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title: Split numbers

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# Split numbers
Split numbers are very **intuitive** and it would be a good start for us to understand **new numbers**.

All real numbers can be written as a **symmetric pair**

$$
1=(1,1)\qquad \pi=(\pi,\pi), \qquad 9=(9,9)
$$

We call this form a **split number**. When we apply addition, subtraction and multiplication, we simply work with pairs
$$
(3,3)+(5,5)=(8,8)\qquad (2,2)\times(3,3)=(6,6)
$$

An **asymmetric split number** is an asymmetric pair for example
$$
(3,2), \qquad (2,3)
$$
The multiplication of asymmetric numbers is simple
$$
(1,2)\times(3,8)=(1\times 3,2\times 8)=(3,16)
$$
$$
2\times(1,2)=(2,2)\times(1,2)=(2,4)
$$
$$
(1,3)+2=(1,3)+(2,2)=(3,5)
$$

**Proposition** A split number $\mu=(a,b)$ is a new number generated by the polynomial
$$
(X-a)(X-b)
$$

Therefore, to formally solve a polynomial $(X-3)(X-5)$, we can use symmetric numbers $(3,5)$

**Proposition** The root for formally solving the polynomial $(X-a)(X-b)$ is given by
$$
\mu_1=(a,b)\qquad \mu_2=(b,a)
$$

>**Ex** Formally solve the polynomial equation $X^2-3X=0$ by split numbers

**Solution:** $X^2-3X=0\implies X(X-3)=0$ So the root are given by
$$
(0,3)\qquad (3,0)
$$

## Galois Conjugation of Split numbers

The **Galois conjugation** of a split number $\mu$ is changing its components. Denoted by $\overline\mu$

For example, if 
$$
\mu = (3,2)\qquad \implies \qquad \overline\mu=(2,3)
$$
<span style="color:red">
**Theorem** An equation of split numbers remain an equation if we apply the Galois conjugation to it

For example if we have the equation
$$
(1,3)\times(2,3)+(2,5)=(4,14)
$$we can apply Galois conjugation and got a new equation
$$
(3,1)\times(3,2)+(5,2)=(14,4)
$$
**This is a very deep philosophy.**
</span>

## Specialization of split numbers

By specialization, we mean we only look at one component. We define specialization as 
$$
p_1(a,b)=a\qquad p_2(a,b)=b
$$
<span style="color:red">
**Theorem** An equation of split numbers remain an equation if we apply the specialization to it
</span>

**Conjugation are invertible, but specialization are not.**

For example, if we have an equation 
$$
(1,3)\times(2,3)+(2,5)=(4,14)
$$we can apply first-specialization and got a new equation
$$
1\times 2+2=4
$$

**Specialization help us to understand split numbers**

## Real part and split part of split numbers
**Definition** The real part of a split number $\mu$ is the average of its components. Denoted by $\mathrm{Re}(\mu)$

For example $\mathrm{Re}(3,2)=2.5$, $\mathrm{Re}(8,4)=6$

**Definition** The split part of a split number $\mu$ is half of the difference between the first and the second component of the number
$$\mathrm{Im}(a,b)=\frac{a-b}2$$

**Corollary** For any split number $\mu$, we have
$$
\mathrm{Re}(\mu)=\frac{\mu+\overline{\mu}}2
$$

$$
\mathrm{Im}(\mu)=\frac{\mu-\overline{\mu}}{2(1,-1)}
$$
**Definition** A split number is called **pure split numbers** if its real part is 0.

**Proposition** If $\mu$ is a pure split number, then 
$$
\overline\mu=-\mu
$$

**Proposition** We have
$$
\mu=\mathrm{Re}(\mu)+\mathrm{Im}(\mu) (1,-1)
$$
$$
\overline\mu=\mathrm{Re}(\mu)-\mathrm{Im}(\mu) (1,-1)
$$
$$
\overline\mu=2\mathrm{Re}(\mu)-\mu
$$

**Pure split numbers are like** $(0,0)$, $(2,-2)$, $(9,-9)$, $(\pi,-\pi)$, $\cdots$



## Describe conjugation by symbols.

**Theorem**  Suppose $\mu$ is a split number generated by polynomial $X^2-mX+n=0$, then the real part of $\mu$ is 
$$
\mathrm{Re}(\mu)=\frac m2
$$


>**Excercise** Suppose $\mu$ is the new number generated by polynomial $X^2-6X+8=0$, find the conjugate of $\mu$.

**Solution:** $\overline\mu+\mu=6$, so $\overline \mu=6-\mu$

>**Excercise** Suppose $\mu$ is the new number generated by polynomial $X^2-6X+8=0$, express $\mu$ as a split number.
**Solution:** $\mu=(2,4)$