How can a proton form a neutron?

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This is related to a homework problem but I want to understand it as well. How can a proton break up into a positron and neutron when a neutron clearly has a greater mass than a proton?
 
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Positron emission occurs in proton-rich nuclei, although in many, electron capture is also possible.

An isolated proton would not decay by positron emission, but an isolated neutron will spontaneously decay.
 
Consider 22Na which undergoes β+ decay to 22Ne.

22Na has 11 protons and 11 neutrons, whereas 22Ne has 10 protons and 12 neutrons. Nevertheless, the mass of a 22Na nucleus is greater than the sum of the masses of a 22Ne nucleus and an electron, because of the different binding energies of 22Na and 22Ne.
 
An isolated proton can not decay since is the lightest particle with baryon number equals to one.
A proton in the presence of other stuff like other protons or neutrons so for instance a proton that is making up the nuclei of some atom may ''borrow" energy from this environment and then it can decay (always conserving baryon number and charge and so on) so one (very) possible outcome will be positron and neutron
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
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