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Math Help - Fractional indicies?

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by jebus197 View Post
    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 \sqrt[n]{x} = x^{\frac{1}{n}} .

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

    But again it is general agreement on notation.
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    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.
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    The only thing I recognised from what you wrote was the square root symbol.
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    Quote Originally Posted by jebus197 View Post
    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.
    3^4=81 thus \sqrt[4]{{81}} = 3. That is basic.

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

    \sqrt[3]{{256}} = 4\sqrt[3]{4}. BUT HOW?
    Well \sqrt[3]{{256}} = \sqrt[3]{{2^8 }} = 2^{\frac{8}<br />
{3}}  = 2^{2 + \frac{2}<br />
{3}}  = 2^2  \cdot 2^{\frac{2}<br />
{3}}  = 4\sqrt[3]{4}
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    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. \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 3^4, but I don't get what that specific part means.
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  7. #7
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    Quote Originally Posted by jebus197 View Post
    E.g. the 4 in this example. \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.
    \sqrt[4]{{81}} is the number whose fourth power of is 81.
    The fourth power of 3 is 81.

    \sqrt[3]{{64}}=? That question is asking 'what number has its third power equal to 64?'
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    Well the answer to that last question is obviously 4, since 64 is 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.
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    Quote Originally Posted by jebus197 View Post
    Well the answer to that last question is obviously 4, since 64 is 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.
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    It's not too difficult to show that, for any positive integers, m and n, (a^m)^n= a^{mn}. For example, since a^m is equal to a multiplied m times: a\cdot a\cdot a\cdot a\cdot\cdot\cdot a m times, we can think of (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 (a^{1/2})^2= a^{(1/2)2}= a. That is, we define a^{1/2} to be "the positive number whose square is a", that is, we define 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.
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  11. #11
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    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.
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  12. #12
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    Suppose you wanted to write \displaystyle \sqrt{x} as an indicial form. That would mean \displaystyle \sqrt{x} = x^n for some \displaystyle n.

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

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


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

    So \displaystyle 1 = 2n

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


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