Prove not

by MM » Wed Oct 08, 2008 2:17 pm

Prove that any number of the form $a\sqrt{b}+c\sqrt{d}$ where $a,b,c,d \in \mathbb{Z}$ and $b,d\ge0$ is not transcendental.

by **martin123456** » Wed Nov 25, 2009 4:09 am

MM wrote:Prove that any number of the form $a\sqrt{b}+c\sqrt{d}$ where $a,b,c,d \in \mathbb{Z}$ and $b,d\ge0$ is not transcendental.

multiplying it by $a\sqrt{b}-c\sqrt{d}$ it gives $a^2b-c^2d$ , summing it - $2a\sqrt{b}$ . So the expression is root of
$x^2-2a\sqrt{b}+a^2b-c^2d=0$ . now multiply LHS of this equation by
$x^2+2a\sqrt{b}+a^2b-c^2d=0$ , so u get that $a\sqrt{b}+c\sqrt{d}$ is a root of the $(x^2+a^2b-c^2d)^2-4a^2b$ that is with whole coefficients

Prove not

Prove not

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