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I have recently been relearning algebra, so that I can have a better foundation for studying set theory and other more advanced topics. My true interest is in philosophy of mathematics. I have had many questions about mathematics over the years that have never been answered. In particular, here's one that has been bothering me recently:

Does the algebraic manipulation of an equation generate a new equation, or is it merely the same equation with a different visible form? For algebra to "work," it would seem to have to be the same equation, just with a different form. Yet there are examples of valid algebraic manipulation which do indeed change the actual equation completely. If I multiply both sides of a linear equation by x, then I generate a quadratic equation. Also, in solving a rational equation, we eliminate the denominators from the equation by multiplying by a common denominator. Are we still dealing with the same rational equation after doing this procedure? If I were to take a snapshot of the equation after eliminating the denominator, and present this picture to somebody, and ask them if this were a rational equation, they would say of course not.

I guess here I am employing a distinction between the form a thing takes, and what that actual thing is, ontologically. Two equations can have different forms, but can be identical ontologically. Liquid water and ice are both H2O, for example, but have different visible forms.

Thanks, Kyle

Kyle V
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  • We first need to define what "different" means for an equation. Your example was, that we have, say, $x+1=0$ and you want to multiply by $x$ to obtain $x^2+x=0$. This is not an equivalent equation, because we multiplied with a new equation $x=0$ and created new solutions. The theory of linear algebra settles this question for linear equations. – Dietrich Burde Mar 22 '20 at 15:00
  • Note: $x=y\implies x^2=y^2$, but if $x^2=y^2$ it could be $x=-y$ – J. W. Tanner Mar 22 '20 at 15:01
  • When you manipulate equations, there is an equality between {set of solutions to equation 1} and {set of solutions to equation 2}, where equation 2 has been transformed from equation 1 by a "legal operation". Note that one operation might be legal in one set, for example the set of real numbers, but not always legal in another set, for example taking a square root in the set of rationals or taking an inverse in the set of matrices. – MasB Mar 22 '20 at 15:22
  • I suppose different for me means non-equivalent. If we graph the equation, they will have different graphs. But the heart of my question is this: if some algebraic manipulations - such as multiplying by x in your example - generate non-equivalent equations, then how do we determine which manipulations are valid and which are invalid in trying to find solutions to a given equation? – Kyle V Mar 22 '20 at 15:23
  • then how do we determine which manipulations are valid and which are invalid --- The valid operations are those which preserve equality. Think of the operations as functions you apply to both sides of the equation. For example, to subtract $6$ from both sides you apply the function $f(x) = x - 6$ to both sides. To square both sides, you apply the function $f(x) = x^2$ to both sides. It's not clear what function corresponds to multiplying both sides by $x,$ but for now let's ignore this. Then manipulations that consist of applying (continued) – Dave L. Renfro Mar 22 '20 at 16:35
  • one-to-one functions to both sides of an equation are always valid in your sense, this being a straightforward consequence of their defining property "$a = b$ if and only if $f(a) = f(b)$". Conversely, manipulations that consist of applying non-one-to-one functions to both sides of an equation can be invalid. They are not always invalid, because it can happen that the choices of $a$ and $b$ relevant for the equation still satisfy the "one-to-one" property, an example being applying the function $f(x)=x^2$ to two always positive quantities. – Dave L. Renfro Mar 22 '20 at 16:45
  • Related, possibly helpful: https://math.stackexchange.com/questions/2738360/what-exactly-is-an-equation/2738382#2738382 – Ethan Bolker Mar 25 '20 at 20:43

2 Answers2

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This is just a particular instance of the more general distinction between names and referents: e.g. "Kurt Godel" and "The person who first proved the the consistency of CH" are different names with the same referent. So the short answer to your question is:

As strings of symbols, "$x+1=2$" and "$x=1$" are clearly different. However, they have the same referent. And the relation between name and referent here is no more - or less - complicated than it is in natural language.

EDIT: My interpretation of equations as names as opposed to statements is not unobjectionable - see the comments below - but I do think it's ultimately correct here (a sentence is really just a name for its meaning, after all :P).

That said, it's worth saying a bit about what makes this stand out from general natural language issues (and why this question may be appropriate here as opposed to only philosophy.stackexchange):

There are a couple of important points to make here, one common to mathematics and natural language and the other fairly specific to mathematics. I'll start with the commonality.

  • Just like with natural language, it's often quite hard to pin down what sort of thing the referent of a mathematical expression actually is.

Take "$x^4=12$." The most obvious way to interpret this string of symbols as referring to some object is via solution sets: "$x^4=12$" is a name for the set of all numbers whose fourth power is twelve. However, there's a serious issue here: what's the context? $x^4=12$ defines one solution set over $\mathbb{R}$ and another one over $\mathbb{C}$ (and yet another one over $\mathbb{Q}$, and so on). Moreover, it's often mathematically useful to think of "$x^4=12$" as deliberately ambiguous (we sometimes want to consider the "same" polynomial in different fields).

Now this isn't unique to mathematics. For example, "the president" is similarly domain-vague (president of what?). But where we see some arguably unique behavior is when we look at the interaction between this kind of vagueness and the second important point:

  • Unlike with natural language in general, we have explicit rules for manipulating mathematical names which are given axiomatically as opposed to inferred through experience.

All of language is of course subject to a general "usage/meaning feedback loop," but in mathematics this axiomatic structure leads to a level of clarity (initiallly, at least) which as far as I can tell is unique. Somehow, in mathematics - when we really "look under the hood" - we manage to coherently precisely manipulate vagueries. This is really quite neat!

Noah Schweber
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  • In case an equation were a name ( or a definite description, " the so and so")equations would not be sentences and could not be said logically equivalent ($\equiv$), but should be said identical ( $=$). Ordinarily, logical equivalence does not hold for names, but for sentences ( or open sentences). –  Mar 25 '20 at 20:37
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    @RayLittleRock That distinction between names and sentences is extremely fuzzy, though: a sentence is a name for its meaning :). Whether we think of an equation as primarily an assertion or primarily a description varies from context to context, and in my opinion when we really look under the hood the latter becomes by far more important (and this carries through to logic too; e.g. in my opinion the right way of thinking about a sentence of first-order logic is as a name for its class of models). The $=/\equiv$ distinction from this perspective just highlighting important context distinctions. – Noah Schweber Mar 25 '20 at 20:44
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    That said, I've edited to explicitly point this out. – Noah Schweber Mar 25 '20 at 20:47
  • That makes sense and is consistent with Frege's idea acccording to which a sentence denotes its truth value ( is a name for its truth value). –  Mar 25 '20 at 20:48
  • Noah, thank you for the detailed answer. But my question for you is this: I will admit that "$x+1=2$" and "$x=1$" have the same referent. But is this the same as saying that they are equivalent expressions? To put it another way, are you saying that "$x+1=2$" ≡ "$x=1$", where "≡" means biconditional in propositional logic? Are all algebraic manipulations biconditionals? I think not. Multiplying "$x+1=0$" by $x$ $\supset$ "$x^2+x=0$". The later does not imply the former. It is only a material conditional. But if it's only a material conditional, then they don't have the same referent. – Kyle V Mar 29 '20 at 14:28
  • @KyleV Algebraic manipulations don't all preserve referents. That's what's going on here: "$x+1=0$" and "$x^2+x=0$" don't have the same referent (that is, their solution sets are different). Having the same referent is indeed the same as being equivalent expressions, but algebraic manipulations don't generally preserve referents. Put another way, the answers to "is [having the same referent] the same as [being] equivalent expressions?" and "Are all algebraic manipulations biconditionals?" are yes and no, respectively - they're essentially unrelated. – Noah Schweber Mar 29 '20 at 14:59
  • Note that "manipulating mathematical names" is broader than "manipulating mathematical names in referent-preserving ways." For example, one such rule is "multiplying both sides of an equation by the same term results in an at-least-as-large solution set." Here the point is that the collection of possible referents has some structure (namely, we can talk about set inclusion), and the name manipulation we're looking at interacts with that structure in a certain way. So that's not a counterexample to my final point, just an elaboration. – Noah Schweber Mar 29 '20 at 15:06
  • @Noah Schweber Thank you so much for the reply. This is an enormous help. You've answered all my questions. – Kyle V Mar 30 '20 at 15:11
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Suppose you have a quadratic equation with a negative power

$$n^2-4+\frac{6}{n}=0$$ and you want to solve it with the quadratic equation $$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

as it turns out, this is really a cubic equation with a different face $$n^3-4n+6=0$$

which can be solved.

This is just an example but I think it makes the point that manipulation does not change the basic equation, just the view.

poetasis
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