By Alfred North Whitehead

Whitehead explains in extensive phrases what arithmetic is set, what it does, and the way mathematicians do it.Generations of readers who've stayed with the thinker from the start to the top have chanced on themselves amply rewarded for taking this trip. As

*The long island Times*saw many years in the past, "Whitehead does not popularize or make palatable; he's easily lucid and cogent ... A finely balanced mix of wisdom and urbanity .... may still pride you."

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Deﬁnition 189 If f : U → Rm and g : V → Rm are two functions deﬁned in open neighborhoods U , V of 0 ∈ Rn , then they are called equivalent if there is a neighborhood N ⊂ U ∩ V of 0 such that f |N = g|N . This relation is an equivalence relation, and the equivalence class [f ] of a function f is called the germ of f (at 0). The set F0 of germs of functions f : U → Rm with f (0) = 0 is a real vector space as follows: (1) The sum of germs is [f ] + [g] = [f |W + g|W ], W = U ∩ V being the intersection of the domains U and V of f and g representing the germs [f ] and [g], and (2) the scalar multiplication is λ[f ] = [λf ].

An , bn is continuously diﬀerentiable such that there is a number L with |Dj fi (x)| ≤ L for all x ∈ K o . Then for all x, y ∈ K, we have f (x) − f (y) ≤ n2 L x − y . Proof For x = (x1 , . . xn ), y = (y1 , . . yn ) ∈ K, and for any index i, we have fi (y) − fi (x) = (fi (y1 , . . yj , xj+1 , . . xn ) − fi (y1 , . . yj−1 , xj , . . n and then, by the mean value theorem 267 on each coordinate, fi (y1 , . . yj , xj+1 , . . xn ) − fi (y1 , . . yj−1 , xj , . . xn ) = yj − xj · Dj fi (wij ) for some vector wij .

All this together proves that i · 2π Z is the kernel of exp. Let us ﬁnally concentrate on the real arguments in exp. Since e = exp(1) > 1, there are arbitrary large real numbers exp(n) = e n for real arguments, and by 1 also arbitrary small real values for real arguments. By propoexp(−n) = exp(n) sition 245, every positive real value is taken by exp(x) for x ∈ R. Now, every complex number z ≠ 0 can be written as z = z u, u ∈ U. Therefore there are x, θ ∈ R, such that exp(x) = z and exp(i · θ) = u.