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It seems like a simple question, but is actually quite hard to prove. You don't just simply say:

$a = c$ because $a = b$ and $b = c$.

Like I actually want a proof behind it, something like using formulae, arithmetic, etc.

How is this possible?

Lord_Farin
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Zerium
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3 Answers3

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In any part of mathematics where one uses equality (and I can't think of any parts where one doesn't) one admits as axioms those that express that equality is an equivalence relation. This means that you can always assume that

  • For any value $x$ one has $x=x$,
  • Whenever $x=y$ then also $y=x$,
  • Whenever $x=y$ and $y=z$ then also $x=z$.

The question you ask is the third axiom, so you can use this without needing to prove it.

Any notion of "equality" that does not satisfy these axioms should not be denoted by "$=$", as this would immediately invite erroneous arguments that do use the above properties. An example of such a relation is "approximately equal" among (for instance) real numbers, formulated using any reasonable precise definition you like.

Marc van Leeuwen
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  • In addition to the 3 properties above (equivalence), Wikipedia defines $=$ as being antisymmetric, hence an order and that these 4 properties uniquely define equality. However, I asked about isomorphism: http://math.stackexchange.com/questions/285484/distinguishing-equality-and-isomorphism-as-relations which is also a apparently an equivalence and an order just like equality. – alancalvitti Feb 17 '13 at 15:42
  • @alancalvitti: Saying equality is anti-symmetric is not false but it is disengenious. Literally it means that if $a=b$ and $b=a$ then $a=b$, which cannot be denied. But the final "$=$" is the equality relation built into the definition of anti-symmetric, not the relation that 'anti-symmetric' is talking about. So equality is not on equal footing with other relations that are anti-symmetric. – Marc van Leeuwen Feb 18 '13 at 11:22
  • Understood - however, what exactly is that final "=" equal to, pun intended? And what's the difference if isomorphism is substituted for all of them? – alancalvitti Feb 18 '13 at 15:02
  • @alancalvitti: I don't quite follow the last part of your comment. Being isomorphic is not an anti-symmetric relation (even though $a\cong b$ and $b\cong a$ imply $a\cong b$ trivially, they do not imply $a=b$). – Marc van Leeuwen Feb 18 '13 at 15:16
  • But you haven't completely characterized what $a=b$ means. Suppose "=" and isomorphism are the same relation, then anti-symmetry is also expressed in terms of isomorphism. – alancalvitti Feb 18 '13 at 15:20
  • Goldblatt in Topoi (and I assume a definition in quite general use) writes that a skeletal category is one where isomorphic objects are identical. So in a skeletal cat, if equality is anti-symmetric so is isomorphism, since they are the same thing. – alancalvitti Feb 22 '13 at 05:14
  • @alancalvitti: So you see that if isomorphic objects are identical then two objects cannot be mutually isomorphic unless they are the same. Great. The same implication is true with any relation in the place of "isomorphic". And what is the relation to the current question (the one at the top of the page)? – Marc van Leeuwen Feb 22 '13 at 06:50
  • The Q asks about relational property of equality and we were discussing categorical formalization of antisymmetry in pre-orders. A poset is a skeletal pre-order category. – alancalvitti Feb 22 '13 at 18:43
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I'd just like to point out that at least in ZF set theory, a current standard foundation for mathematics, equality is defined in terms of membership and simple logic, so this becomes a theorem.

Definition:

$$A = B \iff (x \in A \iff x \in B)$$

Theorem (Transitivty of Equality):

$$(A = B \wedge B = C) \implies A = C$$

Proof:
Suppose: $$A = B \wedge B = C$$ Applying the definition of equality: $$(x \in A \iff x \in B) \wedge (x \in B \iff x \in C)$$ By transitivity of logical equivalence (which I believe is a theorem of elementary logic): $$(x \in A \iff x \in C)$$ Which by definition is, $$A = C$$

Jack M
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  • Isn't equivalence an equality as per Marc's definition? – alancalvitti Feb 17 '13 at 15:36
  • Logical equivalence is an equivalence relation on propositions, yes, in some sense. I'm not sure if that statement really makes sense, formally speaking, but intuitively, it's correct. – Jack M Feb 17 '13 at 15:41
  • Check my comment to Marc's answer - in the link I ask about the distinction between equality and isomorphim; you may be interested in, no one has answered it so far. – alancalvitti Feb 17 '13 at 15:46
  • Logical equivalence is usually defined as an abbreviation: $x\leftrightarrow y$ means $(x\to y)\land(y\to x)$. Since $(x\to y)\land(y\to z)\implies(x\to z)$ is either an axiom or easily derivable, one can prove without difficulty $(x\leftrightarrow y)\land(y\leftrightarrow z)\implies(x\leftrightarrow z)$. – Marc van Leeuwen Mar 17 '13 at 09:20
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This is the transitive property. Many relations depend on the transitive property, such as partial order relations and equivalence relations. You can read more about relations here http://en.wikipedia.org/wiki/Binary_relation or for example, Munkres Topology gives a nice little introduction good enough for many practical purposes.

Euler....IS_ALIVE
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