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This question might be more philosophical than mathematical.

In school we are taught how to solve equations such as $x^2 - 1 = 0$ or $\sin(x) - 1= 0$. Solutions to these equations are quite simple. For example $x = 1$ and $x = -1$ are the solutions to the first equation. One could say that solving equation $f(x) = 0$ is same as finding the values of $x$ that satisfy the equation. To me this answer doesn't really tell what does it mean to solve a equation, because the meaning of the verb to find is ambiguous. If someone says that the solutions of the equation are all the numbers in the set $\{y \in \mathbb{R} | f(y) = 0\}$ has he or she solved the equation? I don't think he/she has.

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    $\begingroup$ One might criticize the word "simplify" of the same - but outside of a teaching context, I'm not sure people expect these words to have rigorous meaning (i.e. they might appear as in "We simplify $X$ to $Y$" where they explain the direction of the work, but aren't actually a structural part of it) $\endgroup$ – Milo Brandt Jan 13 '16 at 18:51
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    $\begingroup$ In a math course for engineers, someone asked "is it OK to write $\sqrt2$ or do we have to calculate the exact result". No joke. $\endgroup$ – Gyro Gearloose Jan 13 '16 at 19:35
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    $\begingroup$ It's a good question. I think most mathematicians just agree to call determining the solution set solving the equation. There are equations which cannot be solved analytically; however we can "guess" the right answer and still say we "solved" the equation simply because we determined the solution set. $\endgroup$ – ThisIsNotAnId Jan 14 '16 at 0:59
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    $\begingroup$ I think you may be overthinking this. In most cases an equation represents a problem (or question) of some kind, and when we say that we are "solving" the equation what we really mean is that we are solving the problem (or answering the question). In other words, the meaning of "solving an equation" is determined only by the context. There's no universal definition, nor any need for one. $\endgroup$ – Harry Johnston Jan 14 '16 at 7:54
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    $\begingroup$ @GyroGearloose: It's actually a perfectly fair question. On the one hand, you can't build a bridge that's $2^\pi$ feet long, so $2^\pi$ can be viewed as an abstract approximation and in some engineering problems, the decimal "approximation" is the exact answer. On the other hand, leaving the solution in its analytical form means a) if you plug that solution into another equation, it can be algebraically simplified to potentially require far less computation for future computations, and b) it can be solved to any arbitrary precision. Thus making $\sqrt{2}$ the far better answer in these cases. $\endgroup$ – MichaelS Jan 15 '16 at 5:22
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Interesting question. I'd say that solving $f(x) = 0$ amounts to

  • exhibiting the set S = $\{ x \mid f(x) = 0 \}$, typically by enumerating its members, or giving a sequence whose elements are all the members of $S$

  • demonstrating that the listed or enumerated items are exactly equal to $S$.

So if I say that the solutions of $\sin x = 0$ are $n\pi, n = 0, \pm 1, \pm 2, \ldots$, I've given a purported solution set $S'$. I now need to show that for each element $t$ of $S'$, we actually have $\sin t = 0$, and that no other values of $t$ satisfy $\sin t = 0$.

The method by which I arrive at the set $S$ is not really germane, despite the active verb "solve"; the solution might come from algebraic or geometric manipulations, or it might come to me in a dream. But the second part -- the demonstration that the purported solution set is the actual solution set -- that must follow the rules of logic and mathematics.

This is, however, mostly opinion about common mathematical speech, rather than a fact about mathematics.

PS: For infinite solution sets that are not countable, Christian Blatter's answer starts to get at a good description, although it doesn't take into account things like "the solution set is all irrationals," where a parametrization of the set may be very hard to come up with. Roughly speaking, as the solution sets get more complicated, exhibiting the set gets more and more complicated. No big surprise there...

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    $\begingroup$ Of course, what we accept as "exhibiting" a set is still subjective. If the set is $ S = \{ x\ |\ 0 \leq x \lt 1 \} $, I think this counts as "exhibiting" $ S $, though this is neither enumerating the members nor giving a sequence whose elements are all members of it. $\endgroup$ – cardboard_box Jan 13 '16 at 18:57
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    $\begingroup$ I'd personally be happy to have an uncountable set be a solution. $\endgroup$ – fleablood Jan 14 '16 at 1:18
  • $\begingroup$ This reminds me of $P$ vs. $NP$, where finding the solution may be hard but verifying may be easy. $\endgroup$ – lhf Jan 14 '16 at 10:48
  • $\begingroup$ Good point, cardboard_box. I didn't really want to get into issues of uncountable sets, and the exact nature of "exhibiting" because they seemed to miss what the OP was asking. Maybe I'll edit a tiny bit to address this. $\endgroup$ – John Hughes Jan 14 '16 at 12:39
  • $\begingroup$ Sometimes people give up on solving $f(x) = 0$ and just solve $f(x) \approx 0$ and are happy $\endgroup$ – cactus314 Jan 14 '16 at 17:44
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I agree with John Hughes; this is an interesting question. It is made more interesting by the observation that we often do answer questions like "Solve $f(x) = 0$" with something like $\{x \mid g(x) = 0\}$, which seems on the face of it like answering one question with another. Typically, of course, $g(x)$ is simpler in some sense than $f(x)$: For instance, we might answer

$$ \text{Solve for $x \in \mathbb{N}$ in }8x+1-y^2 = 0, y \in \mathbb{N} $$

with

$$ x \in \left\{\left.\frac{n(n+1)}{2} \,\,\right|\, n \in \mathbb{N}\right\} $$

where we might think of $x-n(n+1)/2$ as the equation $g(x)$ that characterizes the solutions to $f(x) = 8x+1-y^2 = 0$ for $y \in \mathbb{N}$. (There are better examples; this is just the one that comes to mind.)

What makes this characterization of $x$ simpler or better than the one that is posed as the problem? The explanation that I come up with is that we agree that—there is consensus that—the solution is a more immediately transparent description of the $x$ that solve the problem than the problem itself is. In other words, mathematics (like science in this regard) is a social activity, with (often unspoken) agreements about what constitutes progress toward a more primitive characterization of a mathematical object.

When we begin, as students, we get used to the idea of solutions being concrete, like the number $4$ for $3x-12 = 0$. Later, we apprehend that mathematics is a big web or network of relationships, and solutions often merely move from an expression that is less simple or transparent to one that is more simple or transparent. What it is that simplicity or transparency actually represent is not usually made explicit, and I'm not sure that it can be made explicit in any universal way.

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    $\begingroup$ Really nicely put. Yours is the answer I wanted to add to my own, after letting the idea digest for the night. :) $\endgroup$ – John Hughes Jan 14 '16 at 12:40
  • $\begingroup$ Thank you, yours is well thought out, too, +1. $\endgroup$ – Brian Tung Jan 14 '16 at 18:20
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An equation, or a system of equations, defines a solution set $S$ as the set of all members $x$ belonging to some universe $X$ that satisfy certain conditions encoded in a formula, or a "story" ${\cal P}(x)$: $$S:=\{x\in X\>|\>{\cal P}(x)\}\ .\tag{1}$$ This is an implicit description of the set $S$. In most cases it is easy to check whether a proposed $x\in X$ actually belongs to $S$ or not.

Solving $(1)$ means producing an explicit description of $S$. Such an explicit description could consist in a proof that $S$ is in fact empty, it could consist in a finite list $S=\{x_1,\ldots, x_p\}$ of explicitly exhibited elements $x_k\in X$, or it could consist in a parametric representation $$f:\quad I\to X,\qquad \iota\mapsto x_\iota\in X\ ,\tag{2}$$ where $I$ is a certain "standard" set, e.g., $I={\mathbb N}$, $f(I)=S$, and $f$ is injective. In other words: Each element of $S$ is produced by $f$ exactly once in a well understood way.

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  • $\begingroup$ I don't think (2) is sufficient. What if you need an interval on the reals, or some other uncountable thing? $\endgroup$ – Kevin Jan 14 '16 at 0:20
  • $\begingroup$ @Kevin: In that case, I think $I$ would be some "standard" uncountable set like $\mathbb{R}$. $I = \mathbb{N}$ is explicitly only an example. $\endgroup$ – Brian Tung Jan 14 '16 at 17:21
  • $\begingroup$ By $I$, I don't mean me. I mean some other standard set. :-P $\endgroup$ – Brian Tung Jan 14 '16 at 17:21
  • $\begingroup$ @BrianTung: Why not just use the domain of $f$ (assuming we're solving $f(x) = 0$)? That's definitely big enough, even if you have something crazy like $P(\mathbb{R})$ (power set of the reals). $\endgroup$ – Kevin Jan 14 '16 at 19:37
  • $\begingroup$ @Kevin: Why not, indeed. I just thought you had somehow snagged on the notion of $I = \mathbb{N}$. $\endgroup$ – Brian Tung Jan 14 '16 at 20:33
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If someone says that the solutions of the equation are all the numbers in the set {y∈R|f(y)=0}{y∈R|f(y)=0} has he or she solved the equation? I don't think he/she has.

i think you have to understand 'solve' more like simplify in this context. From an arbitrary equation to the solution is like from abstract to specific. You specifie the abstract rules of your model. You have an equation, which represents something in abstract form, you solve it, you have the specific objects.

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    $\begingroup$ I like this line of thinking. Simplifying is easier to define than solving. Though, one has to define a function that tells how simple an expression is. $\endgroup$ – 12345678 Jan 13 '16 at 19:13
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IMO solving an equation is giving the solution set in extension ($S=\{-1,1\}$) rather than in intention ($S=\{x\in\mathbb R|x^2=1\}$).

The elements of the set can be specified via mathematical expressions, preferably that allow effective computation. This makes the solution constructive.

In cases that no mathematical expression is possible, one may consider the equation solved if root isolation has been performed, i.e. the enumeration of intervals guaranteed to contain exactly one root, potentially allowing computation by means of numerical methods.

For instance, I consider that stating "the equation

$$x^7=x+1$$ has a single real root" is sufficient as a solution. Adding "the solution lies in $(1,2)$" is better but is just a hint.

In a nutshell, I see the solution of an equation as the discussion of root separation. In the end, you can refer to every root unambiguously (even without knowing its exact value).

The equation $x^2=1$ has one negative root, $x_-$ and one positive root, $x_+$. Without knowing them, we can anyway affirm $x_-=-x_+$.

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  • $\begingroup$ In most cases one would not consider an equation like y = ln(e^x) solved. $\endgroup$ – TLW Jan 13 '16 at 23:45
  • $\begingroup$ @TLW ln(e^x) is x in the real numbers. so y=ln(e^x) is y=x. in the complex numbers if you solve for x it is x=log(e^y)+2i(pi)n and -(pi)<Im(y)<(pi). $\endgroup$ – Armin Jan 14 '16 at 0:42
  • $\begingroup$ @TLW: I assume you mean that the expression isn't simplified enough to be acceptable. I don't see that as a requirement to the resolution of an equation; what matters is that you gave an explicit formula for the value of the root. Formula simplification is another topic. $\endgroup$ – Yves Daoust Jan 14 '16 at 9:09
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A prerequisite to be able to answer the question "What does it mean to solve an equation?" is to have a definition of the word "equation".

A definition is: an equation is an equality containing one or more unknown(s). Do you agree with this definition ?

If yes, solving the equation consists either :

  • of determining which values, or particular form(s), of the unknown(s) make the equality true,

  • or proving that no value and no particular form of the unknown(s) make the equality true.

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    $\begingroup$ Consider x = ln(2) and x = ln(e) - we generally consider the first solved, but the second not yet solved. $\endgroup$ – TLW Jan 13 '16 at 23:40
  • $\begingroup$ @TLW isn't ln(e)=1 because the question ln(y)=x translates to e^x=y. you asked for e^x=e and there is only one solution, that is x=1 $\endgroup$ – Armin Jan 14 '16 at 0:37
  • $\begingroup$ But what then we shunt the question to what constitutes a "particular" form. why is n(n+1) an acceptable form but "where ln(x) is an integer" is not. $\endgroup$ – fleablood Jan 14 '16 at 1:26
  • $\begingroup$ @TLW: There's a difference between being unsolved (both of your examples have been solved for x) and being unsimplified (ln(e) obviously simplifies to 1). In some cases, the unsimplified form is actually preferred ($0, 3.14, 6.28, 9.42 \approx 0\pi, 1\pi, 2\pi, 3\pi$ rather than $\approx 0, \pi, 2\pi, 3\pi$) so you can more easily see a pattern. In other cases, which version is "simplified" can be hard to see. "Solved" on the other hand, is pretty explicit: An equation is solved for x if x is alone on one side of the equation and there are no xs on the other side of the equation. $\endgroup$ – MichaelS Jan 15 '16 at 5:44
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First of all let's look for what solution means:

The Free Dictionary: To find an answer to, explanation for, or dealing with (a problem, for example).

So.

We have a result on the right side of the equal sign.

If you see there a number then the word solution doesn't make a sense. It is ambitious.

But.

If you see there a result of either an event or events, we can say that:

Wow! Something happened and we got 2, an accident, and so on.

Then we start to find the possible causes to explain why the result is there.

I think the verb, solve, is the best describer of whole finding causes process.

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Your example of {$y \in \mathbb R | f(y) = 0$} as solution for the equation $f(x)=0$. It's simply stating that all solutions for f(x) are all solutions for f(x). A more descriptive solution set would be {$y \in \mathbb R | g(y)=0$}.

And now we solve the solutions recursively. You look at the solution set of $g(x)$ and then at the function which describes it and so on, until you reach a trivial solution, which is a list of numbers, which can be not simplified*, e.g. {$2n | n \in \mathbb N$}. This set can be written down without solving on equation.

*With simplified I mean that there is no function in the solution set anymore and thus does not require anymore recursion.

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