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Question on odd space

conscipost

Member
Jan 26, 2012
39
Someone I talked to this week wanted to define a function from reals to reals that captured the sense that each negative number has an "nth root" if n is odd. We talked about how the standard definition only applies to positive reals, but considered this case if we defined, for instance, f(x)=(-2)^x=-2^x when x is the sum of rationals with odd denominators. If we just deleted/ignored the remaining real x's, and considered the resulting space would f(x) be continuous?

Any feedback appreciated,
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Someone I talked to this week wanted to define a function from reals to reals that captured the sense that each negative number has an "nth root" if n is odd. We talked about how the standard definition only applies to positive reals, but considered this case if we defined, for instance, f(x)=(-2)^x=-2^x when x is the sum of rationals with odd denominators. If we just deleted/ignored the remaining real x's, and considered the resulting space would f(x) be continuous?

Any feedback appreciated,
I'm afraid not, since $(-2)^x=-(2^x)$ does not hold for $x=2/3$.
That is because $(-2)^{2/3} = 2^{2/3} \ne -(2^{2/3})$.
In other words, the sign flips up and down discontinuously.

If you further restrict the domain to only fractions with an odd numerator and an odd denominator, the function becomes continuous, since it is effectively indeed $-(2^x)$.
In any interval there are still infinitely many elements, so the limit-definition of continuous is satisfied everywhere.
However, you're left with almost no algebraic structure.
This set is not a field anymore (which $\mathbb Q$ is), it's not a ring (which the set with odd denominators is), and it's not a group.