$Q$ value of beta plus decay (positron emission) I am unable to understand Q value for positron emission. The general reaction is as follows:
$$p \to n + e^+  + \nu$$
$$ ^A_ZX \to ^A_{Z-1}Y+ e^+ + \nu \tag{1a}$$
This reaction $(1a)$ was giving in my text. First question is that where did one electron go? We began with $Z$ electrons but on the right side it seems only $Z-1$ electrons are present. Maybe this is the cause of confusion.
So I rather took this reaction $(2a)$ for positron emission:
$$^A_ZX \to ^A_{Z-1}Y+ e^+ + e^-+ \nu \tag{2a}$$
Now writing the Q value, we find the mass defect
$$[m_n(^A_ZX)-(m_n(^A_{Z-1}Y)+m_{e^+}+m_{e^-} +m_\nu)]$$
here $m_n(^A_{Z}X)$ are mass of nuclei. Now we can rewrite in terms of atomic mass numbers $m_a(^A_{Z}X)$ as
$$\Delta m=[(m_a(^A_ZX)-Zm_e)-((m_a(^A_{Z-1}Y)-(Z-1)m_e)+m_{e^+} + m_{e^-}+m_\nu)] $$
$$\Delta m= (m_a(^A_ZX)-m_a(^A_{Z-1}Y) - 3m_e -m_\nu) \tag{2b} $$
but as you can see, if we use reaction $(1a)$ then we will get 
$$\Delta m= (m_a(^A_ZX)-m_a(^A_{Z-1}Y) - 2m_e -m_\nu) \tag{1b} $$
Am I wrong somewhere? I have highlighted key points(mistakes) in italic.
 A: 
We began with Z electrons but on the right side it seems only Z−1 electrons are present.

During beta plus decay we consider that electrons do not participate in the decay for easier calculations.(In reality they in fact must be included) Only nuclei participate. So their are no electrons on the left side and just a positron on the right which came from one of the protons.

The equation (2a) their are no electrons on the left but you added an electron on the right(assuming only nuclei participate in the decay), that cannot happen. That also violates the law of conservation of charge as a proton should convert into a neutron and positron but after introducing the electron you have decreased the overall charge by one.
A: You should write the reaction in terms of atom and ions as 
$$^{\rm A}_{\rm Z }\rm X  \Rightarrow ^{\,\,\,\,\,\rm A}_{\rm Z-1 } Y ^-+ e^+$$
noting that the daughter is a negative ion which has a net negative charge because it still has $Z$ elections orbiting the nucleus.  
Think of it as a change happening in the nucleus resulting in the emission of a positron which is nothing to do with the orbiting $Z$ electrons although of course those $Z$ electrons will have to move into new orbitals but there will still be $Z$ electrons.
The masses given in tables are the masses of the neutral atoms.  
So if the equation is to be written in terms of neutral atoms it can be written as follows
$$^{\rm A}_{\rm Z }\rm X  \Rightarrow \left (^{\,\,\,\,\,\rm A}_{\rm Z-1 } Y +e^- \right )+ e^+$$  
with both sides of the equation having a net zero charge.
In essence in order for this reaction to occur two electron masses have to be created and you have to include those two electron masses when you calculate the change in mass.
