Negative numbers to non-interger powers

[scott_alexsk]

Hello,

How would you determine the value of such numbers as -3^1.5?

Thanks,
-scott

 

[Integral]

$$- 3 ^{ 1.5} = - 3^ {\frac 3 2}$$

Do either the square root or the cube first, either way results in the square root of a negative number. So the solution lies in the imaginary plane.

 

[BoTemp]

Have you ever worked with complex numbers before? Define i = (-1)^0.5
-3^1.5 = (3)^1.5 * (-1)^1.5 = 3^1.5 * (-1)^1 * (-1)^0.5 =
$$-i3\sqrt{3}$$. By extension of this procedure one can define negative numbers to any power.

 

[cracker]

hmm... idk i would just put it in the caculator :D but i do know that a negative number squared by an odd nuber is X i idk if this helps

 

[bomba923]

$$- 3 ^{ 1.5} = - 3^ {\frac 3 2}$$

Do either the square root or the cube first, either way results in the square root of a negative number. So the solution lies in the imaginary plane.

As you wrote it, you do not have a square root of a negative number:
$$-3^{1.5} = -3^{\frac{3}{2}} = - \sqrt{27} = - 3\sqrt{3}$$

On the other hand,
$$(-3)^{1.5} = (-3)^{\frac{3}{2}} = \sqrt{-27} = 3i \sqrt{3}$$

Have you ever worked with complex numbers before? Define i = (-1)^0.5
-3^1.5 = (3)^1.5 * (-1)^1.5 = 3^1.5 * (-1)^1 * (-1)^0.5 =
$$-i3\sqrt{3}$$. By extension of this procedure one can define negative numbers to any power.

$$(-3)^{1.5} = (-3)^{\frac{3}{2}} = \sqrt{-27} = \sqrt{27} \sqrt{-1} = \boxed{3i \sqrt{3}}$$
not
$$-3i \sqrt{3}$$

*Also, note that:
$$\forall x < 0, x^k \notin \mathbb{R} \; \text{ if } \, k \notin \mathbb{Q}$$

 

[HallsofIvy]

You posted this 9 times?! I deleted the other 8.

As you wrote it, you do not have a square root of a negative number:
$$-3^{1.5} = -3^{\frac{3}{2}} = - \sqrt{27} = - 3\sqrt{3}$$

On the other hand,
$$(-3)^{1.5} = (-3)^{\frac{3}{2}} = \sqrt{-27} = 3i \sqrt{3}$$

A good point!

$$(-3)^{1.5} = (-3)^{\frac{3}{2}} = \sqrt{-27} = \sqrt{27} \sqrt{-1} = \boxed{3i \sqrt{3}}$$
not
$$-3i \sqrt{3}$$

While "the" square root of a positive real number is define to be the positive root, that is not true for complex numbers, where most functions are "multi-valued". (-3)^1.5 has two values,
$3i\sqrt{3}$ and $-3i\sqrt{3}$

*Also, note that:
$$\forall x < 0, x^k \notin \mathbb{R} \; \text{ if } \, k \notin \mathbb{Q}$$

?? or even if k is rational: 1.5 is certainly rational! Did you mean
$$k \notin \mathbb {I}$$?

 

[bomba923]

You posted this 9 times?! I deleted the other 8.

Not really ~~

The problem was that I couldn't edit the $$\LaTeX$$ just by using the "Edit" or "Go Advanced". I had to post, delete, and repost (for some time, so it "appeared" as though I posted nine times)

While "the" square root of a positive real number is define to be the positive root, that is not true for complex numbers, where most functions are "multi-valued". (-3)^1.5 has two values,
$3i\sqrt{3}$ and $-3i\sqrt{3}$

I see; so the answer is then
$$\pm 3i\sqrt{3}$$

or even if k is rational: 1.5 is certainly rational!

I didn't intend to be an 'if and only if' statement...

Did you mean $$k \notin \mathbb {I}$$?

Not really//
my statement there wasn't necessarily for scott_alesk's question;
it's just a property I like, that's related to complex numbers (~which this thread involves~).

What I meant to say was that any negative real raised to an irrational power cannot be a real number (the imaginary part of all solutions for $x^k$ is always nonzero if $x<0$ and $k \notin \mathbb{Q}$).

(However, is this also the case if $k\notin \mathbb{Z}$ ? But doesn't $(-1)^{1 / 3}$ have one real solution equal to $-1$ ?)

 

[0rthodontist]

Sometimes after you've changed the latex by editing and committed the changes, they don't show until you refresh the page.

 

[HallsofIvy]

My reader is so messed up it doesn't always show the changes even after "refresh"! Sometimes what I do is make the changes, copy it, then delete the post (clicking "physically remove") and paste into a new post.

 

[bomba923]

My reader is so messed up it doesn't always show the changes even after "refresh"! Sometimes what I do is make the changes, copy it, then delete the post (clicking "physically remove") and paste into a new post.

~That's exactly what I did! (..."nine times" )

 

[HallsofIvy]

Note the part about clicking "physically remove"!

 

[Hurkyl]

I often open the image itself in a new window, and reload that. It's easy to do in opera, but I don't know about the other major browsers.