$.

In other words, the sign flips up and down discontinuously.

If you further restrict the domain to only fractions with an odd numerator and an odd denominator, the function becomes continuous, since it is effectively indeed $-(2^x)$.

In any interval there are still infinitely many elements, so the limit-definition of continuous is satisfied everywhere.

However, you're left with almost no algebraic structure.

This set is not a