It seems to me an equation, in an abstract sense, must always involve some varying quantities where the varying quantities belong in some space (set, algebraic structure, what have you). In order to make precise the phrase, "vary quantities", it seems to me that one must have a mechanism for evaluation of each side of the equation. Ultimately, I think solving an equation must always be, in essence, finding the pre-image of a mapping. If the pre-image is empty then there are no solutions.

Consider the equation over $\mathbb{C}$ $$x^2-3x+2=0$$

We have that $x^2-3x+2$ is a polynomial and this polynomial induces a natural map from to $\mathbb{C}$ to $\mathbb{C}$ called evaluation. The equation is really asking for the pre-image of $0$ of this map.

Consider the functional equation $$f(x+y) + f(x-y) = 0$$ where we are looking for solutions that are functions from $\mathbb{R}$ to $\mathbb{R}$. I think ultimately that this equation can be thought of in terms of the pre-image of the map $G$ that takes functions like $f$ and maps them to the function from $\mathbb{R^2}$ to $\mathbb{R}$ by sending $f$ to the function of two variables $f(x+y)+f(x-y)$. And the equation is really asking for the pre-image of the zero function under this map.

Is it correct to view all equations in this manner? That is finding the solutions must always be equivalent to finding the pre-image of some element of some mapping?

Think of basic equations one finds in college algebra books. I tell my students that we take the given equation and apply solution preserving operations to it to transform the equation into a simpler one. The goal is to ultimately end up with a simpler equation whose solutions we can find by inspection. For instance,

$$3x - 2 = 5 \implies 3x = 7 \implies x = \frac{7}{3}$$

At each step we transform the equation to a simpler one whose solution set is the same. We can solve the last equation by inspection. Ultimately, isn't this how all equations are solved? We transform the equation to simpler equation(s) and end up with an equation that can be solved by inspection.

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    $\begingroup$ My first year algebra teacher once made the stunning announcement: “equations don't exist”. :-) $\endgroup$
    – egreg
    Apr 15, 2018 at 14:16
  • $\begingroup$ A restatement of what I mean: Abstractly, an equation always involves one or more varying quantities that live in some space (an example would be the space of continuous functions from $\mathbb{R}$ to $\mathbb{R}$). I'll assume one varying quantity for the sake of simplicity. We have two sides of the equation, LHS and RHS. Both sides must be able to be evaluated at a particular instance of the varying quantity and each such evaluation results in some object. Loosely speaking solving an equation means finding the set {x| (x, LHS(x)) is the same point as (x, RHS(x))}. Is this correct? $\endgroup$
    – sykh
    Apr 15, 2018 at 20:13
  • $\begingroup$ See Equation. $\endgroup$ Apr 16, 2018 at 6:22
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    $\begingroup$ "An equation always involves one or more varying quantities ..." What about $1+1=2$? Isn't that an equation? $\endgroup$ Apr 16, 2018 at 13:31
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    $\begingroup$ An equation is simply a statement that two things are equal. To be more specific than that, you'd probably have to specify the domain under discussion. $\endgroup$ Apr 16, 2018 at 19:13

4 Answers 4


One approach: An equation is a predicate, $P(x)$, of the form $s(x) = t(x)$ where $x$ is a free variable (or vector of free variables) and $s(x), t(x)$ are terms -- expressions which evaluate to elements of the universe (e.g. real numbers if you are doing mathematics over the reals) when values are substituted for variables. This means that $P(x)$ is something which evaluates to either $true$ or $false$ when $x$ is replaced by members of the universe. In the 1-variable case it can be thought of as a function of the form

$$P(x): U \mapsto \{true,false\}$$

where $U$ is the domain of discourse.

To solve an equation is to determine $P^{-1}(true)$, the set of all values which make the predicate true.

  • $\begingroup$ That’s a nice way to put it. Thank you, $\endgroup$
    – sykh
    Apr 15, 2018 at 23:44
  • $\begingroup$ What about undecidable predicates? $\endgroup$
    – Couchy
    Apr 17, 2018 at 3:32
  • $\begingroup$ @Couchy311 I'm not quite sure what you are driving at. Are you suggesting that the existence of undecidable predicates implies that equations can't be viewed as predicates? That wouldn't follow. For one thing, equations are predicates of a special type. Not all predicates are equations. For another thing, by the negative solution to Hilbert's 10th problem, there isn't an algorithm for deciding if an arbitrary Diophantine equation has a solution. Just because you write down an equation, it isn't always possible to solve it. $\endgroup$ Apr 17, 2018 at 13:27
  • $\begingroup$ I think I'm referring to the fact that if an equality predicate is undecidable then it cannot be solved, so I'm wondering if it makes sense to assign a truth value to it. $\endgroup$
    – Couchy
    Apr 17, 2018 at 20:32
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    $\begingroup$ @Couchy311 Just because you can't decide if a predicate is true doesn't mean that the predicate lacks a truth value. For any predicate P and any element x, P(x) or not P(x) is a tautology. $\endgroup$ Apr 18, 2018 at 13:35

Short answer touching on foundations and notation.

Mathematicians use equations to tell their readers that two expressions (the things on either side of the $=$ sign) are actually two (different) names for the same underlying object. But there are contexts in which that simple meaning can be lost or forgotten.

In elementary school kids (and teachers) are uncomfortable writing $$ 3 = 1 + 2 $$ because they want to think of the $+$ and $=$ as operations, analogous to the buttons on a calculator, so want to read them only from left to right.

In beginning algebra the equation $$ 3x + 2 = 8 $$ is meant to be "solved". That is, you are to find the values of the variable $x$ that make the two sides of that equation represent the same number, namely $8$. The rules that say you can "do the same thing to both sides" essentially preserve the fact that the two sides continue to name the same object.

If all the $x$'s are on the same side of the equation you can (but need not) think of this as asking for the preimages of the other side under the given map.

Later on when you encounter $$ f(x) = x^2 $$ you can be confused because there's nothing to "solve". This equation tells us that we will use "$f$" to name the squaring function.

When you encounter a "functional equation" like $$ f(x+y) = f(x) + f(y) $$ you may try to solve it: find all the "values" for the function $f$ that make the equation true (for every $x$ and $y$). In this case (assuming continuity or some other weak regularity) the answer is $$ f(x) = cx \quad \text{for some constant } c $$ but you can't get to that conclusion by transforming

the equation to simpler equation(s) and end up with an equation that can be solved by inspection.

In all these cases the equality tells you two things are really the same. What you do with that information depends on the context.

  • $\begingroup$ It seems to me we can't say "find the values of the variable x that make the two sides of that equation represent the same number" and then say that "equality tells you two things really are the same". At the beginning of the equation we can't say x varies over all real numbers, for instance, and then at the end say x is the number 4/3. The expression 3x+2 is not really the same thing as the expression 8 if we permit ourselves to think of x as a variable (that is a varying quantity). $\endgroup$
    – sykh
    Apr 15, 2018 at 14:40
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    $\begingroup$ How you are supposed to think about variables depends on context. When you're given "an equation to solve", $x$ doesn't stand for a variable number, even "at the beginning" . All along it stands for any of the numbers that satisfy the equation. At the end you've found all the values that work: the set of solutions. $\endgroup$ Apr 15, 2018 at 14:44
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    $\begingroup$ There's where we disagree. An equation sometimes can be interpreted as asking you to find the preimage of a mapping. Many routine algebra exercises use equations this way. But that's not what an equation "really is". $\endgroup$ Apr 15, 2018 at 14:48
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    $\begingroup$ You are indeed asserting that $3x+2$ is the same as $8$ provided you use the right value of $x$, which you are challenged to find. I think we'll just have to agree to disagree from now on. $\endgroup$ Apr 15, 2018 at 15:06
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    $\begingroup$ @EthanBolker -- You might want to add something about quantifiers. The statement "For all $x$, $3x+2=8$" is false, but "For some $x$, $3x+2=8$" is true. The equation can only be "solved" in the latter case, to find that "some $x$". In the former case, the equation is basically asserting that two functions are equal, instead of two numbers being equal. $\endgroup$
    – mr_e_man
    Apr 15, 2018 at 19:38

The universe of equations does not have uniform semantics.

Equations can be tautologies. $$ 1=1, x+x=2x, \tan x = \frac{\sin x}{\cos x}, \frac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^x = \mathrm{e}^x, \dots$$

Equations can be explicit definitions. $$ f(x) = f(f(x-1)), \\\frac{\mathrm{d}}{\mathrm{d}x}g(x) = \lim_{h \rightarrow 0} \frac{g(x+h) - g(x)}{h}, \dots $$

Equations can be implicit specifications. This seems to be the only kind of equations you are discussing. $$ 8x+7 = 15, x^5+x-1 = 0, 0 = \int_{\Omega}f \cdot g^* \,\mathrm{d}\mu, \dots $$

These semantics can be nested. $$ f''(x) = - f(x), \mathrm{Tor}_n^R(A,B) = (L_nT)(A), \dots $$

Frequently, these equations are (implicitly) augmented by specifying the sets from which the variables may be drawn. For instance, \begin{align*} g &\in \{\mathbb{R} \rightarrow \mathbb{R}\}, \\ U &\subset \mathrm{dom}(g), \\ m(U) &> 0, \\ H_0(U; \mathbb{Z}) &= \mathbb{Z}, \\ x &\in \mathrm{int}(U), \\ h &\in \mathbb{R}, \text{[footnote]} \\ x+h &\in \mathrm{dom}(g), \\ \frac{\mathrm{d}}{\mathrm{d}x}g(x) &= \lim_{h \rightarrow 0} \frac{g(x+h) - g(x)}{h} \end{align*} Changing these specifications may alter what the final equation denotes, if it denotes anything.

[footnote]: There's notational abuse here. "$h$" only appears as a dummy variable in the limit -- it is expanded to infinite sequences. We really want to say that every element of such a sequence is real and when you add $x$ to any of them, you still land in the domain of $g$. This is (in some sense) really a defect of our usual notation for limits that we do not get to specify the set from which the sequences are drawn. Some would write these two conditions with the "$h \rightarrow 0$", but this doesn't change that these constraints are really to be applied to the various $h_i$.

  • $\begingroup$ I was only considering what you called implicit specifications. That’s what I had in mind. $\endgroup$
    – sykh
    Apr 15, 2018 at 22:41

I think most of these answers are too complicated. The simple answer to the simple question:

An equation is the statement that two things are equal

"equal", "equate". "equality" and "equation" are different forms of the same word.

In mathematics the "=" sign is put between two things that are being said to be. or required to be, equal.

A formula is a special type of equation that we remember for working out a problem without having to always start from scratch.

A simple example:

Distance traveled at uniform speed = the speed multiplied by the time taken

Using symbols, we might write this as d=st

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    $\begingroup$ the problem is that you have to know for which purpose they are equal (or isomorphic). in a computer for instance, there must be a sequence of mechanical-down-to-the-bit and without-a-human steps that leads to a bit-per-bit identical data structure. finite integer arithmetic is a dead-solved problem for precisely this reason, where equation manipulation tactics get befuddled by assumptions that are only obvious when the human is not supervising the process. $\endgroup$
    – Rob
    Apr 18, 2018 at 1:39

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