# Analysis leicht gemacht by Silvanus P. Thompson

By Silvanus P. Thompson

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Example text

22 [I, §2] REAL NUMBERS We deduce consequences from these axioms. Since e,p 0 and e = e 2 = (_e)2, and since either e or -e is positive, we conclude that e must be positive, that is e E P. By ORD 2 and induction, it follows that e + ... + e (the sum taken n times) is positive. An element x E R such that x ,p 0 and x 1# P is called negative. If x, yare negative, then xy is positive (because -XEP, -yEP, and hence (-x)(-y)=xyEP). If x is positive and y is negative, then xy is negative, because - y is positive, and 2 hence x( - y) = -xy is positive.

As an exercise, prove that for any elements x, Y E R we have (-x)(-y) = xy. Also prove that (-x)y = -(xy). We shall usually write x - y instead of x + (- y). From distributivity, we then see easily that (x - y)z = xz - yz. We can generalize distributivity so as to apply to several factors, by induction, namely x(Y, + ... ) = xy, + ... · As an example, we give the proof. The statement is obvious when n Assume n > 1. Then by induction, X(Y, + ... ) = = = = 1. _,) + xY. xy, + ... _, + xY•. x(Y, Similarly, if X ...

A, is a finite number of statements. each holding for all sufficiently large integers. then they are valid simultaneously for all sufficiently large integers. Indeed, if A,(n) is valid for n ~ N, •... ,A,(n) is valid for n ~ N" we let N be the maximum of N .. •N, and see that each A,(n) is valid for n ~ N. We shall use this terminology of sufficiently large only when there is no danger of ambiguity. } is increasing if x. +' for all positive integers /I. 1. ) be all illcreasillg sequellce, alld assume that it is bounded from above.