Let the arithmetic name the daughter

Label the emitted particle with its own $A$ and $Z$ and the bookkeeping does itself.

1. Alpha: emits a helium-4 nucleus, $A = 4$, $Z = 2$. So the daughter has $A - 4$ and $Z - 2$.
2. Beta-minus: emits an electron, $A = 0$, $Z = -1$. A neutron becomes a proton, so $A$ is unchanged and $Z$ rises by $1$.
3. Beta-plus: emits a positron, $A = 0$, $Z = +1$. A proton becomes a neutron, so $A$ is unchanged and $Z$ falls by $1$.

Worked: carbon-14 has $A = 14$, $Z = 6$. Beta-minus emits an electron with $A = 0$ and $Z = -1$, so the daughter must carry $A = 14$ and $Z = 7$, which is nitrogen-14. Not nitrogen-13: beta decay does not move the mass number, because the emitted electron has essentially no nucleons.

Balancing the top row and leaving the bottom one inconsistent is the other half of this error, and writing both rows out catches it immediately.

A gamma photon carries no charge and no nucleons: $A = 0$, $Z = 0$. Balance the rows and the daughter is the same nuclide, excited before and settled after.

So a gamma emission does not move the nucleus to a different element or a different isotope. What changed is its energy: the nucleus dropped to a lower nuclear energy level, which is the same idea as an atom dropping between electron levels but at MeV rather than eV.

That is also why radioactive decay is defined as transformation into a different nucleus or to a lower energy level of the same nucleus. Arguing that nothing decayed because the element never changed uses only half of that definition.

In practice gamma emission normally follows an alpha or beta decay that left the daughter in an excited state, which is why gamma rays are so often observed alongside the other two.

Lepton number is conserved alongside nucleon number and charge, so a complete beta equation has three products, not two:

1. Beta-minus: a neutron becomes a proton, emitting an electron and an antineutrino.
2. Beta-plus: a proton becomes a neutron, emitting a positron and a neutrino.

Both partners carry zero charge and negligible mass, so leaving them out does not disturb the $A$ and $Z$ balance, which is exactly why they get dropped. Note the pairing too: the electron goes with the antineutrino and the positron with the neutrino, and swapping them is its own error.

The observational reason they have to be there: emitted beta electrons come out with a spread of energies, not one fixed value. A two-body decay would give a single sharp energy, since the shares are fixed by conservation. A third particle taking a variable share is what makes the beta spectrum continuous, and that spectrum is how the neutrino was inferred decades before it was detected.

The method in four steps, and it never requires recall.

1. Write the parent with its $A$ on top and $Z$ below.
2. Write the emitted particle with its $A$ and $Z$: $^4_2$He, $^{\ \ 0}_{-1}$e, $^0_{+1}$e, or $^0_0\gamma$.
3. Sum each row on the product side and require it to match the parent.
4. Look up the element with the resulting $Z$ and name the daughter.

Two checks worth running. The daughter's $Z$ names the element, so if it did not change, no alpha or beta decay happened. And the daughter's $A$ changes only in alpha decay, so a mass number that moved by $1$ or by $3$ means an arithmetic slip rather than a new kind of decay.