First thing you should keep in mind is that in theory of numbers we’ll mostly be talking about numbers from the set of whole or the set of real numbers. One of the most important basics in theory of number is the definition of divisibility and some basic rules.
Definition 1. If there exists such that $ b = d \cdot a$, we say that a divides b, where a and b are whole numbers and $ a \not= 0$. It is said that b is the multiple of a, and a is the divisor of b. If a divides b we denote it as $ a \mid b$ and if not $ a \phi b$.
Proposition 1. If $ a \mid b$ and $b \not= 0$ then $\mid a \mid \le \mid b \mid$.
Proof. If $ a \mid b$ and $b \not= 0 \Rightarrow$ (by the Definition 1.) $\ni d$ such that $ b = d \cdot a$, $d \not= 0$, $\mid d \mid \ge 0$.
Since we know that $ b = d \cdot a$ we can affect this equation with absolute value. We get that $\mid b \mid = \mid d \mid \cdot \mid a \mid$, and we got where we wanted to be.
We wanted to prove that $\mid a \mid \le \mid b \mid$ and we got that $\mid b \mid$ is equal to the product of some positive whole number d and $\mid a \mid$ which proves our proposition.
Proposition 2. If $ a \mid b$, then a divides every multiple of b.
Proof. Let’s say that $ b = a \cdot d_1$ and that c is the multiple of b. According to that, there exists number $ d_2$ such that $ c = b \cdot d_2$ and since $ b = a d_1$ we have that $ c = a \cdot d_1 \cdot d_2$. Since $ d_1$ and $ d_2$ are both whole numbers, so is their product and we got that the multiple of b-c, is the product of some whole number and a. This means that $ a \mid c$.
Proposition 3. If $ a \mid b$ and $ a \mid c$, then a \mid (b \pm c) and $ a \mid (b \cdot c)$.
Proof. If $ a \mid b$ and $ a \mid c$ then exist $ d_1$ and $ d_2$ such that $ b = a \cdot d_1$ and $ c = a \cdot d_2$. Since we want to prove that a divides $ b \pm c$ we’ll simply include b and c into this expression. This means that we’ll have $ b \pm c = a (d_1 \pm d_2)$. Since $d_1$ and $d_2$ are whole number, so are their sum and difference and we got that $ b \pm c$ is equal to the product of number with some whole number. This means that $ a \mid (b \pm c)$.
The same goes for the product of b and c:
$ b \cdot c = a^2 (d_1 \cdot d_2) = a (a \cdot d_1 \cdot d_2)$
and in the same way we got that $ b \cdot c$ is equal to the product of number a and some whole number which means that $ a \mid (b \cdot c)$.
This relation – “to be divisible” is relation of partial order. This means that are fulfilled following three conditions:
1. Reflexive property – This means that a certain number is always divisible by itself or: $ a \mid a$
2. Antisymmetric property – This means that if one number is divisible by another number, and in the same time that other number is divisible by the first number, those two numbers must be equal. $ a \mid b$ and $ b \mid a \Rightarrow a = b$
3. Transitive property – it tells us that if one number divides second number which divides third number, then the first number will also divide third number. $ a \mid b$ and $ b \mid c \Rightarrow a \mid c$
The remainder theorem
For real number a and whole number b exist unique whole numbers q and r such that
$ b = a \cdot q + r$, where $ 0 \le r \le a$.
Proof. We’ll divide this proof into two parts. First part is in which we will prove existence and second in which we will prove uniqueness.
1. Existence: let’s observe rational number $\frac{b}{a}$ and define number q such that $ q \le \frac{b}{a} \le q + 1$
If we subtract q from these inequalities we’ll get: $ 0 \le \frac{b}{a} – q < 1$
Now we can define number r: $ r = a (\frac{b}{a} – q) = b – aq$
This means that $ b = r + aq$
If we multiply the inequality $ 0 \le \frac{b}{a} – q < 1$ with a we get: $ 0 \le b – aq < a$
Since we defined r as $ r = b – aq$ we now have $ 0 \le r < a$ which proves existence.
2. Uniqueness: here we’ll assume there exist whole numbers $ d_1$ and $d_2$ such that $ aq_1 + r_1, 0 \le r < a$
This would mean that $aq + r = aq_1 + r_1$.
$ a(q – q_1) = r_1 – r$. If we assume that $ r_1 \ not= r$, then so is $ q \not= q_1$.
If $ r \not= r_1 \Rightarrow 0 < \mid r_1 – r \mid < a$.
From $ a(q – q_1) = r_1 – r$ (by applying absolute value) we get that
$ \mid r_1 – r \mid = \mid a \mid \cdot \mid q – q_1 \mid \ge \mid a \mid = a$ which is in contradiction with the statement that $\mid r_1 – r \mid < a$.
This means that $ r_1 = r$ if and only if $ q = q_1$. We say that r is the remainder in division of number b with number a, and q is the quotient of integer $ division – q= \frac{b}{a}$.
Polynomials
The Remainder Theorem for polynomials an application of Euclidean division on polynomials. Let’s explain it closer…
When we divide any polynomial f(x) by a linear polynomial $ (x – c)$ we get another polynomial:
$ f (x) = (x – c) q(x) + r$
The point of this theorem is realizing what this remainder r really is. Let’s try to replace x with c and see what happens.
$ f(c) = (c – x) q(c) + r$
$ f(c) = 0 \cdot q(c) + r$
$ f(c) = r$
When we divide a polynomial $ f(x)$ by a linear polynomial $ (x – c)$, where $ c \in \mathbb{R}$, the remainder r is equal to f(c).
This theorem is very useful, especially if we want to know whether one polynomial is divisible by another, linear polynomial. If the remainder is equal to zero, then he is divisible, if it is any other number, he is not.
Example 5. Is a polynomial divisible by $ f(x) = 5x^4 + 3x – 6x + 3$ divisible by $ (x + 2)$.
We have a linear polynomial $ (x + 2)$ which means that our c is equal to – 2. Now we calculate $ f(-2) = 0$.
$ f(-2) = 5 \cdot (- 2) + 3 \cdot (- 2) – 6 \cdot (- 2) + 3 = – 10 – 6 + 12 + 3 = – 1$
$ – 1 \not= 0$ which means that polynomial $f(x)$ is not divisible by $(x + 2)$.
