# When the exponent is negative

1. ### mather

146
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. ### phinds

8,768
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.

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
3. ### mather

146
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?

4. ### phinds

8,768
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.

5. ### symbolipoint

3,175
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)

6. ### gopher_p

575
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.