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1 + 1 = 3

Name: Anonymous 2006-03-06 1:26

Discuss.

Name: Anonymous 2006-03-22 0:59 (sage)

>>6
go back to grade school, equal signs are equivalence relations when used between sets; this entails reflexivity.
the problem is the group of assumptions made on the sets.
C = set of cheap things
E = set of expensive things
if we assume C and E are well defined:
the intersection of C and E is the null set, because things that are expensive are by definition not cheap.
G = set of genuinely cheap things
you could argue that g is not well defined, or that it is in fact equal to C, however assuming some cheap things are not genuinely cheap:
since G is a subset of C, the intersection of G and E is the null set.
apart from that:
R = the set of rare things.
this set is not a subset of E.
if C and E are not well defined because of their subjectivity, we cant use them to make arguments anyway.

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