prove its true: |A union B| = |A| + |B| – |A intersection B|

OSMAN

New member
Joined
May 21, 2019
Messages
1
Prove AB=A+BAB\displaystyle |A\cup B|=|A|+|B| - |A\cap B|
 
Last edited by a moderator:
Divide A and B into three sets: all x that are in A but not B, all y that are in B but not A, all z that are in AB\displaystyle A\cap B. Show that that every member of A and B is in one and only one of those.
 
Prove AB=A+BAB\displaystyle |A\cup B|=|A|+|B|-|A\cap B|
If X, Y, & Z\displaystyle X,~Y,~\&~Z are three pairwise disjoint sets then XYZ=X+Y+Z\displaystyle |X\cup Y\cup Z|=|X|+|Y|+|Z|
(AB)=AAB\displaystyle |(A\setminus B)|=|A|-|A\cap B|

If you understand that the show that (AB)=(AB)(BA)(AB)\displaystyle (A\cup B)=(A\setminus B)\cup (B\setminus A)\cup (A\cap B)
 
If X, Y, & Z\displaystyle X,~Y,~\&~Z are three pairwise disjoint sets then XYZ=X+Y+Z\displaystyle |X\cup Y\cup Z|=|X|+|Y|+|Z|
(AB)=AAB\displaystyle |(A\setminus B)|=|A|-|A\cap B|

If you understand that the show that (AB)=(AB)(BA)(AB)\displaystyle (A\cup B)=(A\setminus B)\cup (B\setminus A)\cup (A\cap B)
Full ans ?
 
Full [answer] ?

 
Prove AB=A+BAB\displaystyle |A\cup B|=|A|+|B| - |A\cap B|
Look up into your text book for "elementary set theory". This is a rudimentary theorem - every text book that I know of states it and proves it.
 
Top