Fractional indicies?

• Apr 5th 2011, 12:00 PM
jebus197
Fractional indicies?
OK I'm a bit stumped. Can anyone explain in as clear and plane terms as possible why 9 to the power of 1/2 = 3? (or more exactly why 9 to the power of 1/2 is equivalent to the square root of 9, which of course again is 3). I don't get it.

PS, sorry, I wish I knew how to use your formula editing script to write my formulas... It would make everything much easier to read.
• Apr 5th 2011, 12:10 PM
Plato
Quote:

Originally Posted by jebus197
OK I'm a bit stumped. Can anyone explain in as clear and plane terms as possible why 9 to the power of 1/2 = 3? (or more exactly why 9 to the power of 1/2 is equivalent to the square root of 9, which of course again is 3). I don't get it.

Because this is posted in pre-algebra/algebra we will keep it simple.
At this level it is just a matter of notation.
If n is a positive integer, we agree that $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$.

As you move on this allows for exponents to retain many of the usual rules.
$\displaystyle \sqrt[n]{{x^m }} = x^{\frac{m}{n}} = \left( {\sqrt[n]{x}} \right)^m$.

But again it is general agreement on notation.
• Apr 5th 2011, 12:18 PM
jebus197
Lol, this is pre pre algebra. I'd appreciate it if you would step back a little further again and use some real numbers and/or worked examples. Running with algebraic lettering at the moment isn't all that helpful. I'm still confused.
• Apr 5th 2011, 12:27 PM
jebus197
The only thing I recognised from what you wrote was the square root symbol.
• Apr 5th 2011, 12:30 PM
Plato
Quote:

Originally Posted by jebus197
Lol, this is pre pre algebra. I'd appreciate it if you would step back a little further again and use some real numbers and/or worked examples. Running with algebraic lettering at the moment isn't all that helpful. I'm still confused.

$\displaystyle 3^4=81$ thus $\displaystyle \sqrt[4]{{81}} = 3$. That is basic.

So how does that work? $\displaystyle \sqrt[4]{{81}} = \left( {81} \right)^{\frac{1}{4}} = \left( {3^4 } \right)^{\frac{1}{4}} = 3$.

$\displaystyle \sqrt[3]{{256}} = 4\sqrt[3]{4}$. BUT HOW?
Well $\displaystyle \sqrt[3]{{256}} = \sqrt[3]{{2^8 }} = 2^{\frac{8} {3}} = 2^{2 + \frac{2} {3}} = 2^2 \cdot 2^{\frac{2} {3}} = 4\sqrt[3]{4}$
• Apr 5th 2011, 01:00 PM
jebus197
Mmm... it makes a vague kind of sense to me. I'll have to revisit the topic when I can get someone to show me and explain it to me on paper I think. The bit I'm getting lost on first is what the number in front of the square root symbol means? I have never seen or been taught this before. E.g. the 4 in this example. $\displaystyle \sqrt[4]{{81}} = 3$ Maybe if you explained this part in words it might help? If it's of any interest I'm a mature student and am trying to learn maths for pretty much the first time. I get that it seems to be the reverse of $\displaystyle 3^4$, but I don't get what that specific part means.
• Apr 5th 2011, 01:07 PM
Plato
Quote:

Originally Posted by jebus197
E.g. the 4 in this example. $\displaystyle \sqrt[4]{{81}} = 3$ Maybe if you explained this part in words it might help? If it's of any interest I'm a mature student and am trying to learn maths for pretty much the first time.

$\displaystyle \sqrt[4]{{81}}$ is the number whose fourth power of is 81.
The fourth power of 3 is 81.

$\displaystyle \sqrt[3]{{64}}=?$ That question is asking 'what number has its third power equal to 64?'
• Apr 5th 2011, 01:22 PM
jebus197
Well the answer to that last question is obviously 4, since 64 is $\displaystyle 4^3$. But I'm not sure that it is always that obvious (to me). I guess I'll have to try to take a little time to consider your answer.
• Apr 5th 2011, 01:31 PM
Plato
Quote:

Originally Posted by jebus197
Well the answer to that last question is obviously 4, since 64 is $\displaystyle 4^3$. But I'm not sure that it is always that obvious (to me). I guess I'll have to try to take a little time to consider your answer.

Well of course it is not obvious to anyone.
That is what calculators are for.
• Apr 5th 2011, 02:16 PM
HallsofIvy
It's not too difficult to show that, for any positive integers, m and n, $\displaystyle (a^m)^n= a^{mn}$. For example, since $\displaystyle a^m$ is equal to a multiplied m times: $\displaystyle a\cdot a\cdot a\cdot a\cdot\cdot\cdot a$ m times, we can think of $\displaystyle (a^m)^n$ as that repeated n times- perhaps n rows, each of m "a"s. That is (think "width times length") a total of mn "a"s.

Because that is such a nice property, we would like it to be true for all powers. In particular, we want $\displaystyle (a^{1/2})^2= a^{(1/2)2}= a$. That is, we define $\displaystyle a^{1/2}$ to be "the positive number whose square is a", that is, we define $\displaystyle a^{1/2}= \sqrt{a}$. We define it to be the positive root in order to be precise- some values of a have two numbers whose square is a.
• Apr 5th 2011, 02:22 PM
jebus197
Whoosh... lol. Over my head mate... lol. I still think it might have worked better with numbers. On the course I'm doing we haven't got around to doing much algebra yet.
• Apr 5th 2011, 07:32 PM
Prove It
Suppose you wanted to write $\displaystyle \displaystyle \sqrt{x}$ as an indicial form. That would mean $\displaystyle \displaystyle \sqrt{x} = x^n$ for some $\displaystyle \displaystyle n$.

So $\displaystyle \displaystyle (\sqrt{x})^2 = \left(x^n\right)^2$

$\displaystyle \displaystyle x = x^{2n}$, since squaring undoes a square root, and taking a power to a power is the same as multiplying the powers.

$\displaystyle \displaystyle x^1 = x^{2n}$ (it should be obvious that $\displaystyle \displaystyle x^1 = x$).

Looking at this equation, since the bases are equal, so are the powers.

So $\displaystyle \displaystyle 1 = 2n$

$\displaystyle \displaystyle n = \frac{1}{2}$.

So that means $\displaystyle \displaystyle \sqrt{x} = x^{\frac{1}{2}}$.