Negative Indices/ Fractional Indices Question

[dobermann3016]

Hello, I am self teaching maths and have a quick question

Why does the denominator and numerator flip from 1/8 to 8/1 in this equation?

The example I encountered earlier with fractional indices

8 1/3 = cube root of 8 = 2

I just don't understand how 1/8 has turned into 8 can someone with greater knowledge than me explain?

 

[Deleted member 4993]

Hello, I am self teaching maths and have a quick question

Why does the denominator and numerator flip from 1/8 to 8/1 in this equation?

The example I encountered earlier with fractional indices

8 1/3 = cube root of 8 = 2

I just don't understand how 1/8 has turned into 8 can someone with greater knowledge than me explain?

Why does the denominator and numerator flip from 1/8 to 8/1 in this equation?

That comes from the Laws of exponents. One of those laws states:

\(\displaystyle a^{-n} \ = \ \frac{1}{a^{n}}\) ..........or

\(\displaystyle a^{-n} \ = \ \left (\frac{1}{a}\right )^{n}\)

 

[dobermann3016]

Why does the denominator and numerator flip from 1/8 to 8/1 in this equation?

That comes from the Laws of exponents. One of those laws states:

\(\displaystyle a^{-n} \ = \ \frac{1}{a^{n}}\) ..........or

\(\displaystyle a^{-n} \ = \ \left (\frac{1}{a}\right )^{n}\)

Thank you Subhotosh, I was overthinking something simple!

Much appreciated!

So 8/1 = 8

 

[Steven G]

At some point you need to start thinking of that negative sign in the exponent as you think of +, - or any other operation. A negative in the exponent means to take the reciprocal. Do you always have to take the reciprocal? No, just like you don't have to add 3 and 2 in 11(3+2)--you can distribute first

 

[yoscar04]

Notice that 1=a⁰=a^n-n=aⁿa^-n. That's the reason why a^-n=1/aⁿ

 

[HallsofIvy]

I've written this before but I have nothing better to do!

You probably saw "\(\displaystyle a^n\)" for n a positive integer as "a multiplied by itself n times". From that definition we can get two important properties:
\(\displaystyle (a^n)(a^m)= a^{n+ m}\) and \(\displaystyle (a^n)^m= a^{mn}\).
We can see the first by thinking of \(\displaystyle a^{n+ m}\) as the product of n+ m copies of a multiplied together. We can separate those into n copies and m copies so \(\displaystyle a^n\) and \(\displaystyle a^m\).

For \(\displaystyle (a^n)^m\) think of this as m copies of \(\displaystyle a^n\) and think of each \(\displaystyle a^n\) as n copies of a multiplied together on one row, m rows. That is a total of mn copies of a multiplied together.

Okay, what in the world could we mean by "\(\displaystyle a^0\)"? We can't multiply "0 copies" of a together! So we need to define \(\displaystyle a^0\) separately. We are free to define it any way we want but it would be nice if \(\displaystyle (a^n)(a^m)= a^{n+ m}\) were true even when n= 0. So we want \(\displaystyle (a^n)(a^0)= a^{n+ 0}\). But 0 is the "additive identity"- \(\displaystyle m+ 0= m\) so that says \(\displaystyle (a^n)(a^0)= a^n\). As long as \(\displaystyle a^n\ne 0\), which means \(\displaystyle a\ne 0\), we can divide both sides by \(\displaystyle a^n\): \(\displaystyle a^0= 1\). That is, in order to have \(\displaystyle (a^n)(a^m)= a^{n+m}\) we must define \(\displaystyle a^0= 1\). Again, that is if \(\displaystyle a\ne 0\). "\(\displaystyle 0^0\)" is NOT defined.

What about negative integers? How should we define \(\displaystyle a^{-n}\)? Again, we would like \(\displaystyle (a^n)(a^m)= a^{n+ m}\) for m= -n. In that case, we have \(\displaystyle (a^n)(a^{-n})= a^{n- n}= a^0= 1\). As long as a is not 0, we can divide by \(\displaystyle a^n\) to get \(\displaystyle a^{-n}= \frac{1}{a^n}\).

That is, for \(\displaystyle a\ne 0\), \(\displaystyle a^{-n}= \frac{1}{a^n}\). In particular, \(\displaystyle 8^{-1/2}= \frac{1}{8^{1/2}}\).

 

[HallsofIvy]

You also asked about "fractional indices". What I did before, up to negative integers, used the fact that \(\displaystyle (a^n)(a^m)= a^{m+ n}\). To deal with fractional indices we need \(\displaystyle (a^n)^m= a^{nm]\).

So look at \(\displaystyle (a^{1/n})^n= a^{n/n}= a\). So \(\displaystyle a^{1/n}\) must be the number that, when raised to the nth power, gives n, the nth root or a, \(\displaystyle \sqrt[n]{a}\).

Of course then, \(\displaystyle a^{n/m}= (a^n)^{1/m}= \sqrt[m]{a^n}\). Equivalently, \(\displaystyle a^{n/m}= (a^{1/m})^n= \left(\sqrt[m]{a}\right)^n\).

Now what about irrational numbers? Irrational numbers cannot be defined algebraically- the must be define analytically, using some kind of limit process. The simplest is this: if \(\displaystyle \gamma\) is any real number, there exist a sequence of rational numbers \(\displaystyle \{\gamma_n\}\) that converges to \(\displaystyle \gamma\): \(\displaystyle \lim_{n\to \infty} \gamma_n= \gamma\). Given any positive number, a, define \(\displaystyle a^\gamma= \lim_{n\to\infty} a^{\gamma_m}\).