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Thread: A weird asymmetrical math

  1. #1
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    A weird asymmetrical math

    I wonder if someone explored this before.

    As you know there are non-commutative branches of math that deal with A+B not equal B+A kind of stuff and they clash our intuition, but they are still here. Imagine a math in which A=B, but it doesn't necessary mean B=A. Is it even possible to build such a math system?
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  2. #2
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    Re: A weird asymmetrical math

    symmetry, i.e.

    $a=b \Rightarrow b=a$

    is a requirement of an equivalence relation such as equals.
    Last edited by romsek; Apr 5th 2018 at 01:33 PM.
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    Re: A weird asymmetrical math

    I see. I'll try to say it another way: knowing B contains all the necessary information to learn A, but knowing A contains no information about B. Is it possible to build such a math system?
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    Re: A weird asymmetrical math

    Quote Originally Posted by Fox333 View Post
    I see. I'll try to say it another way: knowing B contains all the necessary information to learn A, but knowing A contains no information about B. Is it possible to build such a math system?
    You are talking about relations that are not equivalence relations. You can create any type of relation you want. A relation is a subset of the direct product of a set with itself. So, a relation on a set $A$ is $R \subseteq A\times A$. Equality is the relation $R = \{ (a,a) | a \in A \}$. This is the trivial equivalence relation. You can have any other type of relation you want, but it will not have the meaning of an equivalence relation.

    So, basically, an equivalence relation is already a abstraction of equality. You want to take it a step further and use an arbitrary relation (that is known to be non-symmetric) in place of equality and see what happens. Nothing meaningful. Non-symmetric relations are useful in many contexts, but not in place of equality. For example, the total order operator: <. That is extremely useful, and it is non-symmetric. But, if you were to replace equality with it, you would have garbage.
    Last edited by SlipEternal; Apr 5th 2018 at 01:23 PM.
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    Re: A weird asymmetrical math

    Are there some explored math system which based on non-trivial equivalence relations?
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    Re: A weird asymmetrical math

    Quote Originally Posted by Fox333 View Post
    Are there some explored math system which based on non-trivial equivalence relations?
    Certainly. Modular arithmetic. Finite field theory. There is plenty of math that is based on non-trivial equivalence relations. For abstractions of this:

    https://en.wikipedia.org/wiki/Quotient_space_(topology)
    Last edited by SlipEternal; Apr 5th 2018 at 02:25 PM.
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    Re: A weird asymmetrical math

    A "relation" between set A and set B is any way of matching objects in A with objects in B, not necessarily "one to one". The simplest definition of such a relation is 'a set of ordered pairs, (x, y), with x in A and y in B". An "equivalence relation" is a relation between a set, A, and itself such that
    1) if x is in A, (x, x) is in the relation (reflexive property).
    2) if x, y, z are in A, both (x, y) and (y, z) are in the relation then (x, z) is in the relation (transitive property).
    3) if x and y are in A and (x, y) is in the relation, then (y, x) is in the relation (symmetric property).

    Now, exactly what are you using "=" to mean? The strictest definition of "=" (equality) is that
    "A= B" if and only if "A" and "B" are (possibly different) names for the same thing. Of course, with that (strict) definition if A= B, then B= A. It is not uncommon to use "=" to mean some equivalence relation. But that still requires (3) above, that if A= B then B= A. There certainly are relations that are not symmetric: $\displaystyle A\subseteq B$ with A and B sets, for example. But I can't imagine anyone using "=" to denote such a relation.

    (The "possibly different" above is because I once defined "equals" as "A= B if and only "A" and "B" are different names for the same thing and someone complained that that would mean "A= A" would be impossible!)
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    Re: A weird asymmetrical math

    Quote Originally Posted by HallsofIvy View Post
    Now, exactly what are you using "=" to mean?
    I realized my mistake. "=" can't be used in this context. I mean any kind of relation which is not symmetric. I don't know which sign I should use.
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  9. #9
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    Re: A weird asymmetrical math

    Quote Originally Posted by Fox333 View Post
    I realized my mistake. "=" can't be used in this context. I mean any kind of relation which is not symmetric. I don't know which sign I should use.
    How about the less than sign (used for total orders)? Any time a<b, you know that b<a is false. This type of operator is known as anti-symmetric and non-reflexive. The partial order operator $\le$ is reflexive and anti-symmetric.

    A relation that is neither symmetric nor anti-symmetric is the "knows the name of" relation. For instance, in a group of people, aRb means that person a knows person b's name. So, George knows Bob's name, so George R Bob, but Bob is forgetful, so Bob does not remember George's name. That means Bob R George is false. Bob knows Alice's name, so Bob R Alice. But Alice knows Bob's name, so Alice R Bob is true, as well.
    Last edited by SlipEternal; Apr 6th 2018 at 12:36 PM.
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    Re: A weird asymmetrical math

    Yes, that's something like you described.
    I think it would be clearer if I explain why I need it.

    In fact, every interaction in space is mutual. Let's say, the gravity force between two objects. If we know the low of the force and the mass of any of the two objects we can learn the mass of the second object. It can't be one-way road. But time is another thing. Past can impact future, but not vice versa. Therefore, to describe this kind of impact we need a special math, which is non-symmetric. Does it make any sense?
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