Meaning of Irrational Exponents

Date: 03/17/97 at 21:11:51
From: Jonah Knobler
Subject: Meaning of Irrational Exponents

I am a student in an Algebra II class, and I'm wondering a bit about 
exponents and roots.

We know that 2^2 is the product of 2 factors of 2, and that y^x is the 
product of x factors of y.  I also know that 2^(1/2) is the product of 
1/2 a factor of 2 (i.e. it needs to be taken twice to even reach the 
base) and that y^(1/x) is the product of 1/x factors of y (i.e. after 
taking y an x number of times, we reach y^1).  That's understandable.

But where do irrational exponents fit in?  Can you raise 2 to the 
sqrt(2) power?  Is there any definition for this?  I know there must 
be, because Euler's Formula (about which I've only heard vague 
whispers as yet) says something about it. But how can you raise a 
number to an irrational power?  How can you have an irrational number 
of factors?  I can't fathom it.

Similarly, is there a sqrt(2) root of 2?  (i.e. irrational roots)?  
Can you find the pi root of seven?  How?  What is the definition?

A final corollary: what about imaginary and complex roots?  How would 
you evaluate i to the i power?  Is it real or imaginary?  How can you 
have a number multiplied by itself an IMAGINARY or worse, a COMPLEX 
number of times, when you can't even count to i or (i+1)?

- Jonah Knobler

Date: 03/18/97 at 13:45:48
From: Doctor Rob
Subject: Re: Meaning of Irrational Exponents

>Can you raise 2 to the sqrt(2) power?  ... How can you have an 
>irrational number of factors? 

Yes, you can raise 2 to the sqrt(2) power.  You do it by approximating
the irrational exponent by rational exponents that are closer and
closer to the irrational one.  In your example, you want the exponent
sqrt(2).  You know that, to one significant figure, sqrt(2) = 1, so 
the first approximation to 2^sqrt(2) is 2^1 = 2.  To two significant 
figures, sqrt(2) = 1.4 = 14/10 = 7/5, so the second approximation to 
2^sqrt(2) is 2^(7/5) = the 5th root of 2^7 or the 7th power of the 5th 
root of 2, or approximately 2.639.  To three significant figures, 
sqrt(2) = 1.41 = 141/100, so you have to take a 100th root of 2 and 
raise it to the 141st power, or else take the 100th root of 2^141, to 
get approximately 2.65737  Continuing this way, we get the sequence of 
approximations

2, 2.639, 2.6574, 2.66475, 2.665119, 2.6651375, 2.66514310, ...

These converge to a limit (this is a calculus idea) because the 
function 2^x is continuous (another calculus concept).  That limit is 
what we call the value of 2^sqrt(2) ~=~ 2.665144143.  We can't write 
down all the digits in the decimal expansion, but then neither can we 
do so for sqrt(2), so that doesn't bother us.

Presumably you have not yet learned about logarithms, because what you
are asking about is just the question of how do you compute the
antilogarithm of sqrt(2) to the base 2.  Patience, you should see this
before you finish Algebra II.

>Similarly, is there a sqrt(2) root of 2?  (i.e. irrational roots)?  
>Can you find the pi root of seven?  How?  What is the definition?

Roots are just powers with exponents which are reciprocals.  Your 
first quantity would be 

2^(1/sqrt(2)) = 2^(sqrt(2)/2) = sqrt(2^sqrt(2))

which we computed above.  Answer, approximately 1.632526919.  

Your second quantity would be 7^(1/pi), and 1/pi ~=~ 0.318309886.  
Find rational numbers which get closer and closer to this decimal, and 
raise 7 to each of them, and they will get closer and closer to some 
number which is defined to be your second quantity.  In this case, 
about 1.857817549.

Of course, I haven't told you how to take the 100th root of 2^141.
There are various numerical ways to do that.  A simple one 
conceptually is to guess and check:  Try 2.1: raise it to the 100th 
power and compare to 2^141.  I got 1.66*10^32, whereas 2^141 = 
2.78759*10^42, so our guess is too small.  Try 2.8.  2.8^100 = 
5.1976*10^44, so 2.8 is too big.  Split the difference, and try 2.45, 
getting 8.25*10^38, again too small.  Split the difference, and try 
2.625, getting 8.18*10^41, still too small.  Try 2.7, 1.37*10^43, too 
big.  Try 2.66, 3.077*10^42, a hair too big.  Try 2.65, 2.111*10^42, 
too small.  So the answer is between 2.65 and 2.66, as we found above.  
This relies on knowing that the function 2^x is an increasing function 
of x (more calculus ideas).

>A final corollary: what about imaginary and complex roots?  How would 
>you evaluate i to the i power?  Is it real or imaginary?  How can you 
>have a number multiplied by itself an IMAGINARY or worse, a COMPLEX 
>number of times, when you can't even count to i or (i+1)?

This really has to wait to calculus and the theory of infinite series,
unless you are willing to take DeMoivre's Law on faith.  It says

   e^(i*x) = cos(x) + i*sin(x)

where e is 2.718281827459..., the base of natural logarithms, and i is
the square root of -1.  Using this, it is possible to evaluate things
like r = i^i.  If we put x = pi/2 in DeMoivre's Law, we are told that
e^(i*pi/2) = i.  Now just raise both sides to the i-th power, using
the laws of exponents, and find that r = i^i = e^(i^2*pi/2) = 
e^(-pi/2), which is a real number.  Bizarre, no?  This is not the 
whole story, however!  Try using DeMoivre's Law with x = 5*pi/2, and 
do the same calculation over.  <GRIN>

Keep up the good questions!

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/