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Math Help - Algebraic numbers

  1. #1
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    Algebraic numbers

    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.
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  2. #2
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    Quote Originally Posted by MathoMan View Post
    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".
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  3. #3
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    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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