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Is x^fraction one-to-one or not

  • Thread starter intenzxboi
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  • #1
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Can anyone explain to me how x^(fraction) is a one to one or not?

i know if x raised to a even power then its not one-to-one
and x raised to a odd power is one to one

but what if the power is 4/7(even/odd) or 9/8(odd/even)

or (even/even) or (odd/odd)
 

Answers and Replies

  • #2
HallsofIvy
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If the denominator is even, the function is not defined for negative x. If the numerator is even, then the function is not one-to-one. For example, x2/3= (x1/3)2 so if x= 8, (8)2/3= 22= 4 and if x= -8, (-8)2/3= (-2)2= 4.

xodd/odd is defined for all x and is one-to-one. xeven/odd is defined for all x and is not one-to-one. xodd/even is defined only for non-negative x and is one-to-one. Since a number is even if and only if it has a factor of 2, with xeven/even you can cancel 2s until you have one of the first three cases.
 
  • #3
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so basically all we need to do it look at the numerator.. if its even then its not one to one if its odd it is one to one
 
  • #4
arildno
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If the denominator is even, the function is not defined for negative x. If the numerator is even, then the function is not one-to-one. For example, x2/3= (x1/3)2 so if x= 8, (8)2/3= 22= 4 and if x= -8, (-8)2/3= (-2)2= 4.

xodd/odd is defined for all x and is one-to-one. xeven/odd is defined for all x and is not one-to-one. xodd/even is defined only for non-negative x and is one-to-one. Since a number is even if and only if it has a factor of 2, with xeven/even you can cancel 2s until you have one of the first three cases.

Hmm..cancelling of 2's is a very tricky business when we confine ourselves to the reals:

Wheras [tex](x^{\frac{1}{6}})^{2}[/tex] should have the non-negatives as its maximal domain, whereas [tex]x^{\frac{1}{3}}[/tex] has the real numbers as its maximal domain.
Thus, switching from the first expression to the second, by the mechanism of cancelling 2's, does not preserve logical equivalence..
 

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