# Is x^fraction one-to-one or not

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)

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HallsofIvy
Homework Helper

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.

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

arildno
Homework Helper
Gold Member
Dearly Missed

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 $$(x^{\frac{1}{6}})^{2}$$ should have the non-negatives as its maximal domain, whereas $$x^{\frac{1}{3}}$$ 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..