Raising a Negative to an Irrational Power? It DependsDate: 03/16/2011 at 19:10:28 From: Nick Subject: Negative Numbers To Irrational Powers Dear Dr. Math, Hi, my name is Nick. I am a 10th grader attending Lake Central High School. I do things most other normal kids don't. I seem to rather enjoy math a little too much. I can understand many restrictions and rules in functions and stuff, but I came across a strange question that none of my math teachers could answer -- and it still gets me. The question is as follows: "When taking a function to an irrational power, the negative set is excluded even though the possibility of finding a positive number to an irrational power IS possible. Why is finding a negative to an irrational power not?" I have asked this to all of my teachers and my father (they are the best for my math answers) and even other relatives (though they're not really any help) -- but no one has had a clear answer. So, Dr. Math, why can't you find a value for a negative number raised to an irrational exponent? Or maybe you can; does it deal with logarithmic functions? Please try it. For example, type this on your calculator and see what you get: -3^(sqrt3) I seem to continue to get "Error"; and when writing it out, I am stuck. This is not only for the square root of three as an exponent, but other irrationals, such -3^e, and imaginaries, such as -3^i. I know i is the square root of (-1), but these all seem to not have any real value, so I am just truly baffled. Please, don't discard this request. -Nick Date: 03/19/2011 at 01:08:29 From: Doctor Vogler Subject: Re: Negative Numbers To Irrational Powers Hi Nick, Thanks for writing to Dr. Math. I enjoy math more than other adults do, too, and I am proud of that fact. There are a number of different ways to approach your question. There is some discussion of this issue at Base of an Exponential Function http://mathforum.org/library/drmath/view/55604.html Graph of y = (-n)^x http://mathforum.org/library/drmath/view/66708.html Why Is (-n)^fractional Invalid? http://mathforum.org/library/drmath/view/62979.html y to the x Power http://mathforum.org/library/drmath/view/63367.html Basically, it all comes down to: What do you mean when you write an expression like x^y? The first definition of exponents you have likely seen was the algebraic definition, where you define x^n for positive integers n as x times itself n times, then x^(1/n) as a number y such that y^n = x, and x^(p/q) as (x^p)^(1/q). You might have even seen x^y for irrational numbers y defined as a limit (if you've encountered limits; they are usually introduced in calculus or pre-calculus). In this setting, negative numbers to irrational exponents are not defined because the limit I spoke of generally does not exist; when it does, it depends on the choice of sequence approaching y. A different way to define exponents uses calculus and first defines the natural logarithm ln(x) as an integral, or exp(x) (also written e^x) with a differential equation, and the other as the inverse function. In this context, x^y = exp(y*ln(x)) But then you have the problem that the integral for ln(x) is only defined when x is positive, so this definition only works when the base x is positive. A third way uses complex analysis and imaginary numbers, where the function exp(x) can be defined for complex numbers. But then exp(x) has many inverses, so the natural logarithm ln(x) is a multi-valued function. Lucky for us, exp(y*ln(x)) is always the same, no matter which value you pick for ln(x), as long as y is an integer. But if y is rational, then different choices of ln(x) will give different values for x^y, and the number of different values you can get is equal to the denominator of y. For example, the four values for 1^(1/4) are 1, -1, i, and -i. When y is irrational, then there are infinitely many different values for x^y. If x is a positive real number, and y is a real irrational number, then exactly one of those infinitely many different values for x^y is a real number. But if x is a negative real number, and y is a real irrational number, then none of those infinitely many different values for x^y is a real number; all of them are complex numbers. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/