Is it possible to have a number system without the symmetry of equality axiom?
e.g
a=b;
b!=a;
Name:
Anonymous2007-03-05 20:32 ID:WjFvfjTF
>>1
= is traditionally an equivalence relation, and equivalence relations are symmetric by definition. That's not to say you couldn't use = to mean something other than an equivalence relation of course, but you'd probably succeed only in confusing your readers.
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Anonymous2007-03-05 20:40 ID:QuDqw2Su
Define "number system". Generally when we refer to a number system, we mean a set along with certain axioms relating its elements; at the very least, we mean a group, and the equivalence relation (which requires symmetry) is a fundamental requirement of a group.
>>1
You probably asking this:
A is same as B :: without:: A is equal to B
=A is in the same place in the same set as B
Name:
Anonymous2007-03-11 6:56 ID:QF7B11O7
If you're talking about something akin to first-order Peano arithmetic, no; axiom #2 is "every natural number is equal to itself", which is roughly the corollary of saying that every natural number has only one unique successor. (If you're talking about _non_-Peano arithmetic, then you're in crazyland, populated by utter loons.)
Non-commutative _operations_, on the other hand, are quite spiffy.
Like how you can make a set that isn't a vector space (in linear algebra). Is there a way that you can define everything so that 'a = b', but 'b =/= a'?
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Anonymous2007-03-18 19:10 ID:RDN1/GA/
>>21
No. It's just how the = sign is defined. It is symmetric.
Name:
Anonymous2007-03-18 19:13 ID:RDN1/GA/
>>22 continued:
for any number system or non number system or whatever, the = sign is symmetric.
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Anonymous2007-03-19 0:16 ID:+/jyj6PP
paraconsistent logics lol
Name:
Anonymous2007-03-19 0:24 ID:ey9eEqTV
>>1
The definition of "equals" ("=") says that the property is 1)symmetric, 2)reflexive, and 3)transitive. Therefore, no, it is not possible to have a equal b but b not equal a.
Name:
Anonymous2007-03-19 0:51 ID:LO2RzdXX
Perhaps if we take the ACTUAL = operator, not the symbol "=", to be an assignment operator.