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Must equality "=" be defined as the identity relation or as "the
same"?
Posted:
Apr 25, 2012 1:23 PM
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In the thread
"Just what is equality in mathematics, anyway?"
http://mathforum.org/kb/thread.jspa?threadID=2367800
that I started, many had a problem with accepting the fact that the equality (x,0) = x for any real number x holds in the complex numbers, even though Walter Rudin proved in his textbook *Principles of Mathematical Analysis* (3rd edition) that the equality
(x,y) = x + yi
holds in the complex numbers for all real x,y (Theorem 1.29), from which we can of course derive the sequence of equalities
(x,0) = x + 0i = x + 0 = x.
The problem in accepting this equality (x,0) = x as true seems to be based on how one defines "=".
Here below are a set of theorems and proofs where equality "=" is the binary relation. Except for the proof of the first theorem, Theorem U, where its proof is in standard paragraph form, to make the justifications as perfectly clear as possible the proofs will be in a standard vertical form of the propositional calculus where each step is justified by a brief comment to the right identifying the used rule of inference.
After these theorems and proofs I give some notes on this problem of one's not being able to accept what Rudin proved and what we can easily derive from what he proved.
This problem seems to be based on the belief that for the binary relation of equality, a binary relation that is reflexive, symmetric, transitive, and antisymmetric *must* also be the identity relation. These theorems and proofs suggest that it should be perfectly acceptable to reject this belief and thereby solve the problem.
Definitions:
Let "~" be negation.
Let a binary relation be a subset of a Cartesian product on two sets.
Let I be a binary relation called the identity relation.
Let "=" be a binary relation called equality.
Let R be the reflexive property.
Let S be the symmetric property.
Let T be the transitive property.
Let A be the antisymmetric property.
Theorem U. Where "=" is the binary relation, A always holds.
Proof. Where "=" is the binary relation, by applying the definition of antisymmetry, we see that the implication (x=y & y=x) -> x=y is demonstrably a logical tautology (of the form (p & q) -> p), meaning it is always true, meaning that A always holds.
Theorem V. Where "=" is the binary relation, we have the equivalence that conjunction R&S&T holds if and only if conjunction R&S&T&A holds.
Proof. The proof of the equivalence (R&S&T) <-> (R&S&T&A) will be by direct proof, where for each implication in this equivalence we assume the antecedent and derive the consequent and then infer the implication, where this last step we will call Direct Inference. Where "=" is the binary relation:
(1) R&S&T&A Assumption
(2) R&S&T Conjunction Elimination on (1)
(3) R&S&T&A -> R&S&T Direct Inference on (1) and (2)
(1) R&S&T Assumption
(2) A Theorem U
(3) R&S&T&A Conjunction Introduction on (1) and (2)
(4) R&S&T -> R&S&T&A Direct Inference on (1) and (3)
Theorem W. Where "=" is the binary relation, we have the equivalence that conjunction R&S&T&~I holds if and only if conjunction R&S&T&A&~I holds.
Proof. The proof of the equivalence (R&S&T&~I) <-> (R&S&T&A&~I) will be by direct proof, where for each implication in this equivalence we assume the antecedent and derive the consequent and then infer the implication, where this last step we will call Direct Inference. Where "=" is the binary relation:
(1) R&S&T&A&~I Assumption
(2) R&S&T&~I Conjunction Elimination on (1)
(3) R&S&T&A&~I -> R&S&T&~I Direct Inference on (1) and (2)
(1) R&S&T&~I Assumption
(2) A Theorem U
(3) R&S&T&A&~I Conjunction Introduction on (1) and (2)
(4) R&S&T&~I -> R&S&T&A&~I Direct Inference on (1) and (3)
Theorem X. Where "=" is the binary relation, we have the equivalence that implication (R&S&T -> I) holds if and only if implication (R&S&T&A -> I) holds.
Proof. The proof of the equivalence (R&S&T&A -> I) <-> (R&S&T -> I) will be by direct proof, where for each implication in this equivalence we assume the antecedent and derive the consequent and then infer the implication, where this last step we will call Direct Inference. Where "=" is the binary relation:
(1) R&S&T&A -> I Assumption
(2) R&S&T -> R&S&T&A Theorem V
(3) R&S&T -> I Hypothetical Syllogism on (1) and (2)
(4) (R&S&T&A -> I) -> (R&S&T -> I) Direct Inference on (1) and (3)
(1) R&S&T -> I Assumption
(2) R&S&T -> R&S&T&A Theorem V
(3) (R&S&T -> I) & (R&S&T -> R&S&T&A) Conjunction Introduction on (1) and (2)
(4) R&S&T -> (R&S&T&A & I) Definition of Implication; Distributive and Commutative Properties in the Propositional Calculus; all on (3)
(5) (R&S&T&A & I) -> (R&S&T&A -> I) Implication from Conjunction on the Consequent of (4)
(6) R&S&T -> (R&S&T&A -> I) Hypothetical Syllogism on (4) and (5)
(7) (R&S&T & R&S&T&A) -> I Exportation on (6)
(8) R&S&T&A -> I Conjunction Elimination on the Conjunctions of the Same Symbols in the Antecedent of (7)
(9) (R&S&T -> I) -> (R&S&T&A -> I) Direct Inference on (1) and (8)
Notes.
In case one wants to reject any of the above used rules of inference: The rules of inference in the propositional calculus, like some of the above called conjunction elimination, conjunction introduction, and implication from conjunction (which says that from any conjunction p & q we can infer the three implications p -> q, ~p -> q, and ~p -> ~q) are tautologies, as are all the other rules of inference in the propositional calculus like modus ponens. Rejecting them is as insane as rejecting modus ponens.
If one *defines* binary relations such that R&S&T&A implies I, then Theorem X shows that that definition *bans* by definitional decree any use of "=" such that R&S&T&~I. That is, by Theorem X, directly defining "=" such that the implication R&S&T&A -> I must hold is in effect indirectly defining "=" such that the implication R&S&T -> I must hold, which in effect bans by definitional decree any use of "=" such that R&S&T&~I.
One consequence of this banning by definitional decree is that Rudin's "=" *must* be R&S&T&A&I.
But Rudin's "=" can be viewed as R&S&T&A&~I if we define "=" in a way that allows the possibility of viewing "=" as R&S&T&~I, since R&S&T&~I implies R&S&T&A&~I by Theorem W. That is, by Theorem W, all those who do not want to view Rudin's "=" as R&S&T&A&I may know that it is proved that they may view Rudin's "=" as R&S&T&A&~I if they define "=" in a way that allows the possibility of viewing "=" as R&S&T&~I.
There is no reason why we *must* directly or indirectly define "=" such that R&S&T -> I; there is no reason that we cannot directly or indirectly define "=" such that we can view "=" as R&S&T&~I.
Note that I did not in the above address this subjective phrase "the same" in the context of "=" - I did not do so because it is so subjective.
But now I do: If "the same" means the identity relation, then the above answers any concern raised by using this phrase. If "the same" does not mean the identity relation, then the above again answers any concern raised by using this phrase.
However, to play the devil's advocate:
Suppose one were to claim the following: The symbols or objects (x,0) and x are just representations of underlying deeper objects, and so the statement (x,0) = x does not necessarily say that the symbols or objects (x,0) and x are equal or the same or identical or whatever, but only that what they represent are equal or the same or identical or whatever. That is, the statement (x,0) = x can say that the representations (x,0) and x represent a same or identical underlying object.
If one accepts this line of reasoning, then this representation theory of equality "=" provides an additional solution to the problem of not being able to accept the fact that (x,0) = x: With this representation theory one could affirm this equality even as R&S&T&A&I - one would not need to affirm this equality only as R&S&T&A&~I.
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