1. ## 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.

2. 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.

3. 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.

4. The only thing I recognised from what you wrote was the square root symbol.

5. 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 $\displaystyle \sqrt[3]{{256}} = \sqrt[3]{{2^8 }} = 2^{\frac{8} {3}} = 2^{2 + \frac{2} {3}} = 2^2 \cdot 2^{\frac{2} {3}} = 4\sqrt[3]{4}$

6. 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.

7. 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?'

8. 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.

9. 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.

10. 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.

11. 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.

12. 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 $\displaystyle \displaystyle \sqrt{x} = x^{\frac{1}{2}}$.