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MaPhy-&Atilde;œ4

MaPhy-&Atilde;œ4 - Aufgabe 1 Ist Aquivalenzrelation...

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Aufgabe 1 Ist ¨ Aquivalenzrelation? D.h. a b n | a - b (i) Reflexivit¨ at: a a ? a - a = 0 n | a - a ( a - a n Z ) a - a = 0 n Z a a (ii) Symmetrie: a b b a ? a - b n Z b - a = - 1 ( a - b ) | {z } n Z n ( - Z ) = n Z (iii) Transitivit¨ at: a b b c a c ? a - c = a - b | {z } n Z + b - c | {z } n Z n Z + n Z n Z Sei x, y n Z ⇒ ∃ ˜ x, ˜ y Z | x = ˜ xn, y = ˜ yn x + y = ˜ xn + ˜ yn =) ˜ x + ˜ y | {z } n Z ) n n Z {- 17 n, ... - 2 n, - n, 0 , n, 2 n, ... } “ ist ¨ Aquivalenzrelation. b) Def. ¯ x := { z Z | x z } = x + n Z Z / n Z wir zeigen: × : Z / n Z × Z / n Z Z / n Z : ¯ x + ¯ y = x + y ist wohldefiniert. Sei x x 0 y y 0 Beh.: x + y x 0 + y 0 Bew.: ( x + y ) - ( x 0 + y 0 ) = ( x - x 0 | {z } n Z ) + ( y - y 0 | {z } n Z ) n Z + n Z n Z x + y x 0 + y 0 x + y = x 0 + y 0 q.e.d. Sei x x 0 y y 0 Beh.: x · y x 0 · y Bew.: x · y - x 0 · y 0 = x |{z} Z ( y - y 0 | {z } n Z ) + ( x - x 0 | {z } n Z ) y 0 |{z} Z n ZZ + n ZZ n Z + n Z n Z x · y x 0 · y q.e.d.

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MaPhy-&Atilde;œ4 - Aufgabe 1 Ist Aquivalenzrelation...

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