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Rh attention; it is of a particular nature which distinguishes it even at this stage from the purely logical definition; the equality (1), in fact, contains an infinite number of distinct definitions, each having only one meaning when we know the meaning of its predecessor.

Associative.—I say that a + (b + c) = (a + b) + c; in fact, the theorem is true for c = 1. It may then be written a + (b + 1) = (a + b) + 1; which, remembering the difference of notation, is nothing but the equality (1) by which I have just defined addition. Assume the theorem true for c = &gamma;, I say that it will be true for c = &gamma; + 1. Let (a + b) + &gamma; = a + (b + &gamma;), it follows that [(a + b) + &gamma;] + 1 = [a + (b + &gamma;)] + 1; or by def. (1)—(a + b) + (&gamma; + 1) = a + (b + &gamma; + 1) = a + [b + (&gamma; + 1)], which shows by a series of purely analytical deductions that the theorem is true for &gamma; + 1. Being true for c = 1, we see that it is successively true for c = 2, c = 3, etc.

Commutative. (1) I say that a + 1 = 1 + a. The theorem is evidently true for a = 1; we can verify by purely analytical reasoning that if it is true for a = &gamma; it will be true for a = &gamma; + 1. Now, it is true for a = 1, and therefore is true for a = 2, a = 3, and so on. This is what is meant by saying that the proof is demonstrated "by recurrence."

(2) I say that a + b = b + a. The theorem has just