What is the atomic number of an electron, and why is its symbol $\sideset{_{-1}^{\phantom{-}0}}{}{\text{e}}$ in $\beta^{-}$ decay equations? I'd assume that the atomic number, i.e. the proton number, of an electron is $0$ given that it doesn't have any protons.
However, to balance beta minus decay equations, this source seems to claim that the atomic number an electron is $-1 :$

Beta Particle (electron)

Symbols for a beta negative particle. Note that the first is preferable when balancing equations as it has an atomic number of -1 shown.

I understand that charge is conserved in these equations as well as baryon/lepton numbers, but I don't understand how/why "atomic number" would be conserved.
Questions:


*

*What is the atomic number of an electron?

*Why does this source claim that the atomic number of an electron is $-1 ?$
 A: The electron doesn't have an atomic number.  If it did, would you say that a neutral atom, with equal numbers of protons and electrons, has atomic number zero?  That's just not what people say.
Physicists (well, nuclear physicists)  refer to the "proton number" of a nucleus, rather than to the atomic number, even though the two numbers are identical for all nuclei. The electric charge quantum number for a nucleus with $Z$ protons is also always the same as the proton number. 
For a nucleus, it makes sense to talk about the atomic number, because the nucleus will eventually find enough electrons to become neutral and behave chemically in a certain way.
For the electron, it makes sense to say that its electric charge quantum number is $-1$, and that its proton number is zero.    But a free electron can make its way into any kind of atom, so it doesn't make sense to give it a label like "atomic number" that suggests particular chemical properties.
A: The equation for a beta decay might be written in terms of atoms (and ions) as follows
$$\rm ^{14}_{\;6}C \to {}^{14}_{\;7}N^+ + {}^{\;\;0}_{-1}\beta^- + {}^{0}_{0}\bar \nu$$ 
A reason for doing this is that when evaluating the Q-value of such a decay the masses of atoms (and electrons) rather than nuclei are used.
That being the case it is important to keep tracks of both the constituents of the nuclei and the numbers of orbiting electrons.
$\rm N^+$ indicates that it is a single positively charged ion of nitrogen and together with the the negatively charged $\beta^-$ the right had side of the equation has a net zero charge commensurate with the neutral carbon atom on the right hand side.
That labeling also helps keep track of the numbers of electrons on the right hand side of the equation so that in terms of mass, the nitrogen ion and the beta can be assumed to have the mass of a nitrogen atom (assuming that the binding energy of the seventh electron in the nitrogen atom is very small compared with the nuclear binding energies).  
The $-1$ label for the beta can be thought of as being useful in evaluating the  number of protons in the resulting nitrogen nucleus.
This is done by saying that a beta minus emission results in an increase of one proton in the nucleus which in terms of an algebraic calculation can be written as $6=7+(-1)+0$.
That $-1$ label can be thought of as an aid to help "balance" a beta minus decay equation. 

Think of the $-1$ label as a useful aid which by extension can be used to predict what happens when the label is $+1$ as in a ${}^{0}_{1}\beta^+$ decay?
A: The atomic number is -1 because as you have said; it is more convenient if we can balance the equation. Since in a Beta decay a neutron decays to a proton and releases an electron and its anti neutrino, total number of charge and lepton number is conserved. But then to balance the numbers of the mass and atomic number, the atomic number of electron is -1 for convenience.
