Understanding Beta Decay
For all radioactive decay (or other nuclear reaction) of a nuclide into other nuclides, the atomic number $Z$ and mass number $A$ need to be conserved. 

$$\ce{_{Z_1}^{A_1}X -&gt; _{Z_2}^{A_2}Q + _{Z_3}^{A_3}R }$$
$$Z_1 = Z_2 + Z_3$$
$$A_1 = A_2 + A_3$$

Additionally, the charges must be conserved. If $\ce{_{Z_3}^{A_3}R }$ is an alpha particle $\ce{_2^2\alpha^{2+}}$ or $\ce{_2^2He^{2+}}$, then:

$$\ce{_{Z}^{A}X -&gt; _{Z-2}^{A-2}Q^{2-} +_2^2He^{2+}}$$

Neutrons decay into protons and electrons (beta particles). Both particles are needed for the math to work. And in order for the math to work, you need to know that the atomic number $Z$ is $0$ for a neutron (no protons), and $1$ for a proton (1 proton, or the $\ce{H+}$ ion). The electron does not have an atomic number. In fact, it is not that $Z$ and charges must be conserved separately. The sum of atomic number and charge is conserved.

$$\ce{_0^1n -&gt; _1^1p+ + _{0}^0e-}$$

For the decay of $\ce{^{14}_6C}$ into $\ce{_7^{14}N}$, we first have a neutron decay into a proton and an electron, which is captured by the newly formed $\ce{_7^{14}N^+}$. The nitrogen nuclide is initially a cation because carbon only had six electrons, and nitrogen needs seven.

$$\ce{_6^{14}C -&gt; _7^{14}N+ +e- -&gt; _7^{14}N}$$