Being true for c = 1, we thus see successively that so it is for c = 2, for c = 3, etc.
Commutativity.—1º I say that
a + 1 = 1 + a.
The theorem is evidently true for a = 1; we can verify by purely analytic reasoning that if it is true for a = γ it will be true for a = γ + 1; for then
(γ + 1) + 1 = (1 + γ) + 1 = 1 + (γ + 1);
now it is true for a = 1, therefore it will be true for a = 2, for a = 3, etc., which is expressed by saying that the enunciated proposition is demonstrated by recurrence.
2º I say that
a + b = b + a.
The theorem has just been demonstrated for b = 1; it can be verified analytically that if it is true for b = β, it will be true for b = β + 1.
The proposition is therefore established by recurrence.