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March 3rd, 2009, 12:55 PM   #1
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(x^2 = x * x) (x^0.5 = ??)

Help me!

I know that the x^0.5 is the same as sqrt(x).. the square root of x...
But my question is...

if "x^2" is "x * x".. and "x^5" is "x * x * x * x * x".. What is "x^0.5" ??
How can one do "x * x" half a time??

I know I can solve the square root with the Babylonian method.. And I know stuff like that..
The only thing I am asking is. How can I do "x*x" half a time!?.. I mean "x^0.5"!?!?

So... what is the proper definition of "x^0.5" ?

This thing is driving me crazy! please help me!!


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March 3rd, 2009, 02:35 PM   #2
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Re: (x^2 = x * x) (x^0.5 = ??)

It's defined that way so that the rules of arithmetic are consistent.

For whole numbers, it is clear that

If you assume this rule holds for fractions as well, you find that .

It turns out that this generalisation of the idea leads to consistent results, and so we can use it. You can even generalise further to irrational exponents by finding the limit of a sequence of approximating fractions.

One way of thinking about it is to say that if I start with the number 10 and "half-add" 4, I get 12. If I "half-add" 4 again, I get 14, which is the same as one whole addition of 4. If I start with the number 10 and "half-multiply" 4, I get 20. If I "half-multiply" 4 again, I get 40, which is the same as one whole multiplication by 4. The starting-place for addition is 0, and the starting-place for multiplication is 1.
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March 3rd, 2009, 03:08 PM   #3
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Re: (x^2 = x * x) (x^0.5 = ??)

I know 3 * 4 = 4 + 4 + 4, but what is 0.5 * 4? I know it's supposed to be equal to two by the rational multiplication formula, but how do I fill in the blank in "0.5 * 4 = 4 + ??"?
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March 3rd, 2009, 03:28 PM   #4
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Re: (x^2 = x * x) (x^0.5 = ??)

I like this kind of questions because in mathematics, the most important thing, Iīts how you SEE things.

x*x=x^2... but what happens if we dismantle this identity, and we ask for which number: y*y gives x... x^0.5*x^0.5 = x (you need to add the exponents)

I think these things implies to see things from the image set instead of the origin set: thatīs inverse functions.

I apologize if this is considered an stupidity.
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