Is x^fraction one-to-one or not

[intenzxboi]

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)

 

[HallsofIvy]

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, x^2/3= (x^1/3)² so if x= 8, (8)^2/3= 2²= 4 and if x= -8, (-8)^2/3= (-2)²= 4.

x^odd/odd is defined for all x and is one-to-one. x^even/odd is defined for all x and is not one-to-one. x^odd/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 x^even/even you can cancel 2s until you have one of the first three cases.

 

[intenzxboi]

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]

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, x^2/3= (x^1/3)² so if x= 8, (8)^2/3= 2²= 4 and if x= -8, (-8)^2/3= (-2)²= 4.

x^odd/odd is defined for all x and is one-to-one. x^even/odd is defined for all x and is not one-to-one. x^odd/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 x^even/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..