[go back](https://hackmd.io/@LVcpSNoxQAim3tB5Xaj-gg/rk6e99Sdv) # 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:** $\mu=(2,4)$