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How the distinction between problems (find/describe such values of x that… ) and universal truths (identities) is taught to secondary-school students and higher? Especially in English-speaking countries.

Ī mean the following thing (sorry if Ī present an obvious stuff in a confused way): if we have a relation, equality or else, and a proposition with this relation at the top and some variable(s), then we have two distinct useful cases, although notation might be the same. In the first case the proposition is not necessarily true and we have to find values of the variable where it is true. In the second case the proposition is necessarily true (law/identity/“theorem”) and variables become bound by universal quantification. Universal truths can be used in subsequent equivalent transformations of expressions, equations, inequalities, or something alike. Students must learn to distinguish these cases, haven’t they? Examples:

Binary relations:         “=”  (equals)     “>” (greater than)

   Problems:   “$x^2 - x - 1 = 0$” (equation)   “$x^2 - x - 1 > 0$”

Universal truths:   “$(a+b)(a-b) = a^2 - b^2$”    “$\exp(x) > 0$”

How may be used: “α² = δ² ⇒ α = δ ∨ α = −δ ”      exp(α)⋅ℓ > 0 ⇔ ℓ > 0
          (transformed equation)       (transformed inequality (problem))
          “(α + 1)(α − 1) = α² − 1 ”       cosh α > 0
          (transformed expression)      (derived inequality (truth))

Why am Ī preoccupied with it? A year ago, the only person happened to cooperate with me about precise equation–identity distinction in English Wikipedia was a mathematician from France; was a coincidence he wasn’t taught in English? Today Ī argued on this topic (against two presumedly native English speakers) at a neighbouring site and become frustrated. Could an English speaker be correct in insisting that something like an+1 = a × an is an “equation” and calling the “=” symbol in it “equation sign”?

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  • $\begingroup$ I do see this as a very valid question, but I think you are not presenting it correctly. The problem is not human language, the problem is the looseness in mathematical language. I answered assuming this was for general education which I think is highly inappropriate. However, this is very appropriate to mathematical notations which affect both mathematicians and computer scientists. I have always found mathematical proofs to be too loose for my liking--there was never any overarching structure enforced. This is a problem for computer scientists trying to study theorem proving. $\endgroup$ – Jared Oct 30 '14 at 8:19
  • $\begingroup$ And in an even broader sense I have seen that there is a wide range of notations across the mathematical disciplines (including mathematics, physics, computer science, etc.). This includes not only mathematical notations but even terms used. This is a real problem because it makes it difficult to communicate between the disciplines. $\endgroup$ – Jared Oct 30 '14 at 8:22
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I am comfortable saying "Solve the equation $x+2$=4" and also saying "Using the equation $(a+b)(a-b)=a^2-b^2$, we see that...".

On other other hand I would only ever speak of solving an equation, not an identity, and I would be willing to use the word "identity" in my second example above.

So I think I would say that an identity is a kind of equation, one which is universally quantified.

In general, one can avoid confusion jut by saying what you mean, as in "Solve the equation $x+2=4$", or "For all numbers $a$ and $b$, $(a+b)(a-b)=a^2-b^2$"

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  • $\begingroup$ Of course, boys and girls can be told about what does the teacher mean in each particular case. But how do they learn to understand the difference when they see a formula on a blackboard? $\endgroup$ – Incnis Mrsi Oct 25 '14 at 10:21
  • $\begingroup$ @IncnisMrsi What exactly are you hoping for? I think, in general, people pick these things up from context, in exactly the same way they learn to distinguish the two uses of the word "lead" in "The lead developer on the project" and "The alchemist had finally succeeded in turning lead into gold". Also, I think generally one should write "for each" on the board where it is appropriate. Sometimes the extra symbols $\equiv$ or $:=$ can be helpful as well. $\endgroup$ – Steven Gubkin Oct 25 '14 at 14:30
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    $\begingroup$ @IncnisMrsi I really do not appreciate this level of hostility. You asked a question, and I am telling you that, as far as I know, there is no common English language distinction between "equality for a particular $x$" and "Equality for all $x$", unless you explicitly state that somehow. If a student is given $x^2 = 4$ on a homework assignment, none of them will say "False: $1^2 \neq 4$ is a counterexample". Equality does have a meaning all on its own, after all. $\endgroup$ – Steven Gubkin Oct 25 '14 at 15:15
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    $\begingroup$ I would say that an "equation" is any statement expressing equality of two expressions. Both $x^2=1$ and $x(x)=x^2$ are equations, but they are (generally) quantified differently. An identity is a specific kind of equation which is generally universally quantified somehow. So $x(x)=x^2$ is an identity. So it is okay to call $x(x)=x^2$ both an equation and an identity. $\endgroup$ – Steven Gubkin Oct 25 '14 at 15:23
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    $\begingroup$ Interesting. I'm from Poland, and Polish for "equation" (in the sense: something you solve for x) is "równanie". Not only I'd never use this word for an identity, but if a student of mine called e.g. $a^2-b^2=(a-b)(a+b)$ a "równanie", I would correct his erroneous usage of the word. (And I usually spend some time with first year students explaining how the innocently-looking equality sign can mean different things.) $\endgroup$ – mbork Oct 28 '14 at 20:19
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In my experience (being taught in a US school), there is no explicit distinction between an identity and an equation. The reason being is that to make the distinction you have to introduce first order logic which is not generally appropriate for secondary (high school) mathematics education. Students can usually ascertain an identity (a formula) from an equation to be solved so this is not usually an issue. The students that have trouble using formulas or have trouble solving equations are unlikely to benefit from a more detailed explanation (the detailed explanation will likely only further confuse them and make the topic of mathematics far less tangible and far more abstract).

Examples:

  1. Quadratic equation.

We are generally taught that given $ax^2 + bx + c = 0$ such that $a \neq 0$ that $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. This is an identity. It's a first order equation. Formally it would state the following:

$$ \forall x \in \mathbb{C}.\forall a, b, c\in \mathbb{R}:\left( \left(a\neq 0\wedge ax^2 + bx + c = 0\right) \rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\right) $$

  1. Solving an equation:

Solve the following equation: $10x^2 + 6x + 8 = 0$ wher $x$ must be a real number.

The very first thing to do when solving an equation is to determine whether or not it can be solved. This too is a first order logic equation. The question (which must be answered) is whether or not the following is true:

$$ \exists x\in \mathbb{R}: 10x^2 + 6x + 8 = 0 $$

In this particular instance, the answer would be that the above is false therefore there are no solutions. But how do you know that? How do you prove that?

Here is a different example: Solve the following: $x^2 + 6x + 8 = 0$. Again, we must first decide whether or not the following first order equation is true or false:

$$ \exists x\in \mathbb{R}: x^2 + 6x + 8 = 0 $$

In this case it is true (how do we know?). Once we know the equation can be satisfied, we next attempt to find the particular solutions. This is not easy and in fact, "impossible" in general. Here is an example:

Find real values of $x$ that satisfy the following equation

$$ e^x = x + 2 $$

It can be proved that a solution exists:

$$ f(x) = e^x - x - 2 \rightarrow \text{find } f(x) = 0 \\ f'(x) = e^x - 1 \rightarrow f'(x) = 0 \rightarrow e^x = 1 \rightarrow x = \ln(1) = 0 $$

Since the derivative changes from negative to positive at $x = 0$, this represents a relative minimum (and since this is the only extrema this is the global minimum). $f(0) = 1 - 0 - 2 = -1$. Further since $\lim_{x\rightarrow-\infty} e^x = 0$ and $\lim_{x\rightarrow-\infty} = -\infty$ (thus subtracting negative infinity creates a positive infinity) and that $e^x$ dominates $-x$ such that $\lim_{x\rightarrow\infty} e^x - x = +\infty$, we know that this function, $f(x)$, goes towards $+\infty$ on either side, there must be two spots where $f(x) = 0$ and therefore there are exactly two real solutions to $e^x = x + 2$.

Yet even though we know the equation to be satisfiable, it is not clear how (or analytically possible) to solve for those two values.

So when solving equations there are two separate issues that we usually do not identify in secondary education (largely because it's beyond the scope). The first is, whether or not there is a solution and the second is how to find the solution once you can show that one exists--note that the first step becomes tedious to the point of being prohibitive towards learning for the vast majority of problems encountered in elementary to high school level mathematics courses.

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  • $\begingroup$ «⋯ = 1 − 0 − 2 = −2 ». ORLY? $\endgroup$ – Incnis Mrsi Oct 27 '14 at 12:15
  • $\begingroup$ By the way, to prove no more than two real solutions you have to make some more analysis beyond differentiability and limits. $\endgroup$ – Incnis Mrsi Oct 27 '14 at 12:27
  • $\begingroup$ @IncnisMrsi "By the way, to prove no more than two real solutions you have to make some more analysis beyond differentiability and limits." If the function is continuous over all of the reals (which it is) and has only a single (real) critical point (which it does) and we know that the function ends up at $+ \infty$ on both sides, then the fact that its absolute minimum is $-2$ means that it must cross the x-axis to the left and to the right of that critical point. Further, since there are no other (real) critical points, it could not possibly turn back around towards the x-axis. $\endgroup$ – Jared Oct 27 '14 at 17:32
  • $\begingroup$ The fact that its absolute minimum is −2 means that it’s easier to count extrema of a function than perform two arithmetic operations on integers ☺ $\endgroup$ – Incnis Mrsi Oct 27 '14 at 18:51
  • $\begingroup$ The mistake in elementary arithmetic was not fixed for two days, so author’s deviation from the correct value of $f(0)$ is added to the score of the posting. $\endgroup$ – Incnis Mrsi Oct 30 '14 at 6:10

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