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This is not a "Haha gotchu mathematicians!" question. I'm seriously trying to learn. Please don't take my question to be insincere.

IMPORTANT EDIT: I've noticed that people are focussing too much on my examples that they're distracted by my real question which is logic based. I use various examples to explain my "logic based" confusion, I'm not confused about the examples. I do not require explanations for my examples.


This question stems from a question I asked long long long time ago and someone answered that it is incorrect to write $|x|=±x$ and I took their word for it because well... I wasn't experienced enough to ask the right questions.

Since then my instinct to deal with $|x|$ has been to use: $$ |x|= \begin{cases} x, \ x≥0 \\\\ -x, \ x <0 \end{cases}$$

because well... that is the definition of $|x|$.

However I was going through my questions yesterday, when I realised, "Wait hold on, why is the equation $|x|=±x$ incorrect? Because "$±$" means "plus OR minus", you're not insisting that $|x|$ is $x$ AND $-x$. You're only saying it is either $x$ OR $-x$.


But hold on, there's more. With that argument in mind, you can always write $\sqrt{9}=±3$ even though it's just $3$. You can even go more bonkers with this logic by writing $$\sqrt{9} = 3 \text{ or } -3 \text{ or } -193e^2$$ as long as one of them is true. You get the point, right?

You can keep adding on nonsense using the fact that $T \equiv T \vee F$ like so:

$$( \sin x = 0) \equiv (\sin x =0 \text{ or } \cos x = 0) $$ and then get $x \in \{\frac{nπ}{2} : n \in \mathbb{Z}\}$ as the solution which is absurd.


Question: Where is the logical error here?

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    $\begingroup$ When using the symbol $\sqrt{}$, you have to define what it means. Commonly, it is defined as the function $\sqrt : \quad \mathbb R \to [0, +\infty)$, $x \, \mapsto \sqrt{x} = \{ y\in [0, +\infty) \,\text{ s. t. } \,y^2 = x \} $. One can define $\sqrt{}$ differently, but then $\sqrt 9 = \pm 3$ wouldn't be true for both results. $\endgroup$ Aug 22, 2021 at 11:13
  • $\begingroup$ @MiguelMars I know that. But I request that you read my question through to the end. That's not my question. I know that. My question is supposed to be about "logic". $\endgroup$
    – William
    Aug 22, 2021 at 11:15
  • $\begingroup$ I think it is better to interpret $y = \pm a$ as $y \in \{-a, a\}$ which eliminates the possibility of adding nonsense. $\endgroup$ Aug 22, 2021 at 11:19
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    $\begingroup$ I think people are focussing too much on the $|x| = \pm x$ that they're not recognising my question is logic based really. Not calculus. I'd be grateful if someone could recommend suggestions to improve my question. $\endgroup$
    – William
    Aug 22, 2021 at 11:19
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    $\begingroup$ Ok, I understand now the question. I think in Mathematics, we use the equality symbol in a way that $x = \{\text{Set of solutions}\}$ means that $ x = s $ is true for every $s$ in the set of solutions. $\endgroup$ Aug 22, 2021 at 11:29

11 Answers 11

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I will focus on your problem with $T\iff T\lor F$.

Sure, under assumption that some formula $A(x)$ depending on a parameter $x$ is true (independent of the value for $x$), you may use that to deduce that $A(x)\iff A(x)\lor B(x)$ (since $T\iff T$).

However, in your examples, the formula $A(x)$ only holds true for some special values of $x$. It could thus happen that for some values of $x$, $A(x)$ fails to hold, while $B(x)$ is true. In this case, you can clearly not write $A(x)\iff A(x)\lor B(x)$ (since $F$ is not equivalent to $T$).

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    $\begingroup$ Finally a sensible answer. $\endgroup$ Aug 22, 2021 at 12:07
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    $\begingroup$ This is such an important answer. 'paramterized truths' it all makes sense now. $\endgroup$ Aug 22, 2021 at 12:12
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    $\begingroup$ @William Sure! As commented by others, one just needs to make sure that people don't misinterprete it. $\endgroup$
    – Zuy
    Aug 22, 2021 at 12:42
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    $\begingroup$ @Zuy Thanks! I guess this answer is going to be it. I'll probably accept this answer but I'll wait a while. "One just needs to make sure that people don't misinterpret it", I honestly can't see what could be the misinterpretation of that equation. It does pretty much what it says. Care to enlighten? $\endgroup$
    – William
    Aug 22, 2021 at 14:57
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    $\begingroup$ @William What I actually wanted to say is that writing $|x|=\pm x$ (that is, $|x|=x \lor |x|=-x$) is logically correct, but of course doesn't tell the whole story (for which $x$ is it $+x$? $-x$?). It's similar to saying that $-2\leq\sin x\leq 2$ for real $x$: this statement is true, but could be made more precise. $\endgroup$
    – Zuy
    Aug 22, 2021 at 16:16
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  1. $$f(x)=|x|\tag1$$ can be translated as $$\Big(x<0\implies f(x)=-x\Big)\quad\text{and}\quad \Big(x\geq 0\implies f(x)=x\Big),$$ whereas $$f(x)=\pm x\tag2$$ can be translated as $$ x\in\mathbb R\implies\Big(f(x)=-x\quad\text{or}\quad f(x)=x\Big).$$ Statement $(1)$ fully describes the function (in which each input has a certain output) $f,$ whereas statement $(2)$ conveys only partial information about it.

  2. \begin{align} &\forall x{\in}\mathbb R&\bigg(\,f(x)=|x|&\implies f(x)=\pm x\,\bigg)\\ &\forall x{\in}\mathbb R{\setminus}\{0\}&\bigg(\,f(x)=\pm x&\;\kern.6em\not\kern-.6em\implies f(x)=|x|\,\bigg)\\ &\forall (x,y){\in}\mathbb R^2 &\bigg(|x|=|y|&\iff x=\pm y\bigg) \end{align}

  3. Therefore, \begin{align}|x| &:= \begin{cases}-x &\text{ if }x<0; \\x &\text{ if }x\geq0\tag A\end{cases} \\\\ |x| &:\not=\;\pm x \\\\ |x|&=\;\pm x\tag B \\\\ \pm x&=\;|x|.\tag C\end{align} Definition $(\text{A})$ fully specifies $|x|,$ while statement $(\text{B})$ means $$|x|=-x\quad\text{or}\quad |x|=x.$$ So, $\pm x$ is less informative than $|x|.$ Thus, $$|x|=\pm x$$ is not an identity!

    Notice that, unlike statement $(\text{B}),$ statement $(\text{C})$ never gets conflated with definition $(\text{A}).$ This is because the statement feel less definitive when the properties/possibilities of the subject $|x|$ is displayed before the subject itself.

  4. Compare:

    • $$\begin{align}&\lvert2x\rvert=x-1\\\iff&\bigg(x<0 \;\text{ and }-2x=x-1\bigg) \:\text{ or }\: \bigg(x\geq0 \;\text{ and }\; 2x=x-1\bigg)\\\iff&\bigg(x<0 \;\text{ and }\; x=\frac13\bigg) \:\text{ or }\: \bigg(x\geq0 \;\text{ and }\; x=-1\bigg)\\\iff& x\in\emptyset\end{align}$$
    • $$\begin{align}&\lvert2x\rvert=x-1\\\iff&\pm2x=x-1 \;\text{ and }\; x-1\geq0\\\iff& x\in\emptyset\end{align}$$
    • $$\begin{align}&\color{red}{\lvert2x\rvert}=x-1\\\color{red}{\implies}&\color{red}{\pm2x}=x-1\\\iff& x\in\left\{-1,\frac13\right\}.\end{align}$$ (Here, both solutions being extraneous is due to the equation being inconsistent.)
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  • $\begingroup$ Aren't (B) and (C) the exact same thing? $\endgroup$ Jan 25 at 4:37
  • $\begingroup$ @tryingtobeastoic Yes, and the point is that because $(B)$ seems more declarative than $(C),$ it is easier for a human to carelessly conflate $(B)$ and $(A)$ (as both the OP above and yourself did) whereas one is less likely to likewise misinterpret $(C).$ $\endgroup$
    – ryang
    Jan 25 at 6:25
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The term $\pm$ (or sometimes $\mp$), like any notation, carries a lot of meaning and connotations, because we see certain notations in certain places. To use $\pm$ is to say that the choice of either plus or minus should make sense, and perhaps depending on further context one is preferred over the other. To say $|x|$ is to unambiguously say that the quantity is positive and equal in absolute value to $x$.

$|x|\neq\pm x$ because the left hand side is a uniquely determined positive value, whereas $\pm$ is an ambiguous statement suggesting both states of plus or minus are valid unless further context to the question says otherwise. If you are doing algebraic manipulation, and you write $\pm x$ instead of $|x|$, you will find yourself in a nightmare of superimposed states, where you must deal with both cases of plus and minus and as such it is an inferior notation to $|x|$. There is neither reason nor motivation to write $|x|$ as $\pm x$, and $|x|$ is always only one value, is always unique; $\pm x$ is not.

You say one can go bonkers with this logic, saying $\sqrt{9}=3\vee -193e^2$. There is a good quote from somewhere, I don't remember exactly where, saying that a good notation frees the mind to focus on the problem at hand. $\sqrt{9}=3\vee -193e^2$ is not a good use of the $\vee$ notation, and is indeed bonkers as you say. I don't think it is so much a logical error but more a semantic error: logical conjunctions like "or" generally signal that either state is possible, and in any further working or proof we must account for all the states. You can chain "or"s and other conjunctions, and the point of doing this is to logically determine one or more solutions to whatever problem you're facing - introducing absurdities serves absolutely no purpose. It is not correct to write $3\vee-193e^2$ because to take mathematical notation to such a highly pedantic level is to undermine the purpose of notation in the first place, and it will trip you up to write like that if you ever write a proof or work on a harder problem, because littering the working with absurdities and ambiguous notation is not how we do maths.

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The logical flaw in your thought, is that you thought that $T \equiv T \lor F$ implies that $\forall x [A(x) \equiv A(x) \lor B(x)]$; and this is incorrect. This flaw made you think that the set for solutions of $A(x)$ which is $\{x: A(x)\}$ is the set union of itself with $\{x: B(x)\}$ and of course caused the absurdity you referred to.

The correct implication is:

$(T\equiv T \lor F) \implies \forall x [A(x) \implies (A(x) \iff A(x) \lor B(x))]$

This shows that the set of solutions for $A(x)$ is $\{x:A(x)\}$ because any object substituting $x$ would only be a solution if it satisfies $A(x)$, now if it satisfies $B(x)$ but not satisfy $A(x)$ then we'll have $F \equiv F \lor T$ which is not true and so not a solution, and of course if it neither satisfy $A(x)$ nor $B(x)$ then it is not a solution. Of course if $x$ satisfies $A(x)$ then we get $T \equiv T \lor F$ always, so it is always a solution despite whether $B(x)$ is true or not. So logically speaking we have the set of solutions equal to $$\{x: A(x)\} \cap \{x: B(x)\} \cup \{x:A(x)\}$$ which is $\{x:A(x)\}$ itself.

I think this is the answer to your logical question which is basically Zuy's answer.

However, your question is also interesting in another sense, that of proper symbolism of relations and functions. And although this was not your question, but its related to it in some sense. To answer that aspect I'd say that everything depends on how do we define the expressions we write. For instance $F(X)= \pm Y$ is a doubious expression, it's confusing, because generally the expression $F(X)=Y$ is reserved for when $F$ is a one place function symbol, so it assigns ONE value $Y$ to each argument $X$, while $\pm Y$ is usually taken to denote two distinct values that are $+Y$ and $-Y$, so this will supply the impression that the expression $F(X)= \pm Y$ means that $F$ is some relation symbol that sends $X$ to $+Y$ and also sends $X$ to $-Y$, i.e. a One-to-Many relation, but this is confusing because of the use of $=$ which entails that $F$ must be a function. To properly write matters in a logical language one must first see how to write formulas using $\sf One-to-Many$ relation symbols, one better avoid using the equality $=$ symbol, so we better write $F(X,Y)$, so doing that we can for example define:

$ F(X, \pm Y) \iff [F(X,+Y) \land F(X,-Y)]$

So for example if we intend to use the symbol $\sqrt \ $ to designate the converse relation of the square function, then we better write it this way:

$\sqrt \ \ {}(X, \pm Y) \iff [(+Y)^2=X \land (-Y)^2=X] $

However, we can abuse notation and insert the equality symbol and define:

$\sqrt x = \pm y \iff [(+y)^2=x \land (-y)^2=x] $

Which is a wrong way of writting matters because logically it would be read as $\sqrt x = + y \land \sqrt x = -y$ and this clearly leads to $+y = -y$ by identity theory, which is contradictory! But unfortunately this is the intended meaning many times.

Now if we intended for example for the square rooting to be a function, so we can fix one value to each argument, like by arranging it to be the positive value only, so this way $\sqrt 9 =3 \land \sqrt 9 \neq -3$, but if we stick to this interpretation, then $\sqrt 9 = \pm 3$ is a FALSE statement!.

The OP is under the impression that $F(X) = \pm Y$ is a disjunctive expression, to mean that $F(X)=+Y \lor F(X)= - Y$, and this is a wrong capture, the reason is because it would lead to the explosion he is alluding to, so I can for example write $\sqrt 9= \pm 3; 8^2 = \pm 64 , 6/2 = \pm 3, 1+3=\pm 4, |3|=\pm 3,...$ and all would indeed be logically valid under that interpretation, and clearly this is a redundant symbolism, and so it is not the correct capture of what's going on. Yet, if you insist on interpreting it this way, then there is no logical flaw with it, as long as one of the values is true, then you can add as much other values you want and the result remains logically valid, but the possible expressions under that interpretation would of course be REDUNDANT. So you can abuse $T \equiv T \lor F$ as much as you want, there is no logical flaw about it, but when you use this to define expressions, then you'ed expect an explosion of redundant expressions. [Be aware that this abuse must not lead you to the logical flaw about the set of solutions that was pointed out in the begining of this answer]

The reality of the matter is that $F(X)= \pm Y$ is an incorrect symbolism, it is an abuse of notation, and should be just taken in an informal sense to mean that $F(X,+Y)$ is true and $F(X,-Y)$ is true also. The proper way of writing it is as $F(X, \pm Y)$, and the proper definition of it was given above.

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The reason $\lvert x \rvert \neq \pm x$ is that the left hand side, $\lvert x \rvert$ is a function, therefore it produces only a single value for a given argument. On the other hand, $\pm x$ is a $set$ of possible values. Single-valued and multivalued maps are fundamentally different kinds of quantities, so equating them is meaningless.

Here's what I mean, graphically. This is a plot of $y= \lvert x \rvert$,

Whereas this is $y = \pm x$,

(Notice how 'or' translates to superimposition of the different possibilities).

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  • $\begingroup$ This is what I had in mind as well. $x$ or $-x$ represents an entire set of redundant solutions whatever it is below the X axis, (just like the example I have given below with sines and cosines). However, what would you say is the logical error in writing $|x| = \pm x$. (You've only focussed on the first part of the question.) $\endgroup$
    – William
    Aug 22, 2021 at 11:08
  • $\begingroup$ The logical error, I think, would be that $\lvert x \rvert = \pm x$ does not contain enough information to describe the usual piecewise modulus function. Rather, the former equation just puts all the possible outputs into one statement without specifying the finer details, such as what the function looks like in different parts of its domain (which brings us back to the piecewise definition). It is simply a matter of lack of information, if one wishes to see the problem that way. $\endgroup$ Aug 23, 2021 at 15:11
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You are right to state that symbols do not have meaning by themselves without context. When we write $\sqrt{9}=\pm 3$, what we are saying is all this: the equation $x^2-9=0$ has exactly two solutions $x=3$ and $x=-3$. So I would say that it is not exactly that $x=3$ or $x=-3$, that will depend on how $x$ is defined or computed elsewhere.

This is a different usage to writing $|x|=\pm x$ meaning $x$ or $-x$. In fact, I do think you could use it (even incorrect) to consider both possibilities in a single equation, for instance if the sign does not change the final result, e.g. $|x|^2=(\pm x)^2=x^2$. It does not seem very recommendable but I think everyone would understand it. Note that in this case the possibilities are either $x$ or $-x$ but not anything else, so it is not exactly a logical OR, just a shorthand for two equations.

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    $\begingroup$ I think you have missed the point. As the OP knows, $\sqrt 9$ is equal to $+3$; the OP's question is more along the lines of "Is it correct to write $1=\pm 1$, because $1$ is equal to one of the two alternatives $+1$ and $-1$?" $\endgroup$
    – TonyK
    Aug 22, 2021 at 10:57
  • $\begingroup$ @TonyK Yes, thank you. $\endgroup$
    – William
    Aug 22, 2021 at 11:03
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What I think is being understood is the difference between a set and a statement with a truth value. Let me try and explain.

We'll take OP's example of $\sin x=0$ and run with it. What $\sin x=0$ actually tells us is the set of all values of $x$ (in a given domain) such that the statement holds. It has no inherent truth value, it's just a set (of solutions). But, once a $x$ is given, the statement $(\text{For the given }x,\ \sin x=0)$ indeed has a truth value. And here we see what the OP has a confusion in. OP's statement $$(\sin x=0)\equiv (\sin x=0\ \lor \cos x=0)$$ is trying to establish an equivalence between sets, and the logical or is actually just union, and the OP clearly establishes that this equivalence is not true. But what is true is that the following statements can be compared as logical statements : Given an $x$ whether:

$$(\text{For the given }x,\ \sin x=0)\equiv (\text{For the given }x,(\ \sin x=0\ \lor\cos x=0))$$

(Thanks to Hans Lundmark for telling me that I wrote something wrong. I was thinking something else, and wrote something else haha)

Edit : Let me actually add another clarification to a problem OP mentioned in the question, and we'll see how both of them connects.

It has to do with the fact that OP writes $\sqrt 9=3\ \lor -3\ \lor -139e^2$. From a logical point of view, this is absolutely correct, because it's really 3 statements written together :

$$(\sqrt 9=3)\ \lor (\sqrt 9=-3)\ \lor (\sqrt 3=-139e^2)$$

and the first 2 statements are indeed true and hence makes the whole statement true. Being ruthless, we can literally add everything and say $$(\sqrt 9=3)\ \lor (\sqrt 9=-3)\ \lor (\sqrt 9=x,\ x\in\mathbb R\setminus\{3,-3\})$$ This statement is completely true because the first two components are true. This is a problem with "finding the correct solution given a set of statements connected by logical or". Because, if only 1 component is true, the whole statement becomes true, but that means there could be false components hidden in the statement.

An example where this happens is when we find solutions by squaring. Extraneous solutions creep in which doesn't solve the original equation, but the whole solution set as considered in the above fashion is indeed correct (try squaring both sides of $x^2=9$ and see that one has 2 extraneous solutions which doesn't solve this). Hence we are asked to check if indeed everything in the solution set is true, or there's imposters lurking in it (in our case, it would be everything that is not 3 or -3).

tl;dr : The statement OP wrote about $\sqrt 9$ is logically true, but since these are statements connected by logical or, one needs to check which of the solutions are actual solutions and which are not (because throwing in one solution makes the whole statement true).

In the case of the trigonometric statements, we again have 2 statements connected by logical or, and hence one needs to check which component is true and which is not for particular values of $x$.

Hope this helps in seeing where the problem lies, and the OP can transfer the arguments for the other cases as well.

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  • $\begingroup$ I think you're the first person here who has actually tried to understand my question and form an answer around that. This answer is very helpful actually. I'll need some time to think about it and then I'll respond. Thank you. $\endgroup$
    – William
    Aug 22, 2021 at 11:43
  • $\begingroup$ Glad to be helpful! Just let me know if you need any clarification. $\endgroup$ Aug 22, 2021 at 11:44
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    $\begingroup$ But what if the given $x$ is $x=\pi/2$? Dont you get F on the left and T on the right then? $\endgroup$ Aug 22, 2021 at 11:47
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    $\begingroup$ @YuzurihaInori: Thank you for elevating me to the status of Landmark. :-) But I'm not sure I understand what you are trying to say with the edited statement either... $\endgroup$ Aug 22, 2021 at 12:47
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    $\begingroup$ Oh man I laughed too much at your comment @HansLundmark 😂😂 $\endgroup$ Aug 22, 2021 at 12:52
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There is nothing "abusive" about it.

If $A$ is true, then $A \leftrightarrow A\lor B$ is true, regardless of the truth value of $B$. We have the tautology:

$~~~~~~A \to [A \leftrightarrow A\lor B]$

enter image description here

We also have the tautology:

$~~~~~~A\to A\lor B$

enter image description here

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    $\begingroup$ I'm sorry but umm... did you even read my question? $\endgroup$
    – William
    Aug 22, 2021 at 16:46
  • $\begingroup$ You asked where is the logical flaw in your argument. There is none. It is a valid argument. Sorry. $\endgroup$ Aug 22, 2021 at 16:52
  • $\begingroup$ Perhaps you are confused about my using logical propositions instead of the less commonly used T/F notation? $\endgroup$ Aug 22, 2021 at 17:01
  • $\begingroup$ Are you not familiar with truth tables? $\endgroup$ Aug 22, 2021 at 17:03
  • $\begingroup$ @William don't be mean to answerers man. Even if he didn't solve your problem, he still had the generosity to at least post an answer :-) $\endgroup$ Jan 25 at 4:49
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$|x|$ has one definite value for a given $x$. $\pm x$ has two values for a given $x$ (except when $x = 0$). Because of this, it's not clear what $|x| = \pm x$ should mean.

In mathematics, we define something called the pre-image of a function. Given a function $f : \mathbb{R} \to \mathbb{R}$ and a subset $A \subset \mathbb{R}$, we define $$ f^{-1}(A) = \{x \in \mathbb{R} : f(x) \in A\} $$ With this definition in hand, we can give some precise meaning to $|x| = \pm x$. Let $g(x) = |x|$. Then $$ g^{-1}(\{x\}) = \{c \in \mathbb{R} : g(c)= x\} = \{x, -x\} $$ If we think of $\pm x$ as the set $\{x, -x\}$, then we may (though no mathematician would) write $g^{-1}(x) = \pm x$.

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I'll focus on your logic related question. I think that your irritation is due to a conflict between your constructive intuitions (which many mathematicians rightly have) and the non-constructive logic commonly used in classical mathematics.

There is no logical error in your reasoning, if we presuppose that logic means classical first-order logic. This is because classically disjunction is a supremum operation on the ordinal 2. Hence, a classical proof of a disjunction simply consists in a proof of one of the disjuncts without indicating which disjunct is proved.

However, from very early on intuitionists like Brouwer held that disjunction proper obeys a requirement of effective existence: A proof of a disjunction always must indicate which of the disjuncts is true. So, a proof of $\varphi_1 \lor \varphi_2$ must select an $i \in \{1, 2 \}$ and yield a proof of $\varphi_i$. These intuitions were spelled out in the form of the so-called BHK-semantics for first-order logic. This informal semantics recursively defines a relation $\rho$ of realizability between proof objects $c$ and first-order formulas. The salient clause for our disjunctive business is:

  • $c \space \rho \space (\varphi \lor \psi)$ iff $(c=\langle 0, c' \rangle $ and $c' \space \rho \space \varphi$) or $ c = \langle 1, c'' \rangle$ and $c'' \space \rho \space \psi$

This informal semantics can be made precise as a so-called program interpretation (see G. Mints: A short introduction to intuitionist logic, chap. 4)

Now, under the BHK-semantics proofs of disjunctions involve injection operations that give information on what disjunct is proved. So, under this constructive treatment of disjunction, your reasoning is not sound.

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Writing $x=\pm3$ does not simply mean "$x=+3$ or $x=-3$". It usually means that both alternatives are possible. So writing $x=\pm3$ for the solutions to $x^2=9$ is correct, but not for the solution to $x+2=5.$

Also, the symbols $\pm$ and $\mp$ are primarily used when there are two expressions that only differ by a sign. One can then treat them both at the same time by using $\pm$ or $\mp$ where the signs differ.

Examples:

  1. $(x\pm y)^2=x^2+y^2\pm 2xy$ instead of $(x+y)^2=x^2+y^2+2xy$ and $(x-y)^2=x^2+y^2-2xy$,
  2. $-(x\pm y)=-x\mp y$ instead of $-(x+y)=-x-y$ and $-(x-y)=-x+y$,
  3. $x_\pm=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}-q}$ instead of $x_+=-\frac{p}{2}+\sqrt{\frac{p^2}{4}-q}$ and $x_-=-\frac{p}{2}-\sqrt{\frac{p^2}{4}-q}$.
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  • $\begingroup$ But why would you use "and"? Question is not about the "$\pm$" symbols. Let's forget that symbol. Let's say why would it be incorrect to write $T \equiv T\vee F$, see the second half of my question. I'm not confused about $|x|$ or about $±$ or about any of the examples. My question is logic based. $\endgroup$
    – William
    Aug 22, 2021 at 11:30
  • $\begingroup$ It is not logically incorrect to write $T \equiv T \lor F$. It is neither logically incorrect to write $\sqrt{9} = 3$ \lor $\sqrt{9} = -3 \lor \sqrt{9} = -193e^2$. But it can be contextually incorrect and very confusing to write that. $\endgroup$
    – md2perpe
    Aug 22, 2021 at 11:49

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