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akademika

Member

Registered: 2006-08-09

Posts: 9

The power of (-2)^Sqrt(2)

I was wondering, is it possible to calculate the value of (-2)^Sqrt(2), its neither real nor imaginary ?!

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: The power of (-2)^Sqrt(2)

according to google:

(-2)^sqrt(2) = -0.709608865 - 2.56893919 i

its a complex number (addition of a real and imaginary number)

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MathsIsFun

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Registered: 2005-01-21

Posts: 7,711

Re: The power of (-2)^Sqrt(2)

I agree. my Complex Number Calculator gets -0.709608865301661-2.56893918954912i

(I used (-2)^(2^0.5) because my calculator does not understand functions well)

Yes, a Complex Number has a real and imaginary part. Either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers.

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: The power of (-2)^Sqrt(2)

If you think about it, what is an irrational power?

Forgetting about complex numbers for the moment, even just taking something to the power of root 2 doesn't make sense.

Integer powers make sense, because x^n just means that you multiply x by itself x times. Fine.
Rational powers are fine as well, because for x^(m/n), you just multiply x by itself m times and then take the nth root of the result.

But when the power becomes irrational, there's no way to calculate the result (or at least, there shouldn't be). And yet these numbers exist. What are they?

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: The power of (-2)^Sqrt(2)

To answer this, we must first answer what exactly a real number is.  A real number is simple a set of all rational numbers less than it.  At least that's how Dedekind defined them.  So for example, the square root of two is a set of all rational numbers less than the square root of two.  As it turns out, this is very useful in many different ways.

So what is a number to an irrational power?  It's exactly the same thing as a real number.  Let the irrational number I be the set S of all rational numbers less than it.  Then x^I is the set of all elements in x raised to all elements in I.  The greatest lower bound is what x^I equals.

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akademika

Member

Registered: 2006-08-09

Posts: 9

Re: The power of (-2)^Sqrt(2)

If you think about it, what is an irrational power?

Forgetting about complex numbers for the moment, even just taking something to the power of root 2 doesn't make sense.

Integer powers make sense, because x^n just means that you multiply x by itself x times. Fine.
Rational powers are fine as well, because for x^(m/n), you just multiply x by itself m times and then take the nth root of the result.

But when the power becomes irrational, there's no way to calculate the result (or at least, there shouldn't be). And yet these numbers exist. What are they?

That is the main reason. My calculator (mathematica 5) gives me  -0.709608865301661-2.56893918954912i. I don't agree with that, because number is only real, or imaginary. And I asked this question, because I want to understand why is imaginary, because I dont know the divisor N, so I don't know wheter is odd or even.

Last edited by akademika (2007-02-07 19:18:51)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: The power of (-2)^Sqrt(2)

because number is only real, or imaginary.

hm? why does a number need be only real or imaginary? real and imaginary numbers are subsets of the complex numbers.

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MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,711

Re: The power of (-2)^Sqrt(2)

Yes, see: Common Number Sets (particularly illustration at bottom)

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Patrick

Real Member

Registered: 2006-02-24

Posts: 1,005

Re: The power of (-2)^Sqrt(2)

I'm still not convinced ricky, maybe I just didn't understand it. Can you please try to explain it again?

edit: Okay, I think I got it now (reread it) - but how does this help us? Sure, we can approximate the answer, but it still isn't quite the actual number, is it?

Last edited by Patrick (2007-02-08 09:31:38)

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luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: The power of (-2)^Sqrt(2)

Yes, see: Common Number Sets (particularly illustration at bottom)

i think perhaps, you should change that illustration, since technically, the naturals circle should be intersecting every single circle in the hierachy. i.e.

something more like this, where the uppermost circle is the super set, of any circle that is behind it:

O is the zero set.

for example:

the rationals, are a subset of both the real and the complex.
the zero set, is a subset of all of the sets (although you might object for the naturals, depending on whether you accept 0 as a member of the naturals or not)

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MathsIsFun

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Registered: 2005-01-21

Posts: 7,711

Re: The power of (-2)^Sqrt(2)

Is the only reason for this to allow for zero (because zero is also imaginary) ... ?

And what N, Z or Q is not in R?

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: The power of (-2)^Sqrt(2)

i think perhaps, you should change that illustration, since technically, the naturals circle should be intersecting every single circle in the hierachy. i.e.

I think you are just misinterpreting the diagram on the page.

If a circle is interior to another circle, it is read as being a subset.  So for example, the natural numbers are a subset to the integers, rationals, reals, and complex.

Your diagram however, is incorrect.  It implies the existence of elements in the zero set which are not elements of the complex set.  Similarly, it implies the existence of natural numbers that are not integers.

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Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: The power of (-2)^Sqrt(2)

edit: Okay, I think I got it now (reread it) - but how does this help us? Sure, we can approximate the answer, but it still isn't quite the actual number, is it?

Short answer: decimal expansions of irrational numbers are always just approximate.

Long answer:

Sure it is.  Ok, to see this, take a look at the set:

First note that the square root of 2 is not in that set.  But, every single rational number when squared that is less than 2 is.  Now imagine we order the numbers from least to greatest.  As we start looking at the greater and greater numbers, we start noticing that we are getting closer and closer to the square root of two.  In fact, we can get arbitrarily close to it.  It is because of this, that we call S the square root of 2.

Now take two irrational numbers E and F.  They are sets of rational numbers in exactly the same way that S is.  Now we take every element of E and every element of F, and we calculate e^f, and put it in a new set:

And just like we defined the square root of two, we simply call this number E^F.  The greatest number of this set is how we think of it when doing calculations.  But how we constructed E^F is exact.

[hr]

Edit: When writing this post, I didn't realize that I was so close to explaining something that is so incredibly cool.

When you start learning about real analysis, one of the first properties they go over is called completeness.  The following works for any set, but as it turns out, any set with this property must contain the real numbers as a subset (which is in itself really cool).  So for simplicity, I will be using the real numbers.

Take any subset of the real numbers such that a least upper bound exists.  A least upper bound of a set S is a number s such that:

So for my set above:

The least upper bound is the square root of 2.  You can't find any number which is great than or equal to all the numbers in that set, but less than the square root of two.

We call a set S (the reals in this case) complete if for every subset with a least upper bound, that least upper bound is in the S.

So the first thing to show is that the rationals are not complete.  Simple.  Just come up with a subset of the rationals that has a least upper bound which is not rational:

The least upper bound in the square root of 2, which is not rational.  Now going back to my definition of the real numbers, we see that how we define real numbers is entirely because of this property known as completeness.  In fact, the real numbers are the smallest complete set.  This simply means that any complete set must contain the real numbers as a subset.

Whew!

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luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: The power of (-2)^Sqrt(2)

my diagram:

C is on top: then I and R: then Q: then Z: then N: then O:

so its saying that:

I,R,Q,Z,N,O are subsets of C
Q,Z,N,O are subsets of R
Z,N,O are subsets of Q
N,O are subsets of Z
O is a subset of N
O is a subset of I

you're meant to look at which sets are overlapping and in what order.

but yes, looking at the diagram on the page in another way, it is fine :p and mine is overly complicated

Last edited by luca-deltodesco (2007-02-08 10:51:16)

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: The power of (-2)^Sqrt(2)

but yes, looking at the diagram on the page in another way, it is fine and mine is overly complicated

Also, the way MathsIsFun did it is the standard way of drawing sets.

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luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: The power of (-2)^Sqrt(2)

ive never studied sets, so i wouldnt know

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