When the exponent is negative

  1. hello!

    I want to know what happens, what means, what the properties are etc of a negative exponent

    I read on wikipedia "When n is a negative integer and b is non-zero, b^n is naturally defined as 1/b^−n"

    So based on the above, 3^-2 = 1/3^2

    1) Is this correct?

    2) Why does this happen? How can the number of multiplication of a base be a negative number? Also, how can it be a none natural number? Since we are talking about an amount of times of multiplication of a base, isn't by definition only 0, 1, 2, 3, etc? (not sure how this set of number is called, naturals maybe?)

    3) Also, what about x^0? Does it equal always (ie. with every real number x) with 0 ?

    4) Last, in x^n, I expect x to be any real number, is this correct? Or there are any limitations?

    Thanks
     
  2. jcsd
  3. phinds

    phinds 8,347
    Gold Member

    Yes, assuming you understand that exponentiation has precedence over division, other wise you need parentheses around the (3^2)

    It is a convention. The number of multiplications does NOT become negative. The negative sign simply mean that we are now DIVIDING by the number of multiplications

    0^n is always 1 regardless of n
    EDIT: this is corrected down below. I was asleep when I wrote it. SOUND asleep. Very deep. That's my story and I'm sticking with it. :smile:

    No, this is not correct. n can be a complex number (with an imaginary part). Such constructs are useful in many areas of math and engineering.
     
    Last edited: Sep 3, 2014
  4. Why? If I raise 0 to the square, I have 0x0=0, not 1. That's the logical result.

    I was talking about x, not n. It can be any number, right?

    As for n, I expect it to be an integer number. (now that I know about the minus convention in exponents).

    You cannot raise a number in 1.3 or in 5/7 or in π. Am I right? There is no meaning, since n is an amount of times of multiplication. So it need always to be a natural number or their negatives.

    Last, x^0, I expect it NOT to exist. Am I right?
     
  5. phinds

    phinds 8,347
    Gold Member

    OOPS! My bad. It's early and I'm not awake. I meant that n^0 is always 1. You are of course correct about 0^n always being 0.

    In the construct x^n, neither x nor n is constrained. Each can be real or imaginary or complex terms

    . No, as I just said above, n can be integer/real/imaginary/whatever

    No, you are misunderstanding. For example n^(1/2) is what we normally call "the square root of n" but x^in, where "I" is the square root of -1 is MUCH less obvious but is still a valid construct and as I said in my original answer, that kind of construct is used in engineering (electrical engineering in particular). Google "Euler's Identity".

    No, you misunderstand. x^0 is always 1 regardless of x. I KNOW it doesn't make sense but without it as the definition of x^0, lots of math doesn't work properly. Actually, it's been 50+ years since I studied the basics of this so I don't remember clearly but I THINK that x^0 being 1 isn't just a definition, it falls out of a more general method for arriving at the value of x^n.
     
  6. symbolipoint

    symbolipoint 3,077
    Homework Helper
    Gold Member

    This comes from simple arithmetic.


    Real number, a.


    a*a*a=a^3.
    (a*a*a)/a=(a/a)*a^2=a^1.
    (a*a*a)/(a^2)=((a*a)/(a*a))*a=1*a^1.


    (a*a*a)/(a*a*a)=a^3/a^3=1, and note this is the same as a^0=a^(3-3).

    (a^3)/a^4=(a*a*a)/(a*a*a*a)=((a*a*a)/(a*a*a))(1/a)=1*(1/a), and note this is a^(3-4)=a^(-1)
     
  7. Among the first properties of natural number exponents that you likely learned are ##a^m\cdot a^n=a^{m+ n}## and ##(a^m)^n=a^{m\cdot n}##. The way we've chosen to interpret exponentiation by all other real numbers is in large part related to preserving these properties.

    For every natural number ##m>1##, ##a^m\cdot a^{-1}##, should mean ##a^{m+(-1)}=a^{m-1}=\frac{a^m}{a}## if the first property above is to be preserved. This leads us to realize that ##a\cdot a^{-1}## should equal ##1##, so that ##a^{-1}## is the reciprocal ##\frac{1}{a}## of ##a##. If the second property above is also going to hold, then it must be true that ##a^{-m}=a^{m\cdot(-1)}=(a^m)^{-1}=\frac{1}{a^m}## no matter what ##m## is.

    For fractional exponents, see that ##x=x^1=x^\frac{m}{m}## which should equal both ##(x^\frac{1}{m})^m## and ##(x^m)^\frac{1}{m}## (note that, despite our best efforts, we can't actually make this work for all ##x## and all ##m##). Thus ##f(x)=x^\frac{1}{m}## should be the compositional inverse for ##g(x)=x^m##. When ##m## is a positive integer, we already have a name for that compositional inverse (on appropriate intervals when ##m## is even); ##g^{-1}(x)=\sqrt[m]{x}##. So that's why we decide to interpret ##a^\frac{1}{m}=\sqrt[m]{a}## when ##m## is a positive integer. And then we can say that ##a^\frac{p}{q}=(\sqrt[q]{a})^p=\sqrt[q]{a^p}## for rational ##\frac{p}{q}## as long as we're careful when ##q## is even.

    Exponentiation by irrational numbers is often defined in terms of some kind of limit. For instance, to say what ##a^\pi## means, we could take a sequence ##(r_i)_{i\in\mathbb{N}}## of rational numbers convergent to ##\pi## and say that ##a^\pi=\lim_{i\rightarrow\infty}a^{r_i}##. There are other ways to get the job done - for instance ##a^b=e^b\ln a## where the natural exponential and logarithmic functions are defined whatever way you prefer - but the limit version was the most satisfying version for me.
     
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