11
$\begingroup$

what is the definition of $=$?

Above is the question that I would like to be answered, below are some of my thoughts.


I've been thinking about what it means to say

$A = B$

I came to this from thinking about why it makes no sense to say something like

$5 = \begin{pmatrix}1, 2, 3\end{pmatrix}$

My answer to myself was that $=$ is context dependent, it is a relation on a certain set and so only makes sense when the elements on either side of the $=$ are in the said set.

Now, say we are in $\mathbb{N}$, then how do we define $=$? My initial thought was to say that $A = B$ if, given any proposition regarding A, $P(A)$, with truth value $\alpha$, the truth value for $P(B)$ will always be $\alpha$.

However, what happens now if we take $P(\star) $ as $ \star = B$? Do we then need a definition for $=$ in order for $P(\star)$ to make sense? Is this an issue?

$\endgroup$
25
  • 5
    $\begingroup$ It makes sense to say $5=(1,2,3)$. It just isn't true. $\endgroup$ – Chris Eagle Mar 2 '13 at 0:01
  • 5
    $\begingroup$ Your proposal to define $A = B$ as $P (A) \leftrightarrow P (B)$ for all predicates $P$ is basically the same as Leibniz's definition of equality, i.e. the identity of indiscernibles. This is not the usual definition. $\endgroup$ – Zhen Lin Mar 2 '13 at 0:06
  • 2
    $\begingroup$ $=$ is a reflexive, symmetric, transitive and antisymmetric relation. It's governed by the logical axioms for equality. $\endgroup$ – Benedict Eastaugh Mar 2 '13 at 0:09
  • 2
    $\begingroup$ @Hayeder Well, we build the real numbers in terms of the rationals numbers which in turn are built in terms of the integers which in turn are built in terms of the naturals, which can be defined in terms of sets (or axiomatically). However, it can get messy. $\endgroup$ – Pedro Tamaroff Mar 2 '13 at 0:13
  • 8
    $\begingroup$ Relevant comic (from SMBC): i.stack.imgur.com/p5zgM.gif $\endgroup$ – Zev Chonoles Mar 2 '13 at 0:13
9
$\begingroup$

In logic, equality is not something we define, it's a primitive notion. If $A=B$ then $A$ has property $P$ if and only if $B$ has property $P$—this is probably the most important thing to know about equality and although it doesn't define equality, it is assumptions like this that shape our understanding of what equality "is", just like the axioms of set theory shape our understanding of what it means to say $A \in B$ even though membership is a primitive notion that is not defined in terms of other things.

The converse statement "if for every property $P$, $A$ has property $P$ if and only if $B$ has property $P$, then $A=B$" is not true in general. (Unless you allow the property $P$ to have a parameter—then as you point out it's trivially true because $P(x)$ can say $x =B$.) It's true in structures like the natural numbers, where every object is definable, but other structures may have indiscernible elements that cannot be distinguished by any logical property.

$\endgroup$
5
  • 1
    $\begingroup$ To give a finishing touch, $A=B$ if and only if $A$ and $B$ are the same object. It's not a mathematical definition per se, but that's how it is. $\endgroup$ – Asaf Karagila Mar 2 '13 at 1:43
  • 2
    $\begingroup$ @Asaf I like how you say "that's how it is", which presumes the notion of identity once again :) $\endgroup$ – Trevor Wilson Mar 2 '13 at 1:50
  • $\begingroup$ I admitted this not being a mathematical definition... :-) $\endgroup$ – Asaf Karagila Mar 2 '13 at 1:52
  • 1
    $\begingroup$ Well, P(X) = "X is the same object as A" is a property that A has, so if B=A then B also has this property. It may or may not be expressible in the object language, but it is certainly a "property" in the abstract sense. $\endgroup$ – Carl Mummert Mar 2 '13 at 3:06
  • 3
    $\begingroup$ This is sometimes called Leibniz's law, and identity of indiscernibles. en.wikipedia.org/wiki/Identity_of_indiscernibles $\endgroup$ – Andrés E. Caicedo Mar 2 '13 at 3:07
4
$\begingroup$

From a reductionist foundational viewpoint, equality in mathematics is equality of sets, as defined by the (ZFC) axioms of set theory. Recall that the axiom of extensionality states that two sets are equal iff they have precisely the same elements. Let's examine how the equality of sets percolates up to equality on other mathematical objects, e.g. number systems, algebraic structures, etc.

The natural numbers $\,\Bbb N\,$ may be represented by sets, e.g. von-Neumann ordinals. Then equality of naturals is the same as set-theoretic equality. The integers $\,\Bbb Z\,$ may be constructed from $\,\Bbb N\,$ by a standard difference semiring construction, where integers are represented by equivalence classes of pairs $\rm\,(j,k)\,$ of naturals modulo the equivalence relation $\rm\,(j,k) \equiv (m,n)\,$ iff $\rm\,j+n = m+k.\:$ The idea is that the equivalence class $\rm\,[(j,k)]\,$ represents the solution $\rm\,x\,$ of $\rm\, x + k = j,\,$ i.e. the integer $\rm\: j-k.\:$ Thus, set-theoretically, integers are certain subsets of $\,\Bbb N^2,\,$ and, hence, integer equality is set-theoretic equality in $\,\Bbb N^2.\,$

Similarly, rationals (fractions) are constructed as pairs of integers, modulo the well-known equivalence relation for fractions, i.e. if a fraction $\rm\,a/b\,$ is represented by the pair $\rm\,(a,b),\,$ then $\rm\, (a,b) \equiv (c,d)\,$ iff $\rm\,ad = bc,\:$ e.g. $\rm\,1/2 = [(1,2)] = \{\ldots,(-2,-4),(-1,-2)(1,2),(2,4),\ldots\}.$ Again, fractions are certain subsets of $\rm\,\Bbb Z^2,\,$ and fraction equality is the same as set-theoretic equality of these sets (equivalence classes). The same holds true for all other numbers systems (e.g. Hamilton's construction of $\,\Bbb C\,$ as pairs of reals). In every case, their equality relations are equivalence relations, so equality boils down to set-theoretic equality of equivalence classes.

Moving up to the structural level, rings are represented by equivalence classes of isomorphic rings, so ring equality is again set-theoretic equality. Similarly for other algebraic structures.

In analogy with programming languages, this is the "assembly language" view of mathematical objects, where everything has been disassembled to its primitive machine-level data-types and operations. When working at a high-level, we don't conceptualize the objects in terms of these primitive representations. But to be rigorous, they must be built up this way from the primitive set-theoretical foundations. And it is set-theoretic equality that percolates up to the higher-level equivalence relations that define equality on these higher-level composite mathematical objects.

$\endgroup$
3
  • 2
    $\begingroup$ So how do you define $\in$? $\endgroup$ – Asaf Karagila Mar 2 '13 at 2:32
  • 1
    $\begingroup$ @Asaf However you like, as long as you obey the specification, i.e. any model of ZF(C) will do. $\endgroup$ – Math Gems Mar 2 '13 at 2:54
  • 1
    $\begingroup$ +1 for analogy with programming languages, which is how I like to think about this stuff. $\endgroup$ – wj32 Mar 2 '13 at 2:57
1
$\begingroup$

Equality relations are expected to be symmetric, reflexive and transitive. Relations which have these three properties are called equivalence relations. These conditions allow us to reject certain relations from being used as definitions of equality. For instance $\in$ is not an equivalence relation and so it will not serve very well as equality. It is not symmetric: $A\in B$ does not imply $B\in A$. It is also not reflexive: $A\in A$ is not generally true, but only if $A$ is a set which contain itself.

In mathematics, the usual equivalence concerns itself with the sameness of value. If we have a real number $x$ and a real number $y$, then $x = y$ is true if $x$ and $y$ are in fact not distinct, but are the same number. Unlike in computer programming languages, we don't have to worry about the type. That is to say, there is no distinction between 3.0 and 3. In computing, there can be equalities which distinguish those two as unequal; yet, those equalities are still proper equality relations.

Two complex numbers are equal if their corresponding real and imaginary parts, taken as real numbers, are equal, which means that a complex number is only equal to itself and not to any other complex number.

Two vectors are equal if they are of the same dimension, and each member of one vector is equal to the corresponding member of the other.

Two sets are equal if they are of the same cardinality and contain the same elements.

Whether you want to treat objects of different kinds, such the vector or matrix $[1]$ as equal to the real number $1$ or to the set ${1}$, depends on the situation. If such a notion of equality doesn't lead to any flaws in a proof, and adds some kind of convenience, then it is admissible.

$\endgroup$
2
  • $\begingroup$ "Two complex numbers are equal ..." But in that case, you only have one complex number, perhaps identified in two different ways. $\endgroup$ – Peter Smith Mar 2 '13 at 9:52
  • $\begingroup$ @PeterSmith Yes. I tried to avoid that wording as much as possible, but it snuck in! I caught myself writing that and thought that adding "a complex number is only equal to itself and not to any other complex number" was enough to recover. :) $\endgroup$ – Kaz Mar 2 '13 at 10:01
-2
$\begingroup$

Just in terms of math I would say the equality sign of A=B means that A and B have the same displacement from a fixed point of reference (and of course displacement here is a vector quantity).

$\endgroup$
2
  • 4
    $\begingroup$ -1. Many things in mathematics are not elements of normed vector spaces, and your definition would not make sense for them. What is the "displacement" of an exotic smooth structure on the 4-sphere from the standard smooth structure? What is the "displacement" of a set $\mathsf{A}=\{\star,\Box,\oslash\}$ from the set $\mathfrak{B}=\{\circ,\ast\}$? $\endgroup$ – Zev Chonoles Mar 4 '13 at 3:18
  • 1
    $\begingroup$ I'm not really satisfied by this for a few reasons - one of which is that this definition (if/where it makes sense) rests on the notion of two things being 'same', which is the problem in the first place. Thanks for thought, though. $\endgroup$ – user27182 Mar 6 '13 at 2:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.