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We support rendering of the PDF content in our PDF viewer control including: - everything that can be rendered using Apitron Rasterizer can be viewed - various ... For A1, we need to see that C + D is a cut Obviously C + D is not empty If s is an element of C + D and t is a rational number less than s, write s = c + d, where c C and d D Then t c < s c = d D so t c D; and c C Hence t = c + (t c) C + D A similar argument shows that there is an r > s such that r C + D Finally, if x is a rational upper bound for C and y is a rational upper bound for D, then x + y is a rational upper bound for C + D We conclude that C + D is a cut Since addition of rational numbers is commutative, it follows immediately that addition of cuts is commutative Associativity follows in a similar fashion That takes care of A2 and A3 Now we show that if C is a cut then C + 0 = C For if c C and z 0 then C Also, if c C then choose a d C such c + z < c + 0 = c hence C + 0 that c < d Then c d < 0 so c d 0 And c = d + (c d ) Hence C C + 0 We conclude that C + 0 = C This is A4 Finally, for Axiom A5, we let C be a cut and set C to be equal to {d Q : d > d such that c + d < 0 for all c C} If x is a rational upper bound for C then x C so C is not empty It is also routine to check that C is a cut By its very de nition, C + ( C) 0 Further, if z 0 then there is a z 0 such that z < z Choose an element c C such that c + (z z) C (why is this possible ) Let c C be such that c < c Set c = z c Then d = z c > c We claim that c + d < 0 for all c C Suppose for the moment that this claim has been proved Then this shows that c C Then z = c + c C + ( C) so that 0 C + ( C) We then conclude and Axiom A5 is established that C + ( C) = 0, It remains to prove the claim So let d be de ned as above and select c C Then d + c = z + ( c + c) < z + (z z) = z < 0 Here we have used the choice of c This establishes the claim and completes the proof of A5 Having veri ed the axioms for addition, we turn now to multiplication De nition 59 If C and D are cuts then we de ne the product C D as follows: If C, D > 0 then C D = {q Q : q < c d for some c C, d D with c > 0, d > 0 } If C > 0, D < 0 then C D = [C ( D)] If C < 0, D > 0 then C D = [( C) D] If C, D < 0 then C D = ( C) ( D) If either C = 0 or D = 0 then C D = 0. c# wpf document viewer pdf: C# PDF Viewer opensource | The ASP . NET Forums c# open pdf adobe reader A simple PDF viewer windows form - Stack Overflow
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