(a) Let p be a positive integer. Show that the set E={(a0, a1, …, ap) ∈ℕ^p+1:. the polynomial a0

(a) Let p be a positive integer. Show that the set E={(a0,...

(a) Let $p$ be a positive integer. Show that the set $$ E=\left\{\left(a_{0}, a_{1}, \ldots, a_{p}\right) \in \mathbb{N}^{p+1}:\right. $$ the polynomial $a_{0}+a_{1} X+\cdots+a_{p} X^{p}$ has a root in $\left.\mathbb{Z}\right\}$ is primitive recursive. (b) Repeat question (a) replacing $\mathbb{Z}$ by $\mathbb{Q}$. (c) Show that the set $$ \begin{aligned} F=&\left\{\Omega(\sigma): p \text { is an integer, } \sigma=\left(a_{0}, a_{1}, \ldots, a_{p}\right)\right. \text { and the polynomial }\\ &\left.a_{0}+a_{1} X+\cdots+a_{p} X^{p} \text { has a root in } \mathbb{Z}\right\} \end{aligned} $$ is primitive recursive. ( $\Omega$ is defined in the section on codings of sequences.)

(a) Let $p$ be a positive integer. Show that the set
$$
E=\left\{\left(a_{0}, a_{1}, \ldots, a_{p}\right) \in \mathbb{N}^{p+1}:\right.
$$
the polynomial $a_{0}+a_{1} X+\cdots+a_{p} X^{p}$ has a root in $\left.\mathbb{Z}\right\}$
is primitive recursive.
(b) Repeat question (a) replacing $\mathbb{Z}$ by $\mathbb{Q}$.
(c) Show that the set
$$
\begin{aligned}
F=&\left\{\Omega(\sigma): p \text { is an integer, } \sigma=\left(a_{0}, a_{1}, \ldots, a_{p}\right)\right. \text { and the polynomial }\\
&\left.a_{0}+a_{1} X+\cdots+a_{p} X^{p} \text { has a root in } \mathbb{Z}\right\}
\end{aligned}
$$
is primitive recursive. ( $\Omega$ is defined in the section on codings of sequences.)

Key Concepts

Evaluation of Polynomials

The evaluation of polynomials at specific integer or rational inputs is, in fact, a primitive recursive process. Given the coefficients of a polynomial and a candidate root, one can compute the polynomial value using iterative multiplication and addition – all of which are primitive recursive operations. This is important when formulating a primitive recursive predicate that checks if a polynomial has a zero in a given number domain.

Bounded Search and Quantification

Bounded search (or bounded quantification) refers to the process of checking a condition over a finite range of values. Since many classical results (such as bounds on possible integer or rational roots of a polynomial) provide a finite search space, one can use a bounded search to decide if there exists any number in that range satisfying the property. This is a fundamental tool in proving that certain sets are primitive recursive.

Primitive Recursive Sets and Functions

A set or function is called primitive recursive if it can be defined by a process of recursion that uses only basic functions (such as zero, successor, and projection functions), and operations like composition and bounded minimization. This concept is central in computability theory as it defines a class of functions that are computationally very simple and effectively decidable, which plays a key role when one needs to show that a particular set (for example, the set of polynomials having a root in a given domain) is decidable by a primitive recursive process.

Coding of Sequences (Omega function)

The process of coding sequences (or tuples) of natural numbers into a single natural number is a critical idea in computability theory. With a coding function like Omega, one can represent structured data (such as the list of coefficients of a polynomial) as a single natural number. This allows us to address properties about sequences using primitive recursive functions on natural numbers, which is crucial for proving the primitive recursiveness of sets related to polynomials.

Transcript

00:01 Hello, so here to prove that for any polynomial function f and any number a, there is a polynomial function g and a number b such that f of x is equal to x minus a times the function g of x plus b.

00:18 So this is going to be a direct result of the polynomial division algorithm, which says that for any given polynomial f of x and any non -zero polynomial d of x of degree, at least.

00:30 One, there's going to exist unique polynomials q of x and r of x, right? that's, we have, f of x, is going to be equal to d of x times q of x plus the remainder plus r of x, where the degree of r is less than the degree of d.

00:56 So we can let d of x here be equal to x minus a, which has degree one, so degree of r less than which means that r is a constant, say b, and therefore we have that f of x is equal to x minus a times g of x plus b.

01:16 When g of x and q of x here is a polynomial, and b equals r is going to be a constant.

01:22 For part b, the proof here we go, well, from part a, we have our function f of x is equal to x minus a times g of x plus b.

01:31 So for some polynomial g and constant b, we just evaluate both sides at x is equal to a.

01:38 So we have that f of a is equal to, while a minus a times g of a plus b, it's just zero plus b, which is equal to b.

01:48 So if f of a equals zero, then b equals zero, and we have that f of x is equal to x minus a times g of x.

01:59 And then for part c, we prove by induction on degree n.

02:06 So we have our base case would be if n is equal to zero, we have a constant polynomial.

02:12 If it's non -zero, it's going to have zero roots...

Please give Ace some feedback