I would start with something like where I have used " ". Now use the fact that to get rid of that cube root.

You say you know that the "algebraic number form a field" and the field axioms but do you know theCould someone please give me a sketch of proof (or the proof itself) for the following claim. I'd appreciate it.

The claim is:

If is algebraic number prove that also is algebraic number.

Now I know a little bit of higher math/algebra and I know that algebraic numbers form a field and thus a quotient of two algebraic numbers is algebraic number. Is there a proof that does not require the axioms of field? Can someone give me an advice (or two)?

Thx in advance.definitionof "algebraic number". If x is an algebraic number, then there exist in integer, n, and integers, , , ..., and such that . Okay, what do you get if you divide each term in that equation by ?

Or, since you know "a quotient of two algebraic numbers is algebraic number", 1/x is the quotient of 1 and x. Do you know that 1 is an algebraic number.

By the way, "a quotient of two algebraic numbers is algebraic number" is NOT strictly true. The denominator has to be non-zero. In fact, the original claim "If "x" is algebraic number then 1/x also is algebraic number" is NOT true. A counter example is x= 0. What is true is "If "x" is a non-zero algebraic number, then 1/x is also an algebraic number".