[go back](https://hackmd.io/@LVcpSNoxQAim3tB5Xaj-gg/rk6e99Sdv)

# Create Numbers by polynomial

To create a number like $\sqrt{-1}$, we denote it by $i$. Since $i$ is not a real number and $i^2+1=0$. It is clear the lowest-degree polynomial equation with root $i$ is
$$
X^2+1=0
$$
We can just say $i$ is the number **generated by the polynomial** $X^2+1$. Note that the polynomial must use **old numbers** as coefficients.

- $5$ is the number generated by the polynomial $X-5$ because $5$ is a root of it.
- If a number is generated by degree-1 polynomial, it is an **old number**. A new number must be generated by polynomials with degree at least 2.

## Arithmetic of new numbers
We can do addition, subtraction, multiplication and (possibly)division of new numbers. For example, $i=\sqrt{-1}$ is a new number generated by polynomial $X^2+1=0$, we can write this equation as
$$
X^2=-1.
$$
This means, in our calculation, whenever we see forms like $i^2$, we substitute it by $-1$. For example
$$
(1+i)^2=1+2i+i^2=1+2i-1=2i
$$
So in our calculation, we can write all results involving $i$ as $a+bi$ for real numbers $a$ and $b$, because, for example, if we have the form
$$
a_0+a_1i+a_2i^2+a_3i^3
$$
we can substitute $i^2$ by $-1$, and substitute $i^3=i^2\cdot i =-i$, and simplify it as
$$
(a_0-a_2)+(a_1-a_3)i
$$

## Examples of new numbers generated by polynomial of degree 2.

>**Excercise**: Let $\omega$ be the number generated by the polynomial $X^2+X+1$. Compute the following result
>$$(1+\omega)^2$$

**Solution:** $X^2+X+1=0$ means $X^2=-X-1$, so whenever we have the symbol $\omega^2$, wwe substitute it by $-\omega-1$. So
$$(1+\omega)^2=1+2\omega+\omega^2=1+2\omega+(-\omega-1)=\omega$$

>**Excercise**: Let $\mu$ be the number generated by the polynomial $X^2-1$. Compute the following result
>$$(1+\mu)^2$$

**Warning:** although $1$ and $-1$ are solutions of $X^2-1=0$, but $1$ and $-1$ are **NOT** the number generated by this polynomial. <span style='color:red'>This polynomial means we only have $X^2-1=0$ but we do not have either $X-1=0$ or $X+1=0$</span>

**Solution:** The polynomial means $\mu^2=1$ (**warning:$\mu$ is a new number! $\mu\neq \pm 1$**)
$$(1+\mu)^2=1+2\mu+\mu^2=1+2\mu+1=2+2\mu$$

>**Excercise**: Let $\epsilon$ be the number generated by the polynomial $X^2$. Compute the following result
>$$(1+\epsilon)^2$$

**Warning:** although $0$ is the solution of $X^2=0$, but $0$ is **NOT** the number generated by this polynomial. <span style='color:red'>This polynomial means we only have $X^2=0$ but we do not have either $X=0$</span>

**Solution:** The polynomial means $\epsilon^2=0$ (**warning:$\epsilon$ is a new number! $\epsilon\neq 0$**)
$$(1+\epsilon)^2=1+2\epsilon+\epsilon^2=1+2\epsilon$$

## Examples of new numbers generated by higher degree polynomials

If the polynomial has degree $n$, all arithmetic results of new numebers shuold involve terms with at most power index $n-1$. For example.

>**Ex** Let $\beta$ be the new number generated by $X^3+X+1$, compute $(1+\beta)^3$

**Solution**: This means, whenever we see $\beta^3$, we substitute it by $-\beta-1$, but we do not change $1,\beta$ or $\beta^2$. We calculate
$$
(1+\beta)^3=1+3\beta+3\beta^2+\beta^3=2\beta+3\beta^2
$$