Algebraic numbers

**MathoMan** · October 12th 2010, 01:29 AM

Here goes:
I have proven that $\sqrt[3]{2}+\sqrt{3}$ is algebraic number by definition (constructed a polynomial of sixth degree having that number as a root).

Could someone please give me a sketch of proof (or the proof itself) for the following claim. I'd appreciate it.

The claim is:
If $x$ is algebraic number prove that $\frac{1}{x}$ 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.

**HallsofIvy** · October 12th 2010, 04:54 AM

Originally Posted by MathoMan

Here goes:
I have proven that $\sqrt[3]{2}+\sqrt{3}$ is algebraic number by definition (constructed a polynomial of sixth degree having that number as a root).

I would start with something like $(x- \sqrt[3]{2}- \sqrt{3})(x- \sqrt[3]{2}+ \sqrt{3})= (x- \sqrt[3]{2})^2- 3= x^2- 2\sqrt[3]{2}x+ 3/4- 2= x^2- 5/4- 2\sqrt[3]{2}$ where I have used " $(a- b)(a+ b)= a^2- b^2$ ". Now use the fact that $(a- b)(a^2+ ab+ b^2)= a^3- b^3$ to get rid of that cube root.

Could someone please give me a sketch of proof (or the proof itself) for the following claim. I'd appreciate it.

The claim is:
If $x$ is algebraic number prove that $\frac{1}{x}$ 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.

You say you know that the "algebraic number form a field" and the field axioms but do you know the definition of "algebraic number". If x is an algebraic number, then there exist in integer, n, and integers, $a_n$ , $a_{n-1}$ , ..., $a_1$ and $a_0$ such that $a_nx^n+ a_{n-1}a^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$ . Okay, what do you get if you divide each term in that equation by $x^h$ ?

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".

**MathoMan** · October 13th 2010, 03:56 AM

Thanks for the post.

I am aware of all the things you've mentioned. Maybe I wasn't precise enough.

As you said:

What is true is "If "x" is a non-zero algebraic number, then 1/x is also an algebraic number".

I was looking for a way to prove that statement without using the obvious: that 1 is algebraic and that x is assumed to be algebraic and since algebraic numbers form a field hence 1/x must be algebraic number.

I hope we understand each other now. Thanks anyway.

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