Power rules for radical roots and rational exponents.

  • Thread starter Zalajbeg
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  • #1
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Hello everyone,

I am a bit confused about definitions rules. I can have more questions but for now I want to ask only one question:

Let us say I have a number: [itex]\sqrt[6]{3x3x3x3x3x3}[/itex]

3x3x3x3x3x3 is equal to both 27^2 and (-27)^2. But If I write these two expressions separately I can get different results:

[itex]\sqrt[6]{27^2}[/itex]=27^(2/6)=27^(1/3)=3
[itex]\sqrt[6]{(-27)^2}[/itex]=(-27)^(2/6)=(-27^1/3)=-3

I know the true answer is 3 but I wonder which step is accepted wrong and why? Is there a definition avoiding me doing one of the wrong steps I did?
 

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  • #2
hilbert2
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The square root of real number x≥0 is defined to be the positive number y such that y2=x . In other words, [itex]\sqrt{x^{2}}=\left|x\right|[/itex] . This also holds for any n:th root where n is an even number.
 
  • #3
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The square root of real number x≥0 is defined to be the positive number y such that y2=x . In other words, [itex]\sqrt{x^{2}}=\left|x\right|[/itex] . This also holds for any n:th root where n is an even number.
Yes I can see that. If I was solving directly the expression [itex]\sqrt[6]{(-27)^2}[/itex] it would make sense. But I write it as (-27)^(2/6)=(-27)^(1/3)=-3.

This time [itex]\sqrt{x^2}=|x|[/itex] rule definition is not applied. What avoids me doing the operations above?
 
  • #4
hilbert2
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You are taking a square root in that calculation. You are just "hiding" it behind an unusual way to write it. Just following some calculational rules blindly without thinking what one is doing often leads to errors.
 
  • #5
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You are taking a square root in that calculation. You are just "hiding" it behind an unusual way to write it. Just following some calculational rules blindly without thinking what one is doing often leads to errors.
I agree with this. However I want to see some extended definitions for these some calculational rules. If it is a rule I can surely follow them blindly unless there are limitations. I want to see these limitations.
 
  • #6
hilbert2
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Every time you use the rule [itex](x^{a})^{b}=x^{ab}[/itex] and b is the inverse of an even number, it has to actually be [itex](x^{a})^{b}=\left|x\right|^{ab}[/itex]
 
  • #7
Stephen Tashi
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I agree with this. However I want to see some extended definitions for these some calculational rules. If it is a rule I can surely follow them blindly unless there are limitations. I want to see these limitations.
That's a good question and I've never seen an attempt to describe all the situations where
the "law" [itex] (x^a)^b = x^{(a\ b)} [/itex].is correct for real numbers. It works when [itex] a [/itex] and [itex] b [/itex] are both integers. In other situations - "you takes your chances".

Example: [itex] x = -1, a = 4/3, b = 3/4 [/itex]

[itex] {(x^a)}^b = ((-1)^{(\frac{4}{3})})^{(3/4)}= [/itex][itex]{( \sqrt[3]{-1})}^4)^{3/4} [/itex] [itex]= ((-1)^4)^{3/4} = (1)^{3/4} = 1 [/itex]

[itex] x^{(a\ b)} = (-1)^1 = -1 [/itex]

"Laws" for fractional powers in the real number system can be viewed as awkward attempts to write about the properties of the complex numbers without mentioning complex numbers. In the complex number system, a number has two square roots, three cube roots, five fifth roots, etc. This is basically because the number 1 has this property ( See http://en.wikipedia.org/wiki/Root_of_unity). Hence, if you have an expression with a fractional exponent somewhere in it, it isn't surprising that you can evaluate it in more than one way because the notation doesn't stand for a unique number.
 

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