Question

Logic: write Rational Roots Theorem in symbolical form

Answer 1

\[\Large{P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+ a_0=0\\a_i\in \mathbb{Z}(a_n,a_0\ne0) \\\exists p,q \in \mathbb{Z}:((p\perp q) \land (p|a_0) \land (q|a_n))\\P(x_0)=0\to x_0=\pm\frac{p}{q}, x_0\in \mathbb{Q}}\] this is my try, I'm sure I messed up somewhere, to avoid confusion let me be clear that perpendicular symbol here means "is coprime with" and | means "is a factor of"

Answer 2

++++++++++++++

Answer 3

the rational roots theorem says , if f ( p/q) = 0, , then p divides the last number and q divides the first number. also , make sure to factor out x first

Answer 4

so using your notation
if P( xo) = 0 and x0 = p/q, then p | a0 and q | an

Answer 5

you did it wrong , thats not the rational roots theorem

Answer 6

is that so? let me rewrite it and tell me what you think
\[\Large{ P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+ a_0
\\a_i\in \mathbb{Z}(a_n,a_0\ne0) ,\; x_i\in \mathbb{Q}
\\P(x_i)=0 \land x_i=\pm\frac{p}{q} \,\to \, (p\perp q) \land(p|a_0) \land (q|a_n) }\]

Answer 7

if P(xi)=0 and x = p/q for some p,q element of integers Z, then
p | a0 and q | an. you dont have to say x =+ - p/q. , since p and q can be negative anyway