To solve an equation

Question

This might seem as a silly question. The reason why I ask it is basically because I am interested to know the formal and correct way of expressing equations as exercises.

This question arised in a discussion between me and a friend. Consider a textbook and student relationship. We were discussing an exercise problem like this:

1. Solve the equation $$\frac{2x+1}{4x+2} = \frac{1}{2}.$$

We were debating who's responsibility it was to state that $x \neq -1/2$. Is it the responsibility of whoever formulated the equation or the responsibility of whoever tries to solve it.

In my opinion an equation is a predicate, so it is true for some inputs (x in this case) and false for some. To solve an equations is basically to find its truth table. Of course the predicate should be given together with some "universe" $U$ from which x is taken.

So if we consider the above exercise, it is not even a predicate if we do not also state that $x \neq -1/2$. In my opinion then, it is incorrect to say that this is an equation:

$$\frac{2x+1}{4x+2} = \frac{1}{2}$$

without stating that $-1/2$ is not in the universe.

Looking forward to your comments.

Answer 1

I would say that, unless otherwise stated, an expression should be considered on its largest possible domain (within a given universal set). So, in your example, as no domain is given, the largest possible domain is considered (within $\mathbb{R}$ I'm guessing); that is, we take $x \in \mathbb{R}\setminus\{-\frac{1}{2}\}$.

This kind of consideration is not as nitpicky as some people might think. Consider the following DE $$\frac{dy}{dx} = -y^2.$$ When solving a DE, we are trying to find all functions $y$ which satisfy the given equation. Part of the definition of a function is its domain, but this is often not addressed when stating a DE, so one usually assumes we are looking for functions on the largest possible domain for which the DE makes sense (in this case $\mathbb{R}$). Obviously $y = 0$ is a solution, and solving the seperable DE in the usual way, we see that for each $c \in \mathbb{R}$, $y_c(x) = \frac{1}{x+c}$ is also solution. Note that the trivial solution $y = 0$ has largest domain of definition $\mathbb{R}$ while the non-trivial solutions $y_c$ have (the slightly smaller) largest domain of definition $\mathbb{R}\setminus\{-c\} \subset \mathbb{R}$. Furthermore, if we allow ourselves to consider solutions with domain of definition $\mathbb{R}\setminus\{-c\}$, we obtain many more (piecewise) solutions. For example, the function $y : \mathbb{R}\setminus\{1\} \to \mathbb{R}$ given by

$$y(x) = \begin{cases} \dfrac{1}{x-1} & x < 1\\ & \\ \dfrac{1}{x+1} & x > 1 \end{cases}$$

is also a solution to the DE, but can't be extended to a solution on $\mathbb{R}$ (the same is true of the solutions $y_c$). This is very much the point. If we always take unspecified domains to be the largest possible, we sometimes eliminate important and interesting possibilities.

I apologise if this answer has deviated too far from the intention of your question.

Answer 2

I guess there can be several points of view, but perhaps some of the main ones are:

1) Basic equations like the given in the OP usually appear first in junior high school or so, and it usually is part of the exercise for the kids to find out what the "domain of definition of" or "set of possible values to substitute in" the equation is.

Thus, a complete solution to this kind of exercises must contain the above set.

2) From a more purely mathematical point of view, which is also reflected in point (1), one should probably assume that the given exercise contains only mathematical expressions. Since dividing by zero renders an expression meaningless in mathematics(or, at least, in "regular" mathematics. I'm not going to get into possible meanings in this or that realm of what division by zero could be), one could say that "it is obvious" that $\,x\neq-1/2\,$ as otherwise we wouldn't even be dealing with mathematics.

In this case it is assumed the reader understands the above or else it is specifically given with the exercise.

Answer 3

Looking at people's answers so far has given me the impression that it's very possible that there's no consensus as to what counts as a formal answer to that question.

To sum up though, here are the possibilities as I see them. The question could be:

1) "Find all x such that the equation is true"

2) "Find all x such that the equation makes sense, and that it is true"

The 2nd extended question is usually what's being asked in most cases that I observed, but I honestly don't think there's a kind of formal gap between them. You can choose to ask either one and they would both be fine questions.

Answer 4

This was intended as a comment to Michael Albanese's answer. I do not have enough reputation to post it as a comment.

If we always take unspecified domains to be the largest possible, we sometimes eliminate important and interesting possibilities.

Given a predicate $P(x)$ where the domain $x$ belongs to is unspecified, we do not miss solutions by letting the domain (which $x$ belongs to) be the "largest possible". In the case where the predicate is a differential equation $P(f)$ restricting $f$ to total functions on $\mathbb{R}$ is not "letting the domain (which $f$ belongs to) be the largest possible".

As for the original question: the author of the excercise need not state $x \neq -1/2$. In fact, the author could state $x \in \mathbb{R}$ without commiting any formal error. The excercise would then be to find all $x \in \mathbb{R}$ satisfying the equation. That the left-hand side is undefined for $x = -1/2$ does not cause any trouble: this just implies that $x = -1/2$ is not part of the answer to the exercise.

So if we consider the above exercise, it is not even a predicate if we do not also state that $x \neq −1/2$.

If you by predicate mean a formula $P(x)$ in the language of some theory $T$ (of arithmetic), this is not true. It need not be a conjunct of the the formula $P(x)$ that $x \neq -1/2$ for us to have the equivalence:

$T \vdash P(a) \Longleftrightarrow a$ satisfies the equation.

Answer 5

Equations are predicates and can thus define sets. Given a universe, say $\mathbb R$, an equation over a variable $x$ defines a predicate $P(x)$.

The given equation then defines the set

$$S = \{x : x \in \mathbb R \land P(x)\}$$

To solve the equation now means to give a more "useful" representation of this set, such that for example the members can be enumerated.

As for your question: We do not actually have any function symbols in set theory (just sets of pairs with special properties), so e.g. the equation or predicate $P(x) := $ $$x / 0 = 1/2$$ actually just abbreviates $$\exists z (\; ((x,0),z) \in {/} \land z = 1/2\;).$$ In this case we can show that $S = \emptyset$ since the relation $/$ does not contain any members of the form $((\cdot, 0), \cdot)$.

Understood in this way, once a universe is given the equation has a clear meaning. We do not need to be given further restrictions to e.g. make the functions occurring in the equation defined as you were suggesting.